Chemistry Steps

Chemistry Steps

General Chemistry

Chemical kinetics.

This summary practice problem set covers the most common topics of chemical kinetics. You will find questions on the reaction rate, rate constant, rate law, integrated rate law, reaction half-life, and some more.

The links to the corresponding topics are given herein:

  • Reaction Rate
  • Rate Law and Reaction Order
  • How to Determine the Reaction Order
  • Integrated Rate Law
  • The Half-Life of a Reaction
  • Determining the Reaction Order Using Graphs
  • Units of Rate Constant k
  • How Are Integrated Rate Laws Obtained
  • Activation Energy
  • The Arrhenius Equation
  • Chemical Kinetics Practice Problems

Review Questions

How is the rate law of a reaction generally determined?

What do the terms “unimolecular” and “bimolecular” steps mean?

Why are termolecular steps are not common in chemical reactions?

What are the three variables or factors that can affect the rate of reaction?

Show how the rate constant units for a zero, first, and second-order reaction can be derived if the concentrations are expressed in mol/L.

A tiny spark is enough to trigger a highly exothermic explosive reaction between natural gas/gasoline/hydrogen gas, etc.

Explain why a gas container does not explode without this spark.

Sketch a potential-energy diagram for an endothermic, elementary reaction

A + B → C + D

Indicate the reactants , products , the activation energy , the transition state , and the enthalpy change of the reaction.

Molecule A is transformed into molecule D through a multistep reaction. The energy diagram of the reaction is shown below:

chemical kinetics practice problems and solutions pdf

(a) How many intermediates are there in the reaction?

(b) How many transition states are there?

(c) Which is the fastest step in the reaction?

(d) Is the overall reaction exothermic or endothermic?

What is the difference between the rate of a chemical reaction and the rate constant?

Suppose it is determined that the following reaction is second order:

What are three different expressions for the rate law that can be applicable to the reaction? What does each rate law indicate about the mechanism of the reaction?

Consider the reactions:

a) 2NO( g ) + 2H 2 ( g ) → N 2 ( g ) + 2H 2 O( g )

b) 4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( g )

Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

Consider the reaction of HCl forming H 2 and Cl 2 gases:

2HCl( g ) → H 2 ( g ) + Cl 2 ( g )

a) What is the average rate of the reaction during the first 45.0 s if the concentration of HCl dropped from 0.750 M to 0.430 M?

b) If the volume of the reaction is 2.50 L, what amount of Cl 2 (in moles) will be formed during the first 20.0 s of the reaction?

Consider the Haber process for the synthesis of ammonia:

3H 2 ( g ) + N 2 ( g ) ⇆ 2NH 3 ( g )

In the first 25.0 s of this reaction, the concentration of H 2 dropped from 0.600 M to 0.512 M. Calculate the average rate of the reaction during this time interval.

At what rate is ammonia being formed in the previous example?

rate = 1.17 x 10 -3 M/s

Consider the following decomposition reaction of SO 3 :

2SO 3 ( g ) → 2SO 2 ( g ) + O 2 ( g )

During the first 75.0 s, the concentration of SO 3 decreased from 0.265 mol/L to 0.189 mol/L. Determine the average rate of decomposition of SO 3 during this time interval in mol/(L s).

Consider the reaction:

4NH 3 ( g ) + 5O 2 ( g ) → 4NO( g ) + 6H 2 O( g )

Using the rate for the disappearance of NH 3 , fill out the missing data in the table.

Iron(II) ion is oxidized by hydrogen peroxide in an acidic solution.

2Fe 2 + ( aq ) + H 2 O 2 ( aq ) + 2H + ( aq ) → 2Fe 3+ ( aq ) + 2H2O( l )

The rate law for the reaction is determined to be rate = k [H 2 O 2 ][Fe 2 + ]. The rate constant, at certain temperature, is 2.56 x 10 24 / M · s. Calculate the rate of the reaction at this temperature if [H 2 O 2 ] = 0.48 M and [H 2 O 2 ] = 0.070 M .

For the kinetics of the reaction

 2NO( g ) + Cl 2 ( g ) → 2NOCl( g )

The following data were obtained:

a) What is reaction order in Cl 2 and NO?

b) What is the rate law?

c) What is the value of the rate constant?

The date for the initial rate of the following reaction is listed in the table below:

(a) What is the order of reaction with respect to A and to B?

(b) What is the overall reaction order?

(c) What is the value of the rate constant, k ?

Consider the reaction

A(g) + B(g) ⇌ C(g)

The following data were obtained at a certain temperature:

Using the data, determine the order of the reaction and calculate the rate constant:

Carbon dioxide, CO 2 , reacts with hydrogen to give methanol (CH 3 OH), and water.

CO 2 ( g ) + 3H 2 ( g ) ⇆ CH 3 OH( g ) + H 2 O( g )

In a series of experiments, the following initial rates of disappearance of CO 2 were obtained:

Determine the rate law and calculate the value of the rate constant for this reaction.

Integrated Rate Laws

  Dinitrogen pentoxide, N 2 O 5 , decomposes when heated.

2N 2 O 5 ( g ) → 2N 2 O 4 ( g ) + O 2 ( g )  k = 3.4 x 10 -6 /s

What would be the concentration of N 2 O 5 after running the reaction for 3.0 hr if the initial concentration of N 2 O 5 was 0.0465 mol/L?

Hint, first determine the reaction order based on the units of k.

The following decomposition of phosphorus pentachloride (PCl5) was found to be a first-order reaction:

PCl 5 ( g ) → PCl 3 ( g ) + 6Cl 2 ( g )

The half-life of the reaction is 48.0 s at a certain temperature. Calculate (a) the first-order rate constant for the reaction and (b) the time required for 75 percent of the phosphorus pentachloride to decompose.

The following data were collected for the decomposition reaction of hydrogen peroxide:

2H 2 O 2 ( g ) → 2H 2 O( g ) + O 2 ( g )

chemical kinetics practice problems and solutions pdf

Determine rate law, the integrated rate law, and calculate the value of the rate constant.

In a laboratory experiment, the following kinetics data for the concentration of compound A versus time were collected:

A ( g ) → B ( g ) + C ( g )

chemical kinetics practice problems and solutions pdf

Determine the order of the reaction and the value of the rate constant. Predict the concentration of A at 450 s.

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  • Chemistry Concept Questions and Answers
  • Kinetics Questions

Chemical Kinetics Questions

Chemical Kinetics deals with the rates of the reactions. The study of Kinetics involves the factors affecting the rate of the chemical reactions, the mechanisms and the transitions states involved, if any.

Chemical Kinetics Chemistry Questions with Solutions

Q1: What is the difference between the average rate and instantaneous rate?

Answer: Average rate is the rate measured for a long period of time. While the instantaneous rate is the rate measured for an infinitesimally small period of time.

Q2. What does the given graph represent about the nature of reaction? Which of the following expressions are in favour of the graph?

Rate of Reaction

  • Δ[A]/Δt = Δ[B]/Δt
  • -Δ[B]/Δt = Δ[A]/Δt
  • -Δ[A]/Δt = Δ[B]/Δt
  • None of the above

Answer: (c.)

Explanation: The given graph represents that with the decrease in the concentration of A, the concentration of B increases. This implies that A is the reactant and B is the product of the reaction. The reaction can be represented as: A → B.

Since in this reaction, A is the reactant and B is the product; the change in concentration of either of A and B with time gives the rate of the reaction.

Hence, the average rate of formation of B can be written as: Δ[B]/Δt (with a + sign as the concentration of B increases through the reaction).

While the average rate of consumption of A can be written as: -Δ[A]/Δt (with a – sign as the concentration of A decreases through the reaction)

The correct expression for the reaction curves shown in the graph is -Δ[A]/Δt = Δ[B]/Δt.

Q3. The reaction rate of a substance is directly proportional to its_____.

Answer: Active Mass

The active mass is the concentration of the reacting substance in mol l -1 . Hence, the rate of the reaction of the substance is directly proportional to its active mass.

Q4. In the reaction rate expression, the change in concentration of each of the reactants and products are divided by the respective stoichiometric number present in the reaction equation. Why is this division done?

Answer: To understand the above fact, we must look at the following balanced equation:

2N 2 O 5 → 4NO 2 + O 2

The rate of decomposition of N 2 O 5 is -d[N 2 O 5 ]/dt. The rate of formation of NO 2 and O 2 are d[NO 2 ]/dt and d[O 2 ]/dt respectively.

However, these rates are not equal. This is because when 2 moles of N 2 O 5 decompose, 4 moles of NO 2 and 1 mole of O 2 are formed. This implies that the rate of decomposition of N 2 O 5 is twice the rate of formation of O 2 and the rate of formation of NO 2 is 4 times the rate of formation of O 2 .

Hence, the rate of formation of products and the rate of decomposition of the reactants are divided by their stoichiometric coefficients in the reaction in order to get an identical value for the rate of the reaction.

Therefore, the rate of reaction in terms of each of the reactants and the products is :

Rate of the reaction =

Q5. Calculate the rate of the reaction in terms of the different reactants and products for the following reaction.

4NH 3 (g) + 5O 2 (g) → 4NO (g) + 6H 2 O (g)

Given the rate of formation of NO is 3.6 x 10 -3 molL -1 s -1 , calculate the rate of disappearance of NH 3 and the rate of formation of H 2 O.

Answer: The rate of the reaction in terms of each of the reactants and products is given as:

Rate of Reaction =

Since, the coefficients of NO and NH 3 in the balanced chemical equation are the same, both of them have the same rate of formation and disappearance respectively.

Hence, rate of formation of NO = rate of disappearance of NH 3 = 3.6 x 10 -3 mol L -1 s -1

Now, the rate of reaction :

Hence, the rate of formation of water :

Hence, the rate of formation of water is 5.4 x 10 -3 mol L -1 s -1 .

Q6. Calculate the overall order from the given rate expressions.

  • Rate = k[A] ½ [B] 3/2
  • Rate = k[A] 3/2 [B] -1

Answer: The overall orders are calculated as:

  • The order w.r.t. A = 1/2

The order w.r.t. B = 3/2

Overall Order = ½ + 3/2 = 2

  • The order w.r.t. A = 3/2

The order w.r.t. B = -1

Overall Order = 3/2 – 1 = 1/2

Q7. The rate constants of 3 reactions are given. Identify the order of each of the given reactions.

  • k = 2.3 x 10 -5 L mol -1 s -1
  • k = 3.1 x 10 -4 s -1
  • k = 9.3 x 10 -4 mol L -1 s -1

Answer: Since only rate constants for each reaction are given, the order of the reactions will be determined based on the units of the rate constants.

Q8. What are the differences between the rate of the reaction and the reaction rate constant?

Answer: The differences between the two are:

Q9. Give an example of the reaction of the 4th order.

Answer: The dissociation of potassium chlorate to form potassium perchlorate is an example of the reaction of 4th order.

4KClO 3 → 3KClO 4 + KCl

Q10. Derive the general expression of the time taken by the reactant to reduce to its nth fraction in the first order reaction.

Answer: Let us assume the initial amount of the reactant (A) = a

Hence, the nth fraction decreased from “a” in time t = a/n

From the first order reaction,

Hence, the general expression of the time taken by the reactant to reduce to its nth fraction in the first order reaction is represented by t = (2.303/k) logn.

Q11. The half-life period of a 1st order reaction is 60 min. What percentage of the substance will be left after 240 min?

Answer: 1 half-life period = 60 min

Number of half-lives after 240 min = 240/60 = 4 hal-lives i.e. n = 4

Amount of substance left after n half-lives = A o /2 n

Percentage of the amount of substance left after 4 half-lives = A o /2 4 x 100 = A o /16 x 100 = 6.25% of A.

Q12. The dissociation of N 2 O 5 in CCl 4 takes place by the 1st order rate law. The table below shows the concentration of N 2 O 5 measured at different times.

From the given observations, calculate k at t = 410 s and t = 1130 s.

Answer: From the first order rate law:

So, the value of k for the given reaction at t = 410 s is 7.768 x 10 -4 s -1 and that at t = 1130 s is 7.341 x 10 -4 s -1 .

Q13. A 1st order reaction gets 40% completed in 50 minutes. Calculate:

(i.) the rate constant

(ii.) the time in which the reaction will get 80% completed.

Answer: (i.) From the 1st order:

Where 𝑥 = (40/100)a = 0.4a, t= 50 min

The rate constant for the reaction is 0.010216 min -1 .

(ii.)To determine t = ?, 𝑥 = 0.8a

As k is constant for a given reaction, k = 0.010216 min -1

Hence, the time in which the reaction will get 80% completed is 157.58 min.

Q14. Determine the order of a reaction whose rate constant has the same unit as the rate of the reaction.

Answer: Zero Order reaction.

Q15. A reaction is 2nd ordered w.r.t. a reactant. Determine the change in the rate of the reaction when the amount of the reactant is:

  • Reduced to its half

Answer: For a 2nd order reaction, Rate = k[A] 2 = ka 2

  • When A = 2a, Rate = k(2a) 2 = 4.ka 2 , the reaction rate increases 4 times.
  • When A = ½ a, Rate = k(½ a) 2 = ¼ times, the reaction rate decreases by ¼ times.

Practise Questions on Chemical Kinetics

Q1. The given reaction is carried out in a closed vessel:

2N 2 O 5 (g) ⇌ 4NO 2 (g) + O 2 (g)

It was observed that the concentration of NO 2 increased by 2.0 x 10 -2 mol L -1 within 5 seconds of the reaction. Calculate:

(i.) the rate of the reaction.

(ii.) the rate of change of concentration of N 2 O 5 .

Q2. For A → Products, k = 2.0 x 10 -2 s -1 . Calculate the concentration of A after 100 s.

Given [A] o = 1.0 mol L -1

Q3. The equation followed by the composition of a hydrocarbon is:

k = (4.5 x 10 11 s -1 )e -28000 K/T

Calculate the E a .

Q4. A 1st order reaction gets 30% decomposed in 40 min. Calculate its t 1/2 .

Q5. The rate law for a reaction is: Rate = k = [A][B] ½ . Can this be an elementary reaction?

Click the PDF to check the answers for Practice Questions. Download PDF

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Chemistry - Chemical Kinetics: Solved Example Problems | 12th Chemistry : UNIT 7 : Chemical Kinetics

Chapter: 12th chemistry : unit 7 : chemical kinetics, chemical kinetics: solved example problems, molecularity: solved example problems.

Consider the oxidation of nitric oxide to form NO 2

2NO(g) + O 2  (g) → 2NO 2  (g)

(a). Express the rate of the reaction in terms of changes in the concentration of NO,O 2  and NO 2  .

(b). At a particular instant, when [O 2 ] is decreasing at 0.2 mol L −1 s −1  at what rate is [NO2 ] increasing at that instant?

chemical kinetics practice problems and solutions pdf

Evaluate yourself 1

1) Write the rate expression for the following reactions, assuming them as elementary reactions.

i) 3A + 5B 2  →4CD

ii) X 2  + Y 2  →2XY

2). Consider the decomposition of N 2 O 5  (g) to form NO 2  (g) and O 2  (g) . At a particular instant N 2 O 5  disappears at a rate of 2.5x10 -2  mol dm -3 s -1  . At what rates are NO 2  and O 2  formed? What is the rate of the reaction?

1. What is the order with respect to each of the reactant and overall order of the following reactions?

(a). 5Br −  ( aq ) + BrO 3 − (aq ) + 6H +  (aq)

→ 3Br 2  (l ) + 3H 2 O(l)

The experimental rate law is

Rate = k [Br −  ][BrO 3 ][H + ] 2

(b). CH 3 CHO(g ) → Δ → CH 4  (g) + CO(g)

the experimental rate law is

Rate = k [CH 3 CHO] 3/2

a)  First order with respect to Br − , first order with respect to BrO 3 −  and second order with respect to H+ . Hence the overall order of the reaction is equal to 1 + 1 + 2 = 4

b) Order of the reaction with respect to acetaldehyde is 3/2 and overall order is also 3/2

2. The rate of the reaction x + 2y → product is 4 x 10 −3  mol L −1 s −1  if [x]=[y]=0.2 M and rate constant at 400K is 2 x 10 -2  s -1  , What is the overall order of the reaction.

Rate = k [x] n  [y] m

4 x 10 -3  mol L -1 s -1  = 2 x 10 -2  s -1  (0.2 mol L -1  ) n  (0.2mol L -1  ) m

chemical kinetics practice problems and solutions pdf

Comparing the powers on both sides

The overall order of the reaction n + m = 1

Evaluate yourself 2

1). For a reaction, X + Y → product ; quadrupling [x] , increases the rate by a factor of 8. Quadrupling both [x] and [y], increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?

2). Find the individual and overall order of the following reaction using the given data.

2NO(g) + Cl 2  (g) → 2NOCl(g)

chemical kinetics practice problems and solutions pdf

Half life period of a reaction: Solved Example Problems

A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989 ; log10 = 1)

For a first order reaction,

chemical kinetics practice problems and solutions pdf

Let [A 0  ] = 100M

t = t 90%  ; [A]=10M (given that t 90 %  =8hours)

t = t 80%  ; [A]=20M

chemical kinetics practice problems and solutions pdf

Find the value of k using the given data

chemical kinetics practice problems and solutions pdf

Substitute the value of k in equation (2)

chemical kinetics practice problems and solutions pdf

t 80%  = 8hours x 0.6989

t 80%  = 5.59hours

(ii) The half life of a first order reaction x → products is 6.932 x 10 4 s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e 0.06  = 1.06)

Given t 1/2  = 0.6932 x 10 4  s

To solve :2 when t=100 min,

[ [A 0  ] −[A] / [A 0 ]  ] x 100 = ?

We know that

For a first order reaction, t 1/2  = 0.6932 /  k

chemical kinetics practice problems and solutions pdf

k = 10 −5  s −1

Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.

chemical kinetics practice problems and solutions pdf

Evaluate yourself:

1. In a first order reaction A →  products  60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?

2. The rate constant for a first order reaction is 2.3 X 10  −4  s −1  If the initial concentration of the reactant is 0.01M . What concentration will remain after 1 hour?

3. Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

chemical kinetics practice problems and solutions pdf

Show that, the reaction follows first order kinetics.

Arrhenius equation - The effect of temperature on reaction rate: Solved Example Problems

The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s-1 respectively. Calculate the value of activation energy.

According to Arrhenius equation

chemical kinetics practice problems and solutions pdf

Ea = 2305 J mol−1

Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation

chemical kinetics practice problems and solutions pdf

Where Ea is the activation energy. When a graph is plotted for log k Vs 1/T a straight line with a slope of - 4000K is obtained. Calculate the activation energy

chemical kinetics practice problems and solutions pdf

E a  = − 2.303 R m

E a  = − 2.303 x 8.314 J K −1  mol −1  x (− 4000K )

E a  = 76,589J mol −1

E a  = 76.589 kJ mol −1

Evaluate yourself

For a first order reaction the rate constant at 500K is 8 X 10 −4  s −1  . Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol -1  .  

Chemical Kinetics -  Example :  Solved Example Problems

1. The rate law for a reaction of A, B and C hasbeenfoundtobe rate  =  k  [  A ] 2   [ B ][ L ] 3/2

How would the rate of reaction change when

chemical kinetics practice problems and solutions pdf

(i) Concentration of [L] is quadrupled

chemical kinetics practice problems and solutions pdf

(ii) Concentration of both [A] and [B] are doubled

chemical kinetics practice problems and solutions pdf

(iii) Concentration of [A] is halved

chemical kinetics practice problems and solutions pdf

 (iv) Concentration of [A] is reduced to(1/3) and concentration of [L] is quadrupled.

chemical kinetics practice problems and solutions pdf

2.     The rate of formation of a dimer in a second order reaction is 7.5  ×  10 − 3  mol L − 1 s − 1  at 0.05 mol L − 1  monomer concentration. Calculate the rate constant.

Let us consider the dimerisation of a monomer M

Rate= k [M] n

Given that n=2 and [M] = 0.05 mol L -1

Rate = 7.5 X 10 -3  mol L -1 s -1

k = Rate/[M] n

k = 7.5x10 -3 /(0.05) 2  = 3mol -1 Ls -1

3.     For a reaction  x   +   y   +   z   → products the rate law is given by rate  =  k  [   x   ] 3/2   [   y   ] 1/2  what is the overall order of the reaction and what is the order of the reaction with respect to z.

chemical kinetics practice problems and solutions pdf

i.e., second order reaction.

Since the rate expression does not contain the concentration of z , the reaction is zero order with respect to z.

15.     The decomposition of Cl 2 O 7  at 500K in the gas phase to Cl 2  and O 2  is a first order reaction. After 1 minute at 500K, the pressure of Cl 2 O 7  falls from 0.08 to 0.04 atm. Calculate the rate constant in s -1 .

chemical kinetics practice problems and solutions pdf

4.     A gas phase reaction has energy of activation 200 kJ mol -1 . If the frequency factor of the reaction is 1.6  ×  10 13   s − 1  Calculate the rate constant at 600 K.   (  e   − 40 .09   =   3.8   ×   10 − 18   )

chemical kinetics practice problems and solutions pdf

20.     For the reaction 2 x   +   y   →  L find the rate law from the following data.

chemical kinetics practice problems and solutions pdf

5.     The rate constant for a first order reaction is 1.54 × 10 -3  s -1 . Calculate its half life time.

We know that, t 1/2  = 0.693/ k

t 1/2  = 0.693/1.54 x 10 -3  s -1  = 450 s

6.     The half life of the homogeneous gaseous reaction SO 2 Cl 2  → SO 2  + Cl 2  which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO 2 Cl 2  to be reduced to 1% of the initial value?

We know that, k = 0.693/ t 1/2

k = 0.693/ 8.0 minutes= 0.087 minutes -1

chemical kinetics practice problems and solutions pdf

7.     The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?

chemical kinetics practice problems and solutions pdf

8.     A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?

i) Let A = 100M, [A 0 ]–[A] = 20M,

For the zero order reaction

k=([A 0 ]-[A] / t)

k=(20M / 20min) = 1 Mmin -1

Rate constant for a reaction = 1Mmin -1

ii) To calculate the time for 80% of completion

k = 1Mmin -1 , [A 0 ] = 100M, [A 0 ]-[A] = 80M, t = ?

t=([A 0 ]-[A] / k) = (80M / 1Mmin -1 ) = 80min

9.     The activation energy of a reaction is 225 k Cal mol -1  and the value of rate constant at 40°C is 1.8  × 10 − 5  s − 1  . Calculate the frequency factor, A.

Here, we are given that

E a  = 22.5 kcal mol -1  = 22500 cal mol -1

T = 40 ° C = 40 + 273 = 313 K

k = 1.8  ×  10 -5  sec -1

Substituting the values in the equation

chemical kinetics practice problems and solutions pdf

log A = log (1.8) −5 + (15.7089)

log A = 10.9642

A = antilog (10.9642)

A = 9.208 × 10 10  collisions s −1

10.     Benzene diazonium chloride in aqueous solution decomposes according to the equation C 6 H 5 N 2 Cl → C 6 H 5 Cl + N 2  . Starting with an initial concentration of 10 g L − 1  , the volume of N 2  gas obtained at 50 °C at different intervals of time was found to be as under:

chemical kinetics practice problems and solutions pdf

Show that the above reaction follows the first order kinetics. What is the value of the rate constant?

For a first order reaction

chemical kinetics practice problems and solutions pdf

In the present case, V∞ = 58.3 ml.

The value of k at different time can be calculated as follows:

chemical kinetics practice problems and solutions pdf

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0676 min -1

11.     From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

chemical kinetics practice problems and solutions pdf

Where t is the time in minutes and V is the volume of standard KMnO4 solution required for titrating the same volume of the reaction mixture.

chemical kinetics practice problems and solutions pdf

In the present case, V o  = 46.1 ml.

The value of k at each instant can be calculated as follows:

chemical kinetics practice problems and solutions pdf

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.

12.     Where  t  is the time in minutes and V is the volume of standard KMnO 4  solution required for titrating the same volume of the reaction mixture.

i) For the first order reaction k = 2.303/t  log [A 0 ]/[A]

Assume, [A 0 ] = 100 %, t = 50 minutes

Therefore, [A] = 100 – 40 = 60

k = (2.303 / 50) log (100 / 60)

k = 0.010216 min -1

Hence the value of the rate constant is 0.010216 min -1

ii) t = ?, when the reaction is 80% completed,

[A] = 100 – 80 = 20%

From above, k = 0.010216 min -1

 t = (2.303 / 0.010216) log (100 / 20)

t = 157.58 min

The time at which the reaction will be 80% complete is 157.58 min.

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Chemistry LibreTexts

2.E: Chemical Equilibrium (Practice Problems with Answers)

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These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here . In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.

15.1: The Concept of Equilibrium

Conceptual problems.

  • What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the amounts or concentrations of the reactants and the products?
  • Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example.
  • Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of \(NaCl\) in water. What is occurring on a microscopic level? What is happening on a macroscopic level?
  • oxygen and hemoglobin in the human circulatory system
  • iodine crystals in an open beaker
  • the combustion of wood
  • the amount of \(\ce{^{14}C}\) in a decomposing organism

Conceptual Answer

1. When a reaction is described as "having reached equilibrium" this means that the forward reaction rate is now equal to the reverse reaction rate. In regards to the amounts or concentrations of the reactants and the products, there is no change due to the forward reaction rate being equal to the reverse reaction rate.

2. It is not correct to say that the reaction has "stopped" when it has reached equilibrium because it is not necessarily a static process where it can be assumed that the reaction rates cancel each other out to equal zero or be "stopped" but rather a dynamic process in which reactants are converted to products at the same rate products are converted to reactants. For example, a soda has carbon dioxide dissolved in the liquid and carbon dioxide between the liquid and the cap that is constantly being exchanged with each other. The system is in equilibrium and the reaction taking place is: \(CO_{2}\,(g)+2\,H_{2}O\,(l)\rightleftharpoons H_{2}CO_3\,(aq)\).

3. Chemical equilibrium is described as a dynamic process because there is a movement in which the forward and reverse reactions occur at the same rate to reach a point where the amounts or concentrations of the reactants and products are unchanging with time. Chemical equilibrium can be described in a saturated solution of \(NaCl\) as on the microscopic level \(Na^+\) and \(Cl^−\) ions continuously leave the surface of an \(NaCl\) crystal to enter the solution, while at the same time \(Na^+\) and \(Cl^−\) ions in solution precipitate on the surface of the crystal. At the macroscopic level, the salt can be seen to dissolve or not dissolve depending whether chemical equilibrium was established.

a. Exists in a state of equilibrium as the chemical reaction that occurs in the body is: \(Hb\,(aq)+4\,H_{2}O\,(l)\rightleftharpoons Hb(O_{2})_{4}\,(aq)\).

b. Exists in a state of equilibrium as the chemical reaction occurs is: \(H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\)

c. Does not exist in a state of equilibrium as it is not a reversible process. The chemical reaction that takes place is: \(6\,C_{10}H_{15}O_{7}\,(s)+heat\rightarrow C_{50}H_{10}O\,(s)+10\,CH_{2}O\,(g)\).

d. Does not exist in a state of chemical equilibrium as it is not a reversible process. The chemical reaction that takes place is: \(CH_{2}O+O_{2}\rightarrow H_{2}O\,(l)+CO_{2}\,(g)+nutrients\).

15.2: The Equilibrium Constant

  • For an equilibrium reaction, what effect does reversing the reactants and products have on the value of the equilibrium constant?
  • \(2\,HF\,(g)\rightleftharpoons H_{2}\,(g)+F_{2}\,(g)\)
  • \(C\,(s) + 2\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g)\)
  • \(H_{2}C=CH_{2}\,(g) + H_{2}\,(g) \rightleftharpoons C_{2}H_{6}\,(g)\)
  • \(2\,Hg\,(l) + O_{2}\,(g) \rightleftharpoons 2\,HgO\,(s)\)
  • \(NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2NH_{3}\,(g)+CO_{2}\,(g)\)
  • \(C\,(s) + O_{2}\,(g) \rightleftharpoons CO_{2}\,(g)\)
  • \(2\,Mg\,(s) + O_{2}\,(g) \rightleftharpoons 2\,MgO\,(s)\)
  • \(AgCl\,(s) \rightleftharpoons Ag^+\,(aq)+Cl^−\,(aq)\)
  • If an equilibrium reaction is endothermic, what happens to the equilibrium constant if the temperature of the reaction is increased? if the temperature is decreased?
  • Industrial production of \(NO\) by the reaction \(N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)\) is carried out at elevated temperatures to drive the reaction toward the formation of the product. After sufficient product has formed, the reaction mixture is quickly cooled. Why?
  • How would you differentiate between a system that has reached chemical equilibrium and one that is reacting so slowly that changes in concentration are difficult to observe?
  • What is the relationship between the equilibrium constant, the concentration of each component of the system, and the rate constants for the forward and reverse reactions?
  • \(CO\,(g) + H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+H_{2}\,(g)\)
  • \(PCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons PCl_{5}\,(g)\)
  • \(2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)\)

\(2\,NO\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)

\(\dfrac{1}{2}\,H_2\,(g)+12\,I_{2}\,(g) \rightleftharpoons HI\,(g)\)

\(cis-stilbene\,(soln) \rightleftharpoons trans-silbene\,(soln)\)

  • Why is it incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression?
  • \(2\,S\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,SO_{3}\,(g)\)
  • \(C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)\)
  • \(2\,ZnS\,(s)+3\,O_{2}\,(g) \rightleftharpoons 2\,ZnO\,(s)+2\,SO_{2}\,(g)\)
  • \(2\,HgO\,(s) \rightleftharpoons 2\,Hg\,(l)+O_{2}\,(g)\)
  • \(H_{2}\,(g)+I_{2}\,(s) \rightleftharpoons 2\,HI\,(g)\)
  • \(NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\)
  • At room temperature, the equilibrium constant for the reaction \(2\,A\,(g) \rightleftharpoons B\,(g)\) is 1. What does this indicate about the concentrations of \(A\) and \(B\) at equilibrium? Would you expect \(K\) and \(K_p\) to vary significantly from each other? If so, how would their difference be affected by temperature?
  • For a certain series of reactions, if \(\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]} = K_1\) and \(\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]} = K_2\), what is the equilibrium constant expression for the overall reaction? Write the overall equilibrium equation.
  • In the equation for an enzymatic reaction, \(ES\) represents the complex formed between the substrate \(S\) and the enzyme protein \(E\). In the final step of the following oxidation reaction, the product \(P\) dissociates from the \(ESO_2\) complex, which regenerates the active enzyme:

Give the overall reaction equation and show that \(K = K_1 \times K_2 \times K_3\).

Conceptual Answers

1. By reversing the reactants and products for an equilibrium reaction, the equilibrium constant becomes: \(K′ = \dfrac{1}{K}\).

a. This equilibrium is homogenous as all substances are in the same state.

b. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is homogeneous as all substances are in the same state.

d. This equilibrium is heterogeneous as not all substances are in the same state.

a. This equilibrium is heterogeneous as not all substances are in the same state.

c. This equilibrium is heterogeneous as not all substances are in the same state.

4. According to Le Chatelier’s principle, equilibrium will shift in the direction to counteract the effect of a constraint (such as concentration of a reactant, pressure, and temperature). Thus, in an endothermic reaction, the equilibrium shifts to the right-hand side when the temperature is increased which increases the equilibrium constant and the equilibrium shifts to the left-hand side when the temperature is decreased which decreases the equilibrium constant.

5. After sufficient industrial production of \(NO\) by the reaction of \(N_{2}\,(g)+O_{2}\,(g) \rightleftharpoons 2\,NO_\,(g)\) at elevated temperatures to drive the reaction toward the formation of the product, the reaction mixture is cooled quickly because it quenches the reaction and prevents the system from reverting to the low-temperature equilibrium composition that favors the reactants.

6. To differentiate between a system that has reached equilibrium and one that is reacting slowly that changes in concentrations are difficult to observe we can use Le Chatelier’s principle to observe any shifts in the reaction upon addition of a constraint (such as concentration, pressure, or temperature).

7. The relationship between the equilibrium constant, the concentration of each component of a system, and the rate constants for the forward and reverse reactions considering a reaction of a general form: \(a\,A+b\,B \rightleftharpoons c\,C+d\,D\) is \(K=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}=\dfrac{k_f}{k_r}\)

\(K=\dfrac{[CO_2][H_2]}{[CO][H_2O]}\)

\(K_p=\dfrac{(P_{CO_2})(P_{H_2})}{(P_{CO})(P_{H_2O})}\)

\(K=\dfrac{[PCl_5]}{[PCl_3][Cl_2]}\)

\(K_p=\dfrac{(P_{PCl_5})}{(P_{Cl_3})(P_{Cl_2})}\)

\(K=\dfrac{[O_2]^3}{[O_3]^2}\)

\(K_p=\dfrac{(P_{O_2})^3}{(P_{O_3})^2}\)

  • \(K=\dfrac{[NO_2]^2}{[NO]^2[O_2]}\)

\(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{NO})^2(P_{O_2})}\)

\(K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[O_2]}\)

\(K_p=\dfrac{(P_{HI})}{(P_{H_{2}})^{\frac{1}{2}}(P_{O_{2}})}\)

\(K=\dfrac{trans-stilbene}{cis-stilbene}\)

\(K_p=\dfrac{(P_{trans-stilbene})}{(P_{cis-stilbene})}\)

10. It is incorrect to state that pure liquids, pure solids, and solvents are not part of an equilibrium constant expression because they are not reactive enough or cause a change in the concentrations of the ions or the species that exist in the gas phase.

\(K=\dfrac{[SO_3]^2}{[O_2]^3}\)

\(K_p=\dfrac{(P_{SO_3})^2}{(P_{O_2})^3}\)

\(K=\dfrac{[CO]^2}{[CO_2]}\)

\(K_p=\dfrac{(P_{CO})^2}{(P_{CO_2})}\)

\(K=\dfrac{[SO_2]^2}{[O_2]^3}\)

\(K_p=\dfrac{(P_{SO_2})^2}{(P_{O_2})^3}\)

\(K=[O_2]\)

\(K_p=(P_{O_{2}})\)

\(K=\dfrac{[HI]^2}{[H_2]}\)

\(K_p=\dfrac{(P_{HI})^2}{(P_{H_2})}\)

\(K=[NH_3]^2[CO_2]\)

\(K_p=(P_{NH_3})^2(P_{CO_2})\)

\(K=\dfrac{[B]}{[A]^2} \rightarrow 1=\frac{[B]}{[A]^2} \rightarrow [A]^2=[B] \rightarrow [A]=\sqrt{B}\)

\(K\) and \(K_p\) vary by \(RT\), but it largely depends on \(T\) as \(R\) is constant. A raise or decrease in temperature would cause a difference.

\(K_p=K(RT)^{Δn}=K(RT)^{-1}=\dfrac{K}{RT}\)

\(Δn=(total\,moles\,of\,gas\,on\,the\,product\,side)-(total\,of\,moles\,on\,the\,reactant\,side)=1-2=−1\)

\(K=\dfrac{K_1}{K_2}={\dfrac{\dfrac{[OH^−][HCO_3^−]}{[CO_3^{2−}]}}{\dfrac{[OH^−][H_2CO_3]}{[HCO_3^−]}}=\dfrac{[HCO_3^−]^2}{[CO_3^{2-}][H_2CO_3]}}\)

\(CO_3^{2-}\,(g)+H_2CO_{3}\,(g) \rightleftharpoons 2\,HCO_3^{-}\,(g)\)

\(K = K_1 \times K_2 \times K_3=\frac{[ES]}{[E][S]}\times\frac{[ESO_2]}{[ES][O_2]}\times\frac{[E][P]}{[ESO_2]}=\frac{[P]}{[S][O_2]}\)

\(S+O_2 \rightleftharpoons P\)

Numerical Problems

  • Explain what each of the following values for \(K\) tells you about the relative concentrations of the reactants versus the products in a given equilibrium reaction: \(K = 0.892\); \(K = 3.25 \times 10^8\); \(K = 5.26 \times 10^{−11}\). Are products or reactants favored at equilibrium?
  • \(N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)
  • \(\frac{1}{2}\,N_2O_{4}\,(g) \rightleftharpoons NO_{2}\,(g)\)
  • \(\frac{1}{2}N_{2}\,(g)+\frac{3}{2}H_{2}\,(g) \rightleftharpoons NH_{3}\,(g)\)
  • \(\frac{1}{3}N_{2}\,(g)+H_{2}\,(g) \rightleftharpoons \frac{2}{3}NH_{3}\,(g)\)

How are these two expressions mathematically related to the equilibrium constant expression for

\[N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g) ?\]

  • \(C\,(s) + 2\,H_2O\,(g) \rightleftharpoons CO_{2}\,(g)+2\,H_{2}\,(g)\)
  • \(SbCl_{3}\,(g)+Cl_{2}\,(g) \rightleftharpoons SbCl_{5}\,(g)\)
  • Give an equilibrium constant expression for each reaction.

a. \(2\,NO\,(g) + O_{2}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\)

b. \(\frac{1}{2}H_{2}\,(g)+\frac{1}{2}I_{2}\,(g) \rightleftharpoons HI\,(g)\)

c. \(CaCO_{3}\,(s) + 2\,HOCl\,(aq) \rightleftharpoons Ca^{2+}\,(aq) + 2\,OCl^−\,(aq) + H_2O\,(l) + CO_{2}\,(g)\)

6. Calculate \(K\) and \(K_p\) for each reaction.

  • \(2\,NOBr\,(g) \rightleftharpoons 2\,NO\,(g)+Br_2\,(g)\): at 727°C, the equilibrium concentration of \(NO\) is 1.29 M, \(Br_2\) is 10.52 M, and \(NOBr\) is 0.423 M.
  • \(C\,(s) + CO_{2}\,(g) \rightleftharpoons 2\,CO\,(g)\): at 1,200 K, a 2.00 L vessel at equilibrium has partial pressures of 93.5 atm \(CO_2\) and 76.8 atm \(CO\), and the vessel contains 3.55 g of carbon.

7. Calculate \(K\) and \(K_p\) for each reaction.

  • \(N_2O_4\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\): at the equilibrium temperature of −40°C, a 0.150 M sample of \(N_2O_4\) undergoes a decomposition of 0.456%.
  • \(CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)\): an equilibrium is reached at 227°C in a 15.5 L reaction vessel with a total pressure of \(6.71 \times 10^2\) atm. It is found to contain 37.8 g of hydrogen gas, 457.7 g of carbon monoxide, and 7,193 g of methanol.

8. Determine \(K\) and \(K_p\) (where applicable) for each reaction.

  • \(2\,H_2S\,(g) \rightleftharpoons 2\,H_{2}\,(g)+S_{2}\,(g)\): at 1065°C, an equilibrium mixture consists of \(1.00 \times 10^{−3}\) M \(H_2\), \(1.20 \times 10^{−3}\) M \(S_2\), and \(3.32 \times 10^{−3}\) M \(H_2S\).
  • \(Ba(OH)_{2}\,(s) \rightleftharpoons 2\,OH^−\,(aq)+Ba^{2+}\,(aq)\): at 25°C, a 250 mL beaker contains 0.330 mol of barium hydroxide in equilibrium with 0.0267 mol of barium ions and 0.0534 mol of hydroxide ions.

9. Determine \(K\) and \(K_p\) for each reaction.

  • \(2\,NOCl\,(g) \rightleftharpoons 2\,NO\,(g)+Cl_{2}\,(g)\): at 500 K, a 24.3 mM sample of \(NOCl\) has decomposed, leaving an equilibrium mixture that contains 72.7% of the original amount of \(NOCl\).
  • \(Cl_{2}\,(g)+PCl_{3}\,(g) \rightleftharpoons PCl_{5}\,(g)\): at 250°C, a 500 mL reaction vessel contains 16.9 g of \(Cl_2\) gas, 0.500 g of \(PCl_3\), and 10.2 g of \(PCl_5\) at equilibrium.

10. The equilibrium constant expression for a reaction is \(\dfrac{[CO_2]^2}{[SO_2]^2[O_2]}\). What is the balanced chemical equation for the overall reaction if one of the reactants is \(Na_2CO_{3}\,(s)\)?

11. The equilibrium constant expression for a reaction is \(\dfrac{[NO][H_{2}O]^{\dfrac{3}{2}}}{[NH_3][O_2]^{\dfrac{5}{4}}}\). What is the balanced chemical equation for the overall reaction?

12. Given \(K =\dfrac{k_f}{k_r}\), what happens to the magnitude of the equilibrium constant if the reaction rate of the forward reaction is doubled? What happens if the reaction rate of the reverse reaction for the overall reaction is decreased by a factor of 3?

13. The value of the equilibrium constant for \[2\,H_{2}\,(g)+S_{2}\,(g) \rightleftharpoons 2\,H_2S\,(g)\] is \(1.08 \times 10^7\) at 700°C. What is the value of the equilibrium constant for the following related reactions

  • \(H_{2}\,(g)+12\,S_{2}\,(g) \rightleftharpoons H_2S\,(g)\)
  • \(4\,H_{2}\,(g)+2\,S_{2}\,(g) \rightleftharpoons 4\,H_2S\,(g)\)
  • \(H_2S\,(g) \rightleftharpoons H_{2}\,(g)+12\,S_{2}\,(g)\)

Numerical Answers

1. In the given equilibrium reaction where \(K = 0.892\approx1\) has a concentration of the reactants that is approximately equal to the concentration of the products so neither formation of the reactants or products is favored. In the given equilibrium reaction where \(K = 3.25 \times 10^8>1\) has a concentration of the products that is relatively small compared to the concentration of the reactants so the formation of the products is favored. In the given equilibrium reaction where \(K = 5.26 \times 10^{−11}<1\) has a concentration of products that is relatively large compared to that of the concentration of the reactants so the formation of the reactants is favored.

a. \(K=\dfrac{[NO_2]^{2}}{[N_2O_4]}\)

b. \(K=\dfrac{[NO_2]}{[N_2O_4]^{\dfrac{1}{2}}}\)

Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the \(K\) value for a. We would need to square it to get the \(K\) value for b.

\(K’=\dfrac{[NH_3]}{[N_2]^{\dfrac{1}{2}}[H_2]^{\dfrac{3}{2}}}\)

\(K’’=\dfrac{[NH_3]^{\dfrac{2}{3}}}{[N_2]^{\dfrac{1}{2}}[H_2]}\)

\(K=\dfrac{[NH_3]^2}{[N_2][H_2]^3}\)

\(K'=K^{\dfrac{1}{2}}\)

\(K''=K^{\dfrac{1}{3}}\)

a. \(K=\dfrac{[H_2]^2[CO_2]}{[H_{2}O]^2}\)

b. \(K=\dfrac{[SbCl_5]}{[SbCl_3][Cl_2]}\)

c. \(K=\dfrac{[O_2]^3}{[O_3]^2}\)

  • \(K=\dfrac{[HI]}{[H_2]^{\dfrac{1}{2}}[I_2]^{\dfrac{1}{2}}}\)
  • \(K=\dfrac{[Ca^{2+}][OCl^−]^2[CO_2]}{[HOCl]^2}\)

\(K=\dfrac{[NO]^2[Br]}{[NOBr]^2}=\frac{[1.29\,M]^2[10.52\,M]}{[0.423\,M]^2}=97.8\)

\(K_p=K(RT)^{Δn}=(97.8)((0.08206\frac{L\cdot atm}{mol\cdot K})((727+273.15)K))^{3-2}=8.03x10^{4}\)

\(K_p=K(RT)^{Δn}=K(RT)^{2-1}=K(RT) \rightarrow K=\dfrac{K_p}{RT}=\frac{63.1}{(0.08206\frac{L\cdot atm}{mol\cdot K})(1,200\,K)}=6.41\)

\(K=\dfrac{(P_{CO})^2}{P_{CO_2}}=\frac{(76.8\,atm)^2}{93.5\,atm}=63.1\)

\(K=\dfrac{[NO_{2}]^2}{[N_{2}O_{4}]}=\frac{[0.001368\,M]}{[0.149316\,M]}=1.25x10^{-5}\)

\([NO_2]=(2)(0.150\,M)(0.00456)=0.001368\,M\)

\([N_2O_4]=(0.150\,M)(1-0.00456)=0.149316\,M\)

\(K_p=K(RT)^{Δn}=(1.25x10^{-5})((0.08206\frac{L\cdot atm}{mol\cdot K})((-40+273.15)K))^{2-1}=2.39x10^{-4}\)

\(K=\dfrac{[CH_{3}OH]}{[CO][H_{2}]^2}=\frac{[14.5\,M]}{[1.05\,M][1.21\,M]^2}=9.47\)

\([CH_{3}OH]=7,193\,g\,CH_{3}OH \times \frac{1\,mol\,CH_{3}OH}{32.04\,g\,CH_{3}OH} \times \frac{1}{15.5\,L}=14.5\,M\)

\([CO]=457.7\,g\,CO \times \frac{1\,mol\,CO}{28.01\,g\,CO} \times \frac{1}{15.5\,L}=1.05\,M\)

\([H_{2}]=37.8\,g\,H_{2} \times \frac{1\,mol\,H_{2}}{2.02\,g\,H_{2}} \times \frac{1}{15.5\,L}=1.21\,M\)

\(K_p=K(RT)^{Δn}=(9.47)((0.08206\frac{L\cdot atm}{mol\cdot K})((227+273.15)K))^{1-3}=5.62x10^{-2}\)

\(K=\dfrac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}=\frac{[1.00 \times 10^{-3}\,M]^2[1.20 \times 10^{-3}\,M]}{[3.32 \times\ 10^{-3}\,M]^2}=1.09 \times 10^{-4}\)

\(K_p=K(RT)^{Δn}=(1.09 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})((1065+273.15)K))^{3-2}=1.20 \times 10^{-2}\)

\(K=[OH^{-}]^2[Ba^{2+}]=[0.2136\,M]^2[0.1068\,M]=4.87 \times 10^{-3}\)

\([OH^{-}]=0.0534\,mol\,OH^{-} \times \frac{1}{0.25\,L}=0.2136\,M\)

\([Ba^{2+}]=0.0267\,mol\,Ba^{2+} \times \frac{1}{0.25\,L}=0.1068\,M\)

\(K_p=K(RT)^{Δn}=(4.87 \times 10^{-3})((0.08206\frac{L\cdot atm}{mol\cdot K})((25+273.15)K))^{3-1}=2.92\)

\(K=\dfrac{[NO]^2[Cl_{2}]}{[NOCl]^2}=4.59 \times 10^{-4}\)

\(K_p=K(RT)^{Δn}=(4.59 \times 10^{-4})((0.08206\frac{L\cdot atm}{mol\cdot K})(500K))^{3-2}=1.88 \times 10^{-2}\)

\(K=\dfrac{[PCl_5]}{[Cl_{2}][PCl_{3}]}=\frac{[9.80 \times 10^{-2}\,M]}{[4.77 \times 10^{-1}\,M][7.28 \times 10^{-3}\,M]}=28.2\)

\([PCl_{5}]=10.2\,g\,PCl_{5} \times \frac{1\,mol\,PCl_{5}}{208.2388\,g\,PCl_{5}} \times \frac{1}{0.5\,L}=9.80 \times 10^{-2}\,M\)

\([Cl_{2}]=16.9\,g\,Cl_{2} \times \frac{1\,mol\,Cl_{2}}{70.9\,g\,Cl_{2}} \times \frac{1}{0.5\,L}=4.77 \times 10^{-1}\,M\)

\([PCl_{3}]=0.500\,g\,PCl_{3} \times \frac{1\,mol\,PCl_{3}}{137.33\,g\,PCl_{3}} \times \frac{1}{0.5\,L}=7.28 \times 10^{-3}\,M\)

\(K_p=K(RT)^{Δn}=(28.2)((0.08206\frac{L\cdot atm}{mol\cdot K})(250+273.15)K)^{1-2}=6.57 \times 10^{-1}\)

10. \(2\,SO_{2}\,(g)+O_{2}\,(g)+2\,Na_{2}CO_{3}\,(s) \rightleftharpoons 2\,CO_{2}\,(g)+2\,Na_{2}SO_{4}\,(s)\)

11. \(NH_{3}\,(g) + \frac{5}{4}\,O_{2}\,(g)⇌NO \,(g)+\frac{3}{2}\,H_{2}O\,(g)\)

a. \(K= \dfrac{[H_{2}S]}{[H_{2}][S_{2}]^\frac{1}{2}}=K’^{\frac{1}{2}}=(1.08 \times 10^{7})^{\frac{1}{2}}=3.29 \times 10^{3}\) b. \(K= \dfrac{[H_{2}S]^4}{[H_{2}]^4[S_{2}]^2}=K’^{2}=(1.08 \times 10^{7})^{2}=1.17 \times 10^{14}\) c. \(K= \dfrac{[H_{2}][S_2]^\frac{1}{2}}{[H_{2}S]}=K’^{-\frac{1}{2}}= (1.08 \times 10^{7})^{-\frac{1}{2}}=3.04 \times 10^{-4}\)

15.3: Interpreting & Working with Equilibrium Constants

  • Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known.
  • Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when \(K\) is (a) very large and (b) very small? Illustrate this technique using the system \(A+2B \rightleftharpoons C\) for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used?

1. The magnitude of the equilibrium constant for a reaction depends on the form in which the chemical reaction is written. For example, writing a chemical reaction in different but chemically equivalent forms causes the magnitude of the equilibrium constant to be different but can be related by comparing their respective magnitudes.

a. When \(K\) is very large the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion.

\(K= \dfrac{[C]}{[A][B]^2}=\frac{[C]}{very\,small}=\frac{1}{0}= \infty \rightarrow [C]= \infty\)

b. When \(K\) is very small the reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of the reactants.

\(K=\dfrac{[C]}{[A][B]^2}=\frac{very\,small}{[A][B]^2}=\frac{0}{1}=0 \rightarrow [C]=0\)

Simplifying assumptions should not be used if the equilibrium constant is not known to be very large or very small.

Please be sure you are familiar with the quadratic formula before proceeding to the Numerical Problems.

  • In the equilibrium reaction \(A+B \rightleftharpoons C\), what happens to \(K\) if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction \(2\,A \rightleftharpoons B+C\)?
  • The following table shows the reported values of the equilibrium \(P_{O_2}\) at three temperatures for the reaction \(Ag_{2}O\,(s) \rightleftharpoons 2\,Ag\,(s)+ \frac{1}{2}\,O_{2}\,(g)\), for which ΔH° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not?
  • Given the equilibrium system \(N_2O_{4}\,(g) \rightleftharpoons 2\,NO_{2}\,(g)\), what happens to \(K_p\) if the initial pressure of \(N_2O_4\) is doubled? If \(K_p\) is \(1.7 \times 10^{−1}\) at 2300°C, and the system initially contains 100% \(N_2O_4\) at a pressure of \(2.6 \times 10^2\) atm, what is the equilibrium pressure of each component?
  • At 430°C, 4.20 mol of \(HI\) in a 9.60 L reaction vessel reaches equilibrium according to the following equation: \[H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\] At equilibrium, \([H_2] = 0.047\;M\) and \([HI] = 0.345\;M\) What are \(K\) and \(K_p\) for this reaction?
  • Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: \[CO\,(g)+2\,H_{2}\,(g) \rightleftharpoons CH_3OH\,(g)\] with \(K_p = 1.3 \times 10^{−4}\). If 56.0 g of \(CO\) is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions?
  • Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction \(A\,(s) \rightleftharpoons 2\,B\,(g)+C\,(g)\), what is \(K_p\)?
  • The decomposition of ammonium carbamate to \(NH_3\) and \(CO_2\) at 40°C is written as \(NH_4CO_2NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\). If the partial pressure of \(NH_3\) at equilibrium is 0.242 atm, what is the equilibrium partial pressure of \(CO_2\)? What is the total gas pressure of the system? What is \(K_p\)?
  • What is \(K\) for the reaction at each temperature?
  • If a sample at 375 K has 0.100 M \(Cl_2\) and 0.200 M \(SO_2\) at equilibrium, what is the concentration of \(SO_2Cl_2\)?
  • If the sample given in part b is cooled to 303 K, what is the pressure inside the bulb?
  • For the gas-phase reaction \(a\,A \rightleftharpoons b\,B\), show that \(K_p = K(RT)^{Δn}\) assuming ideal gas behavior.
  • For the gas-phase reaction \(I_2 \rightleftharpoons 2\,I\), show that the total pressure is related to the equilibrium pressure by the following equation: \[P_T=\sqrt{K_{p}P_{I_{2}}} + P_{I_{2}}\]
  • Experimental data on the system \(Br_{2}\,(l) \rightleftharpoons Br_{2}\,(aq)\) are given in the following table. Graph \(Br_{2}\,(aq)\) versus moles of \(Br_{2}\,(l)\) present; then write the equilibrium constant expression and determine K.
  • Data accumulated for the reaction (\n-butane(g) \rightleftharpoons isobutane(g)\) at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of n-butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of n-butane and isobutane?
  • Solid ammonium carbamate (\(NH_{4}CO_{2}NH_{2}\)) dissociates completely to ammonia and carbon dioxide when it vaporizes: \(NH_{4}CO_{2}NH_{2}\,(s) \rightleftharpoons 2\,NH_{3}\,(g)+CO_{2}\,(g)\) At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is \(K_p\)? If the concentration of \(CO_2\) is doubled and then equilibrates to its initial equilibrium partial pressure +x atm, what change in the \(NH_{3}\) concentration is necessary for the system to restore equilibrium?
  • The equilibrium constant for the reaction \(COCl_{2}\,(g) \rightleftharpoons CO\,(g)+Cl_{2}\,(g)\) is \(K_p = 2.2 \times 10^{−10}\) at 100°C. If the initial concentration of \(COCl_{2}\) is \(3.05 \times 10^{−3}\; M\), what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations?
  • Aqueous dilution of \(IO_{4}^{−}\) results in the following reaction: \[IO^−_{4}\,(aq)+2\,H_{2}O_(l)\, \rightleftharpoons H_4IO^−_{6}\,(aq)\] with \(K = 3.5 \times 10^{−2}\). If you begin with 50 mL of a 0.896 M solution of \(IO_4^−\) that is diluted to 250 mL with water, how many moles of \(H_4IO_6^−\) are formed at equilibrium?
  • Iodine and bromine react to form \(IBr\), which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation: \[I_{2}\,(g)+Br_{2}\,(g) \rightleftharpoons 2\,IBr\,(g)\] with \(K_p = 1.2 \times 10^2\). If you begin the reaction with 7.4 g of \(I_2\) vapor and 6.3 g of \(Br_2\) vapor in a 1.00 L container, what is the concentration of \(IBr\,(g)\) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system?
  • For the reaction \[2\,C\,(s) + \,N_{2}\,(g)+5\,H_{2}\, \rightleftharpoons 2\,CH_{3}NH_{2}\,(g)\] with \(K = 1.8 \times 10^{−6}\). If you begin the reaction with 1.0 mol of \(N_2\), 2.0 mol of \(H_2\), and sufficient \(C\,(s)\) in a 2.00 L container, what are the concentrations of \(N_2\) and \(CH_3NH_2\) at equilibrium? What happens to \(K\) if the concentration of \(H_2\) is doubled?

1. In both cases, the equilibrium constant will remain the same as it does not depend on the concentrations.

2. No, the data is not consistent with what I would expect to occur because enthalpy is positive indicating that the reaction is endothermic thus heat is on the left side of the reaction. As the temperature is raised \(P_{O_2}\) would be expected to increase to counteract the constraint.

If the initial pressure of \(N_2O_4\) was doubled then \(K_p\) is one half of the original value.

\(K_p=\dfrac{(P_{NO_{2}})^2}{(P_{N_{2}O_{4}})} \rightarrow 1.7 \times 10^{-1} = \frac{(2x)^2}{2.6 \times 10^2 -x} \rightarrow 44.2-0.17x=4x^2 \rightarrow x \approx 3.303\,atm\)

\(P_{N_{2}O_{4}}=2.6 \times 10^2-x=260-3.303=2.6 \times 10^2\,atm\)

\(P_{NO_{2}}=2x=(2)(3.303)=6.6\,atm\)

\(K=\frac{[HI]^2}{[H_2][I_2]}=\frac{0.345\,M}{(0.047\,M)(0.047\,M)}=157\)

\(K_p=K(RT)^{Δn}=(157)((0.08206\frac{L\cdot atm}{mol\cdot K})(430+273.15)K)^{2-2}=157\)

\(Maximum\;Percent\;Yield=\frac{Actual}{Theoretical} \times 100\%=\frac{212.593}{376.127} \times 100\%=56.52\% \approx 57\%\)

\(PV=nRT \rightarrow P=\frac{(1.999\;mol)(0.08206\frac{L\cdot atm}{mol\cdot K})(300+273.15)K}{0.250\;L}=376.127\;atm\)

\([CO]=56.0\;g\;CO \times \frac{1\;mol\;CO}{28.01\,g\,CO}=1.999\,mol\,CO\)

\(K_p=\frac{P_{CH_{3}OH}}{(P_{CO})(P_{H_2})^{2}} \rightarrow 1.3 \times 10^{-4}= \frac{x}{(376.02-x)(100^2)} \rightarrow 1.3= \frac{x}{376.07-x} \rightarrow 488.965-1.3x=x \rightarrow 488.965=2.3x \rightarrow x=212.593\,atm\)

\(K_p=\frac{(P_{CH_{3}OH})}{(P_{CO})(P_{H_2})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(376.127-357.320)(P_{H_{2}})^2} \rightarrow 1.3 \times 10^{-4}= \frac{357.320}{(18.80635)(P_{H_{2}})^2} \rightarrow 0.002444(P_{H_{2}})^2=357.320 \rightarrow (P_{H_{2}})=382.300 \approx 3.8 \times 10^2\,atm\)

\(Minimum\,Percent\,Yield=\frac{Actual}{Theoretical} \times 100 \% \rightarrow 95\%=\frac{x}{376.127} \times 100\% \rightarrow x=357.320\,atm\)

6. \(K_p=\frac{(P_B)^2(P_C)}{P_A}=\frac{[2x]^2[x]}{[0.969-x]}=\frac{4x^3}{0.969-x}\)

\(P_{CO_{2}}=P_{tot}-P_{NH_{3}}=P_{tot}-0.242\,atm\)

\(P_{tot}=P_{NH_3}+P_{CO_2}=0.242\,atm+P_{CO_2}\)

\(K_p=(P_{NH_3})^2(P_{CO_2})=(0.242\,atm)^2(P_{CO_2})\)

\(At\,375\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(375\,K))^{2-1}}=7.80 \times 10^{-2}\)

\(At\,303\,K:K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{2.9 \times 10^{-2}}{((0.08206\frac{L\cdot atm}{mol\cdot K})(303\,K))^{2-1}}=1.17 \times 10^{-3}\)

b. \(K=\frac{[SO_{2}][Cl_{2}]}{[SO_{2}Cl_{2}]} \rightarrow [SO_{2}Cl_{2}]=\frac{[0.200\,M][0.100\,M]}{7.80 \times 10^{-2}}=2.56 \times 10^{-1}\,M\)

c. If the sample given in part b is cooled to 303 \(K\), the pressure inside the bulb would decrease.

\(K_p=\frac{(P_B)^b}{(P_A)^a}=\frac{((\frac{n_B}{V})(RT))^b}{((\frac{n_A}{V})(RT))^a}=\frac{[B]^{b}(RT)^b}{[A]^a(RT)^a}=K(RT)^{b-a}=K(RT)^{Δn}\)

\(PV=nRT \rightarrow P=\frac{n}{V}RT\)

\(K=\frac{[B]^{b}}{[A]^{a}}\)

\(Δn=b-a\)

\(P_T=P_I+P_{I_2}=\sqrt{K_pP_{I_2}}+P_{I_2}\)

\(K_p=\frac{(P_I)^2}{P_{I_2}} \rightarrow (P_I)^2=K_p(P_{I_2}) \rightarrow P_I=\sqrt{K_pP_{I_2}}\)

The graph should be a positive linear correlation.

\([Br_2\,(l)]=1.0\,g\,Br_2 \times \frac{1\,mol\,Br_2}{159.808\,g\,Br_2} \times \frac{1}{0.1\,L}=6.26 \times 10^{-2}\)

\(K=\frac{[Br_2\,(aq)]}{[Br_2\,(l)]}=\frac{[Br_2\,(aq)]}{1}=[Br_2\,(aq)]\)

12. \(K=\frac{[isobutane]}{[n-butane]}=\frac{x}{1-x}\)

\(P_{NH_3}=2x=2(0.0387)=7.73 \times 10^{-2}\,atm\)

\(P_{CO_2}=x=3.87 \times 10^{-2}\,atm\)

\(K_p=(P_{NH_3})^2(P_{CO_2})=(2x)^2(x)=4x^3=4(0.0387)^3=2.32 \times 10^{-4}\)

\(P_{tot}=P_{NH_3}+P_{CO_2} \rightarrow 0.116=2x+x \rightarrow 0.116=3x \rightarrow x=0.0387\)

If the concentration of \(CO_{2}\) is doubled and then equilibrates to its initial equilibrium partial

pressure +x atm, the concentration of \(NH_{3}\) should also be doubled for the system to restore

equilibrium.

\(P_{COCl_{2}}=9.34 \times 10^{-2}-x=9.34 \times 10^{-2}-9.34 \times 10^{-22}=9.34 \times 10^{-2}\,atm\)

\(P_{CO}=x=9.34 \times 10^{-22}\,atm\)

\(P_{Cl_{2}}=x=9.34 \times 10^{-22}\,atm\)

Assume that the equilibrium mainly lies on the reactants side because the \(K_p\) value is less than 1.

\(K_p=\frac{(P_{CO})(P_{Cl_{2}})}{(P_{COCl_{2}})} \rightarrow 2.2 \times 10^{-10} =\frac{x^{2}}{9.34 \times 10^{-2}-x} \rightarrow 2.0548 \times 10^{-11}-2.2 \times 10^{-10}x=x^{2} \rightarrow x^{2}+2.2 \times 10^{-10}x-2.0548 \times 10^{-11}=0 \rightarrow x=9.34 \times 10^{-22}\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(3.05 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(100+273.15)K=9.34 \times 10^{-2}\)

15. \([H_{4}IO_{6}^{-}]=x=1.6 \times 10^{-3}\,mol\)

\(K=\frac{[H_{4}IO_{6}^{-}]}{[IO_{4}^{-}]} \rightarrow 3.5 \times 10^{-2} =\frac{x}{(0.0448-x)} \rightarrow x=1.568 \times 10^{-3}\)

\(IO_{4}^{-}:50\,mL\,IO_{4}^{-} \times \frac{1\,L\,IO_{4}^{-}}{1,000\,mL\,IO_{4}^{-}} \times \frac{0.896\,mol\,IO_{4}^{-}}{1\,L\,IO_{4}^{-}}=0.0448\,mol\)

16. \(PV=nRT \rightarrow \frac{P}{RT}=\frac{n}{V} \rightarrow \frac{12.9468}{(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K}=3.5 \times 10^{-1}\,M\)

\(K_p=\frac{(P_{IBr})^2}{(P_{I_{2}})(P_{Br_{2}})} \rightarrow 1.2 \times 10^{-2} = \frac{2x}{(1.096-x)(1.479-x)}=\frac{2x}{x^2-2.575x+1.62098} \rightarrow 0.012x^2-0.0309x+0.0194518=2x \rightarrow 0.012x^2-2.0309+0.0194518=0 \rightarrow x=12.9468\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(2.92 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.096\,atm\)

\([I_{2}]=7.4\,g\,I_{2} \times \frac{1\,mol\,I_{2}}{253\,g\,I_{2}} \times \frac{1}{1.00\,L}=2.92 \times 10^{-2}\,M\)

\(PV=nRT \rightarrow P={nRT}{V}=(3.94 \times 10^{-2}\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})(184.4+273.15)K=1.479\,atm\)

\([Br_{2}]=6.3\,g\,Br_{2} \times \frac{1\,mol\,Br_{2}}{159.808\,g\,Br_{2}}=3.94 \times 10^{-2}\,M\)

\([N_{2}]=0.5-12x=0.5-12(0.000471330)=0.494\,M\)

\([CH_{3}NH_{2}]=2x=2(0.0004713300=9.43 \times 10^{-4}\,M\)

\(If\,the\,concentration\,of\,H_{2}\,is\,doubled\,,then\,K=\frac{[CH_{3}NH_{2}]^{2}}{[N_2][H_2]^5}=\frac{(9.43 \times 10^{-4})^{2}}{(0.494)(1.998)^5}=5.65 \times 10^{-8}\)

\(2 \times [H_{2}]=2(1.0-2.5x)=2(1.0-2.5(0.000471330))=1.998\,M\)

\(K=\frac{[CH_{3}NH_{2}]^2}{[N_2][H_2]^5} \rightarrow 1.8 \times 10^{-6}=\frac{(2x)^2}{(0.5-x)(1.0-5x)^5} \rightarrow x=0.000471330\,M\)

\([N_2]=1.00\,mol\,N_2 \times \frac{1}{2.00\,L}=0.5\,M\)

\([H_2]=2.00\,mol\,H_2 \times \frac{1}{2.00\,L}=1.0\,M\)

15.4: Heterogeneous Equilibria

15.5: calculating equilibrium constants, 15.6: applications of equilibrium constants.

1. During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using Figure 15.8 and Figure 15.9 as your guides, sketch the shape of each graph using appropriate labels.

  • \(H_2O\,(l) \rightleftharpoons H_2O\,(g)\)
  • \(2\,MgO\,(s) \rightleftharpoons 2\,Mg\,(s)+O_{2}\,(g)\)
  • \(2\,PbS\,(g)+3\,O_{2}\,(g) \rightleftharpoons 2\,PbO\,(s)+2\,SO_{2}\,(g)\)

2. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  • \(2\,NaHCO_{3}\,(s) \rightleftharpoons Na_2CO_{3}\,(s) + CO_{2}\,(g)+ H_2O\,(g)\): \([CO_2]\) is doubled.
  • \(N_2F_{4}\,(g) \rightleftharpoons 2\,NF_{2}\,(g)\): \([NF_{2}]\) is decreased by a factor of 2.
  • \(H_{2}\,(g) + I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\): \([I_2]\) is doubled.

3. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium?

  • \(CS_{2}\,(g) + 4\,H_{2}\,(g) \rightleftharpoons CH_{4}\,(g) + 2\,H_2S\,(g)\): \([CS_2]\) is doubled.
  • \(PCl_{5}\,(g) \rightleftharpoons PCl_{3}\,(g) + Cl_{2}\,(g)\): \([Cl_2]\) is decreased by a factor of 2.
  • \(4\,NH_{3}\,(g) + 5\,O_{2}\,(g) \rightleftharpoons 4\,NO\,(g) + 6\,H_2O\,(g)\): \([NO]\) is doubled.

a. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([H_2O]\,(l)\,(M)\) and y-axis labeled \([H_2O]\,(g)\,(M)\). The graph should have a positive linear correlation. For any equilibrium concentration of \(H_2O\,(g)\), there is only one equilibrium \(H_2O\;(l)\). Because the magnitudes of the two concentrations are directly proportional, a large \([H_2O]\,(g)\) at equilibrium requires a large \([H_2O]\,(l)\) and vice versa. In this case, the slope of the line is equal to \(K\).

b. According to Figure 15.9, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([MgO]\,(M)\). Because \(O_2\,(g)\) is the only one in gaseous form, the graph would depend on the concentration of \(O_2\).

c. According to Figure 15.8, we could obtain a graph with the x-axis labeled \([O_3]\,(M)\) and y-axis labeled \([O2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_2\,(g)\) there is \(2\,O_3\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \([O3]\,(g)\) at equilibrium requires a large \([O_2]\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\).

d. According to figure 15.8, we could obtain a graph with the x-axis labeled \([O2]\,(M)\) and y-axis labeled \([SO2]\,(M)\). The graph should have a positive linear correlation. For every \(3\,O_{2}\,(g)\) there is \(2\,SO_2\,(g)\). Because the magnitudes of the two concentrations are directly proportional, a large \(O_2\,(g)\) at equilibrium requires a large \(SO_2\,(g)\) and vice versa. In this case, the slope of the line is equal to \(K\).

\(K=[Na_{2}CO_{3}][CO_2][H_{2}O]\)

If \([CO_2]\) is doubled, \([H_2O]\) should be halved if the system is to maintain equilibrium.

\(K=\frac{[NF_2]^2}{[N_{2}F_{4}]}\)

If \([NF_2]\) is decreased by a factor of 2, then \([N_{2}F_{4}]\) must also be decreased by a factor of 2 if the system is to maintain equilibrium.

\(K=\frac{[HI]^2}{[H_2][I_{2}]}\)

If \([I_{2}]\) is doubled then \([HI]\) must also be doubled if the system is to maintain equilibrium.

\(K=\dfrac{[CH_4][H_2S]^2}{[CS_2][H_2]^4}\)

If \([CS_2]\) is doubled then \([H_2]\) must be decreased by a factor of 2√4≅ 1.189 if the system is to maintain equilibrium.

\(K=\dfrac{[PCl_3]}{[Cl_2][PCl_5]}\)

If \([Cl_2]\) is halved then \([PCl_5]\) must also be halved if the system is to maintain equilibrium.

\(K=\dfrac{[NO]^4[H_2O]^6}{[NH_3][O_2]^5}\)

If \([NO]\) is doubled then \([H_2O]\) must also be multiplied by 22/3≅1.587 if the system is to maintain equilibrium.

  • The data in the following table were collected at 450°C for the reaction \(N_{2}\,(g)+3\,H_{2}\,(g) \rightleftharpoons 2\,NH_{3}\,(g)\):

The reaction equilibrates at a pressure of 30 atm . The pressure on the system is first increased to 100 atm and then to 600 atm . Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium?

  • For the reaction \(A \rightleftharpoons B+C\), \(K\) at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed?
  • The following two reactions are carried out at 823 K:

\[CoO\,(s)+H_{2}\,(g) \rightleftharpoons Co\,(s)+H_2O\,(g) \text{ with } K=67\]

\[CoO\,(s)+CO\,(g) \rightleftharpoons Co\,(s)+CO_{2}\,(g) \text{ with } K=490\]

  • Write the equilibrium expression for each reaction.
  • Calculate the partial pressure of both gaseous components at equilibrium in each reaction if a 1.00 L reaction vessel initially contains 0.316 mol of \(H_2\) or \(CO\) plus 0.500 mol \(CoO\).
  • Using the information provided, calculate Kp for the following reaction: \[H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_2O\,(g)\]
  • Describe the shape of the graphs of [reactants] versus [products] as the amount of \(CoO\) changes.
  • Hydrogen iodide (HI) is synthesized via \(H_{2}\,(g)+I_{2}\,(g) \rightleftharpoons 2\,HI\,(g)\), for which \(K_p = 54.5\) at 425°C. Given a 2.0 L vessel containing \(1.12 \times 10^{−2}\,mol\) of \(H_2\) and \(1.8 \times 10^{−3}\,mol\) of \(I_2\) at equilibrium, what is the concentration of \(HI\)? Excess hydrogen is added to the vessel so that the vessel now contains \(3.64 \times 10^{−1}\,mol\) of \(H_2\). Calculate \(Q\) and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations?

The system is not at equilibrium at each of these higher pressures. To reach equilibrium, the reaction will proceed to the right to decrease the pressure because the equilibrium partial pressure is less than the total pressure.

\(K_p=\frac{[NH_{3}]^2}{[N_{2}][H_{2}]^3}=\frac{[15.20]^2}{[19.17][65.13]^3}=4.4 \times 10^{-5}\)

\(K_p=\frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^3}=\frac{[321.6]^2}{[56.74][220.8]^3}= 1.7 \times 10^{-4}\)

1. \(K=\frac{[B][C]}{[A]}=\frac{[2.50][2.50]}{[2.50]}=2.50\)

2. \(K_p=K(RT)^{Δn} \rightarrow K=\frac{K_p}{(RT)^{Δn}}=\frac{19.0}{((0.08206\frac{L\cdot atm}{mol\cdot K})(200+273.15)K)^{2-1})}=0.49\)

\(K_p=\frac{(P_B)(P_C)}{(P_A)}=\frac{(1.75)(14.15)}{1.30}=19.0\)

3. \(K=\frac{[B][C]}{[A]^{2}}=\frac{(18.72)(6.51)}{12.61}=9.7\)

Experiment 1 is about the same as the given \(K\) value and thus considered to be about equilibrium. The second experiment has a \(K\) value that is about 1 so neither the formation of the reactants or products is favored. The third experiment has a \(K\) value that is larger than 1 so the formation of the products is favored.

\(K=\frac{[H_{2}O]}{[H_{2}]}\)

\(K=\frac{[CO_{2}]}{[Co]}\)

\([H_2]=[CO]=0.316\,mol\,H_{2} \times \frac{1}{1.00\,L}=0.316\,M\)

\([CoO]=0.5\,mol\,CoO \times \frac{1}{1.00\,L}=0.5\,M\)

Reaction 1:

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(4.65 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=0.314\)

\([H_{2}]=0.316-x=0.316-0.311=4.65x10^{-3}\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.311)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.0\)

\([H_{2}O]=x=0.311\)

\(K=\frac{[H_{2}O]}{[H_{2}]} \rightarrow 67=\frac{x}{0.316-x} \rightarrow x=0.311\)

Reaction 2:

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.001)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=6.75 \times 10^{-2}\,atm\)

\([CO]=0.316-x=0.316-0.315=0.001\,M\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.315)(0.08206\frac{L\cdot atm}{mol\cdot K})(823\,K)=21.3\,atm\)

\([CO_{2}]=x=0.315\,M\)

\(K=\frac{[CO_{2}]}{[Co]} \rightarrow 490=\frac{x}{0.316-x} \rightarrow x=0.315\)

\(H_{2}\,(g)+CO_{2}\,(g) \rightleftharpoons CO\,(g)+H_{2}O\,(g)\)

\(K_p=\frac{(P_{CO})(P_{H_{2}O})}{(P_{H_{2}})(P_{CO_{2}})}=\frac{(6.75 \times 10^{-2})(21)}{(0.314)(21.3)}=0.21\)

d. The shape of the graphs [reactants] versus [products] does not change as the amount of \(CoO\) changes because it is a solid.

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{n}{V}=\frac{0.101798}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.5 \times 10^{-4}\,M\,HI\)

\([HI]=2x=2(0.050899)=0.101798\,atm\)

\(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(3.2 \times 10^{-1}-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.050899\,atm\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(5.6 \times 10^{-3})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=3.2 \times 10^{-1}\,atm\)

\([H_{2}]=1.12 \times 10^{-2}\,mol\,H_{2} \times \frac{1}{2.0\,L}=5.6 \times 10^{-3}\,M\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(9.0 \times 10^{-4})(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=5.16 \times 10^{-2}\,atm\)

\([I_2]= 1.8 \times 10^{-3}\,mol\,I_{2} \times \frac{1}{2.0\,L}=9.0 \times 10^{-4}\,M\)

For excess hydrogen:

\(Q=\frac{[HI]}{[H_{2}][I_{2}]}=\frac{1.8 \times 10^{-3}}{(594.410)(1.09 \times 10^{-3})}=2.8 \times 10^{-3}\)

The reaction will proceed to the right to reach equilibrium.

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{0.103162}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.8 \times 10^{-3}\,M\)

\([HI]=2x=2(0.051581)=0.103162\,atm\)

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{10.375}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=594.410\,M\)

\([H_{2}]=10.427-x=10.427-0.051581=10.375\,atm\)

\(PV=nRT \rightarrow \frac{n}{V}=\frac{P}{RT} \rightarrow \frac{1.9 \times 10^{-5}}{(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)}=1.09 \times 10^{-3}\,M\)

\([I_{2}]=5.16 \times 10^{-2}-x=5.16 \times 10^{-2}-0.051581=1.9 \times 10^{-5}\,atm\)

\(K_p=\frac{(P_{HI})^2}{(P_{H_{2}})(P_{I_{2}})} \rightarrow 54.5=\frac{(2x)^2}{(10.427-x)(5.16 \times 10^{-2}-x)} \rightarrow x=0.051581\,atm\)

\(PV=nRT \rightarrow P=\frac{nRT}{V}=(0.182\,M)(0.08206\frac{L\cdot atm}{mol\cdot K})((425+273.15)\,K)=10.427\,atm\)

\([H_{2}]=3.64 \times 10^{-1}\,mol\,H_{2} \times \frac{1}{2.0\,L}=0.182\,M\)

15.7: Le Chatelier's Principle

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JEE Main Chemical Kinetics Practice Paper FREE PDF Download

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Chemical Kinetics Practice Paper with Solutions and Answer Key For JEE Main

Excelling in JEE Main Chemistry demands a comprehensive understanding of Chemical Kinetics Practice Paper, as it explores the rates of chemical reactions, factors affecting them, and mechanisms, enabling students to understand reaction kinetics and predict outcomes accurately.

Vedantu is assisting students in their JEE Main Chemistry preparation through a complimentary PDF download of the Chemical Kinetics Practice Paper. This distinctive resource is created for chapter-wise practice, ensuring comprehensive coverage of critical topics. By downloading it for FREE, you can enrich your knowledge and boost your confidence in addressing questions with efficiency. Detailed solutions and answer keys are provided to clarify doubts and offer step-wise methods for practicing questions. Furthermore, Vedantu’s Chemical Kinetics Practice Paper for JEE Main aids in enhancing your question-solving speed.

A minimum of 1 question was asked from the Chemical Kinetics since the chapter is an integral part of the JEE Main Exam. For a comprehensive perspective, consult the table elucidating the Chemical Kinetics chapter's weightage in the JEE Main Exam over the last five years.

Practice Papers for JEE Main help you to find and practice the questions that might get asked in the next JEE Main exam. Download the PDF of the Chemical Kinetics Practice Paper today to excel in your JEE Main exams!

Subject-wise Links For JEE Main Practice Paper

In the JEE Main exam, each of the three subjects— Chemistry, Physics, and Maths—holds a weightage of 33%. Hence along with practicing the Chemistry Practice Paper for JEE Main, students have to practice Physics, as well as Maths Practice Papers. This will lead you to score more than 80% in the JEE Main exam. Here are the links for the JEE Main Subject-wise Practice Paper.

Links For JEE Main Chapter-wise Practice Paper

Chemistry Practice Papers need to be worked out after each chapter since the questions from most of the Chemistry chapters can help you to score well in the Chemistry section of the JEE Main exam. This will help you to cover most of the JEE Main Chemistry Syllabus. You can download the Chapter-wise links for the JEE Main Practice Paper .

Important Topics From Chemical Kinetics for JEE Main Practice Paper

It will be easy for you to work out the JEE Main Practice Paper if you have a strong understanding of the Chemical Kinetics. You have to focus more on the important topics to answer most of the questions from the JEE Main Practice Paper on Chemical Kinetics. Here are some of the important topics of Chemical Kinetics.

Reaction Rate:

Reaction rate measures how quickly reactants turn into products. It's crucial to understand the factors affecting it, such as temperature, concentration, and catalysts, for predicting reaction outcomes.

Rate Law and Rate Constant:

The rate law equation describes the relationship between reactant concentrations and the rate of reaction. The rate constant (k) quantifies this relationship and is essential for calculating reaction rates.

Reaction Mechanisms:

Reaction mechanisms explain the step-by-step pathway through which a reaction occurs. Understanding these intricate processes helps in predicting reaction intermediates and controlling reaction outcomes.

Collision Theory:

Collision theory explains how reactant molecules must collide with sufficient energy and proper orientation to lead to successful reactions. It provides insights into reaction kinetics and reaction rate factors.

Arrhenius Equation:

The Arrhenius equation relates the rate constant (k) to temperature, providing a mathematical understanding of the temperature dependence of reaction rates. It's essential for predicting how changing temperature affects reaction kinetics.

Factors Affecting the Rate of Reaction:

Concentration refers to the number of particles involved; more particles generally lead to a faster reaction. Temperature increases the energy available for the reaction. Pressure affects the reaction by influencing the crowdedness of particles. A catalyst is a substance that speeds up the reaction without being consumed.

Order and Molecularity of Reaction:

The order refers to how the concentration of reactants affects the reaction rate. For example, if doubling the concentration doubles the rate, it's a first-order reaction. Molecularity, on the other hand, describes the number of reactant particles coming together in a reaction. For instance, a unimolecular reaction involves a single reactant.

Collision Theory of Biomolecular Gases Reaction:

This theory suggests that for a reaction to occur, reactant molecules must collide with sufficient energy and proper orientation. The collision theory doesn't delve into mathematical derivations but focuses on the idea that successful collisions lead to a reaction.  

Equations To Score More in Practice Paper of JEE Main Chemical Kinetics

Equations are the base to solve the JEE Main Practice Paper. You have to know which equation or formula to use while solving the Practice Paper for JEE Main. Find the important equations you need to learn while working out the Practice Paper of JEE Main Chemical Kinetics.

Rate of Reaction

The rate of reaction is given by:

\[\text{Rate} = \frac{\Delta [A]}{\Delta t} = -\frac{\Delta [B]}{\Delta t} = \frac{\Delta [C]}{\Delta t} = \frac{\Delta [D]}{\Delta t}\]

Rate Law (for a general reaction)

The rate law for a general reaction is given by:

\[\text{Rate} = k[A]^m[B]^n\]

Half-Life of a Reaction

The half-life of a reaction is given by:

\[t_{1/2} = \frac{0.693}{k}\]

Arrhenius Equation

The Arrhenius equation is given by:

\[k = A \cdot e^{-\frac{E_a}{RT}}\]

Integrated Rate Law (First Order Reaction)

For a first-order reaction, the integrated rate law is given by:

\[\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt\]

For more formulas and equations you can refer to Vedantu’s JEE Main Formula page.

What Makes Vedantu’s Practice Paper PDF of JEE Main Chemical Kinetics Different?

The role of Practice Paper for JEE Main is to equip students for the exam by providing questions created in the same pattern as the JEE Main exam. Vedantu’s JEE Main Practice Papers prove indispensable immediately after learning the completion of each Chemistry chapter. Now, let’s uncover the defining qualities of Vedantu’s Chemical Kinetics Practice Paper for JEE Main.

Quality Content: Vedantu's Practice Paper for Chemical Kinetics is curated by experienced educators and subject matter experts, ensuring that the questions are relevant, accurate, and aligned with the latest JEE Main syllabus.

Variety of Questions: They provide a diverse range of questions, covering different difficulty levels and concepts from Chemical Kinetics, allowing students to thoroughly practice and master each topic.

Detailed Solutions: Vedantu offers detailed step-by-step solutions and answer keys for the Chemical Kinetics JEE Main Practice Paper, ensuring that students understand not just the final answer but also the underlying concepts and problem-solving techniques.

User-Friendly Interface: Their platform is designed to be user-friendly, making it easy for students to navigate through Practice Paper and access the content they need efficiently.

Accessibility: Vedantu's Practice Papers are often easily accessible online and can be downloaded for FREE, allowing students to practice from the comfort of their homes.

How To Prepare For JEE Main With Chemical Kinetics Practice Paper?

Vedantu’s Chemical Kinetics JEE Main Practice Paper is composed of MCQs and Subjective type questions. At the end of the FREE PDF you can get the answer keys and detailed solutions for the questions. If you follow the below instructions while working out the Daily Practice Paper you can easily succeed in the JEE Main exam.

Download the Daily Practice Paper of JEE Main Chemical Kinetics.

You can set a timer of 1 hour.

Solve the easy questions first and give time for tough questions.

Note your answers on a sheet of paper and check with the answer key.

Each question carries 4 marks and gives a negative mark of -1 for each question.

Now calculate the score and analyse yourself. 

You can take the help of detailed solutions given in the PDF for better clarity of questions and answers. 

Learn how to do the incorrect answers and practice the questions again.

Make a note of the time you take for each question to practice. 

When to Start Preparing With JEE Main Practice Paper of Chemical Kinetics?

To ensure success in your JEE Main exams, it's wise to commence your JEE Main Practice Paper of Chemical Kinetics preparation early in your academic journey. This approach facilitates deep subject understanding and extensive revision, both critical for exam success. Consider this timeline for optimizing your JEE Main preparation with Daily Practice Paper.

Foundation Building (1-2 Years Prior): Start with foundational studies and build a strong understanding of the core concepts in Chemical Kinetics.

Concept Mastery (6-12 Months Prior): About a year before the exam, begin incorporating the Practice Paper of JEE Main Chemical Kinetics into your routine. 

Intensive Revision (3-6 Months Prior): As the exam date approaches, intensify your Practice Paper usage. Take a full-length Practice Paper to simulate exam conditions, improve time management, and identify weak areas.

Additional Materials To Cover With  JEE Main Chemical Kinetics Practice Paper

After learning with the Chemical Kinetics chapter, you need to make sure that you are mastering the contents you learn so that you can perform well in JEE Main. Practice Papers for JEE Main prepared by Vedantu is the best resource for this. Right after your revision with the Chemical Kinetics, you can practice the JEE Main Practice Paper. But this is not enough if you want to score more than 85% in JEE Main exam. Here are some additional materials that you can choose while preparing for JEE Main.

Try Our Online Practice Paper Test For Chemistry

Once you are done with practicing the JEE Main Practice Paper for Chemical Kinetics, you can test your online skills for JEE Main Chemistry. Vedantu is also providing you with an online practice paper test where you can get a real experience of attempting the JEE Main Exam.

Mastering Chemical Kinetics With JEE Main Practice Paper

The JEE Main Chemical Kinetics Practice Paper is a great way to practice for the exam. It covers a lot of important topics, and the solutions and answer keys help you check your work. By practicing with this JEE Main Practice Paper, you can learn the material better and get better at solving problems. You can also learn how to manage your time better and figure out where you need to focus your studies. So, if you're preparing for the JEE Main, be sure to download and practice Vedantu’s Chemical Kinetics Daily Practice Paper for FREE!

JEE Mains Sample Paper: Chemical Kinetics

JEE Mains is a highly competitive exam, and students need to be well-prepared to secure a good score. By practicing with JEE Main 2023 Sample Papers , students can get a feel of the actual exam and identify their strengths and weaknesses. The Chemical Kinetics chapter is a crucial section of the JEE Mains syllabus , and it is essential for students to have a thorough understanding of the concepts covered in this chapter. Our sample papers include questions from all the important topics in this chapter, helping students to assess their preparation level and identify areas where they need more practice.

With the help of JEE Main Model Papers and JEE Model Question Papers , students can also get an idea of the types of questions that are frequently asked in the exam. This can help them to develop effective strategies for answering different types of questions, which can be beneficial in improving their overall score. By practicing with these sample papers, students can boost their confidence and improve their chances of success in the JEE Mains examination.

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FAQs on JEE Main Chemical Kinetics Practice Paper FREE PDF Download

1. What is Chemical Kinetics, and why is it important for JEE Main Practice Paper?

Chemical Kinetics is the branch of chemistry that deals with the study of reaction rates and mechanisms. It is crucial for JEE Main preparation because it helps you understand the speed of chemical reactions, which is a fundamental concept in chemistry. Many questions in JEE Main involve reaction kinetics, making it an essential topic to master.

2. How can I practice Chemical Kinetics problems for JEE Main Practice Paper?

You can practice Chemical Kinetics problems for JEE Main by using practice papers and solving previous years' question papers. These papers often include questions related to reaction rates, rate laws, and reaction mechanisms. You can also find dedicated study materials and online resources for additional practice.

3. Where can I find a Chemical Kinetics Practice Paper with solutions and an answer key for JEE Main?

You can find Chemical Kinetics practice papers with solutions and answer keys for JEE Main in various JEE Main, online coaching platforms like Vedantu. You can also download the PDF of the Practice Paper for FREE.

4. What are some common topics in Chemical Kinetics that are frequently asked in the JEE Main Practice Paper? 

Common topics in Chemical Kinetics for JEE Main include reaction orders, rate equations, rate constants, half-life, activation energy, and mechanisms of chemical reactions. Questions related to these topics are commonly featured in the exam.

5. How can I improve my speed in solving Chemical Kinetics problems for JEE Main Practice Paper?

To improve your speed in solving Chemical Kinetics problems, practice regularly, and work on your problem-solving techniques. Break down complex problems into smaller steps, use shortcut methods where applicable, and time yourself while practicing to simulate exam conditions.

6. Are there any specific strategies for tackling Chemical Kinetics questions in JEE Main?

Yes, when tackling Chemical Kinetics questions in JEE Main, start by carefully reading the problem statement and identifying the given data. Then, choose the appropriate rate law equation (zero-order, first-order, or second-order) based on the information provided. Substituting values into the correct equation will lead you to the solution.

7. Is it necessary to memorize all the rate laws and equations for Chemical Kinetics in JEE Main Practice Paper?

While it's important to understand the concepts and principles of Chemical Kinetics, you don't need to memorize all the rate laws and equations. However, you should be familiar with the common rate laws (zero-order, first-order, and second-order) and know how to apply them to solve problems.

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COMMENTS

  1. PDF Test1 ch15 Kinetics Practice Problems

    Where you yourself need to first deduce the rate law, then plug in the values to solve for "k". 31. The initial rate data for the reaction 2N2O5(g) → 4NO2(g) + O2(g) is shown in the following table. Determine the value of the rate constant for this reaction. Experiment [N2O5](M) Rate (M/s) 1.28 × 102 22.5.

  2. 1.E: Kinetics (Practice Problems with Answers)

    1.E: Kinetics (Practice Problems with Answers) Page ID. A general chemistry Libretexts Textmap organized around the textbook. Chemistry: The Central Science. by Brown, LeMay, Bursten, Murphy, and Woodward. These are homework exercises to accompany the Textmap created for "Chemistry: The Central Science" by Brown et al. Complementary General ...

  3. PDF KINETICS Practice Problems and Solutions

    5. Please do #18 in chapter 12 of your text. This increases the [H2], which will increase the rate, but has no effect on k Due to Arrhenius equation, changing temperature changes the value of k. Catalyst function by lowering the activation energy, so due to Arrhenius equation, changing the activation energy changes the value of k. 6.

  4. PDF Chapter 14 Chemical Kinetics

    Chemical kinetics: the area of chemistry dealing with the speeds or rates at which reactions occur. Rates are affected by several factors: • The concentrations of the reactants: Most chemical reactions proceed faster if the concentration of one or more of the reactants is increased. • The temperature at which a reaction occurs: The rates of ...

  5. Worksheet 14: Chemical Kinetics

    Worksheet 14: Chemical Kinetics Last updated; Save as PDF ... Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. ...

  6. PDF Chapter 14 Chemical Kinetics

    CHCl (g) + Cl (g) fi CCl (g) + HCl(g) Rate = k[CHCl ] [Cl ] 1/2. 3 2 4 3 2. Answer: The rate of the reaction in Equation 14.9 is first order in N2O5 and first order overall. The reaction in Equation 14.10 is first order in CHCl3 and one-half order in Cl2. The overall reaction order is three halves.

  7. CHM 112 Kinetics Practice Problems Answers

    1. State two quantities that must be measured to establish the rate of a chemical reaction and cite several factors that affect the rate of a chemical reaction. Answer The rate of a reaction is defined as the change in concentration as a function of time.

  8. PDF Chemical Kinetics

    requires practice. Box 2 shows examples of mechanisms that are described by individual arrows. Use arrow pushing to draw the reactions shown in the Boxes and then tackle the practice problems. Box 1 Energy is not distributed evenly among molecules in a system, and if you were to examine each individual

  9. AP Chemistry: Kinetics Practice Problems

    1. Three major methods used to increase the rate of a reaction are adding a catalyst, increasing the temperature, and increasing the concentration of a reactant. From the perspective of collision theory, explain how each of these methods increases the reaction rate. a. adding a catalyst b. increasing the temperature

  10. PDF Chemical Kinetics

    8. The reaction in which NO2 forms a dimer is second order in NO2: 2NO2(g) → N2O4(g) Rate = k [NO2]2 Calculate the rate constant for this reaction if it takes 0.005 s for the initial concentration of NO2 to decrease from 0.50M to 0.25M. 9. The decomposition of hydrogen peroxide is first order in H2O2:

  11. PDF A.P. Chemistry Practice Test: Ch. 12, Kinetics MULTIPLE CHOICE. Choose

    oxygen is a product of combustion. nitrogen is a product of combustion and the system reaches equilibrium at a lower temperature. oxygen is a catalyst for combustion. Of the following, all are valid units for a reaction rate except __________. mol/L B) M/s C) mol/hr D) mol/L-hr E) g/s

  12. 4.9: Exercises on Chemical Kinetics

    Reaction starts fast at high concentrations but will slow down at low [A] 16. One of the following statements is true and the other is false regarding the first-order reaction 2A àB + C. Identify the true statement and the false one, and explain your reasoning. (a) A graph of [A] versus time is a straight line.

  13. Kinetics Practice Problems

    2. What do the terms "unimolecular" and "bimolecular" steps mean? answer This content is available to registered users only. Click here to Register! By joining Chemistry Steps, you will gain instant access to the Answers and Solutions for all the Practice Problems, Quizzes, and the powerful set of General Chemistry 1 and 2 Summary Study Guides.

  14. Chemical Kinetics Questions

    Chemical Kinetics Chemistry Questions with Solutions Q1: What is the difference between the average rate and instantaneous rate? Answer: Average rate is the rate measured for a long period of time. While the instantaneous rate is the rate measured for an infinitesimally small period of time. Q2.

  15. Chemical Kinetics Problems and Solutions

    Chemical Kinetics Problems and Solutions - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. This document provides 20 examples of rate law and rate constant calculations for first-order reactions.

  16. Kinetics questions (practice)

    Course: MCAT > Unit 9. Lesson 18: Kinetics. Kinetics questions. Introduction to reaction rates. Rate law and reaction order. Worked example: Determining a rate law using initial rates data. First-order reaction (with calculus) Plotting data for a first-order reaction. Half-life of a first-order reaction.

  17. PDF Objectives Chemical Kinetics

    97 Chemical Kinetics It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 4.1). So in problem 4.1, r inst at 600s for example, can be calculated by plotting concentration of butyl chloride as a function of time.

  18. PDF KINETICS Practice Problems and Solutions

    8. Consider the following mechanism. A2 B2 R C (slow) A2 + R → C Write the overall balanced chemical equation. 2 A2 + B2 Identify any intermediates within the mechanism. R What is the order with respect to each reactant? A2 1st; B2 1st Write the rate law for the overall reaction. rate = k [A2][B2] (fast) → 2 C 9.

  19. PDF CBSE NCERT Solutions for Class 12 Chemistry Chapter 4

    Class- XII-CBSE-Chemistry Chemical Kinetics Practice more on Chemical Kinetics Page - 2 www.embibe.com =− 1 2 (−0.1) 10 =0.005 mol L−1min−1 =5×10−3M min−1 Concept Insight: For expression the rate of reaction where stoichiometric coefficients of reactants or products are not equal to one, rate of disappearance of any of the reactants or

  20. Chemical Kinetics: Solved Example Problems

    Evaluate yourself 1 1) Write the rate expression for the following reactions, assuming them as elementary reactions. i) 3A + 5B2 →4CD ii) X2 + Y2 →2XY 2). Consider the decomposition of N2O5 (g) to form NO2 (g) and O2 (g) . At a particular instant N2O5 disappears at a rate of 2.5x10-2 mol dm-3s-1 . At what rates are NO2 and O2 formed?

  21. 2.E: Chemical Equilibrium (Practice Problems with Answers)

    b. K = [NO2] [N2O4]1 2. Although the equilibrium constant expressions have a 2:1 ratio of concentration for the products to the concentration of the reactants for the same species involved to get the K value for a. We would need to square it to get the K value for b. 3. K ′ = [NH3] [N2]1 2[H2]3 2.

  22. JEE Main Chemical Kinetics Practice Paper FREE PDF Download

    Download the Daily Practice Paper of JEE Main Chemical Kinetics. You can set a timer of 1 hour. Solve the easy questions first and give time for tough questions. Note your answers on a sheet of paper and check with the answer key. Each question carries 4 marks and gives a negative mark of -1 for each question.

  23. NCERT Solutions For Class 12 Chemistry Chapter 4 Chemical Kinetics

    Topics and Subtopics in NCERT Solutions for Class 12 Chemistry Chapter 4 Chemical Kinetics: 4.1.For the reaction R—>P, the concentration of reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. 4.2.In a reaction, 2A —-> Products, the concentration of A ...