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## Sudoku for Beginners: How to Improve Your Problem-Solving Skills

Are you a beginner when it comes to solving Sudoku puzzles? Do you find yourself frustrated and unsure of where to start? Fear not, as we have compiled a comprehensive guide on how to improve your problem-solving skills through Sudoku.

## Understanding the Basics of Sudoku

Before we dive into the strategies and techniques, let’s first understand the basics of Sudoku. A Sudoku puzzle is a 9×9 grid that is divided into nine smaller 3×3 grids. The objective is to fill in each row, column, and smaller grid with numbers 1-9 without repeating any numbers.

## Starting Strategies for Beginners

As a beginner, it can be overwhelming to look at an empty Sudoku grid. But don’t worry. There are simple starting strategies that can help you get started. First, look for any rows or columns that only have one missing number. Fill in that number and move on to the next row or column with only one missing number. Another strategy is looking for any smaller grids with only one missing number and filling in that number.

## Advanced Strategies for Beginner/Intermediate Level

Once you’ve mastered the starting strategies, it’s time to move on to more advanced techniques. One technique is called “pencil marking.” This involves writing down all possible numbers in each empty square before making any moves. Then use logic and elimination techniques to cross off impossible numbers until you are left with the correct answer.

Another advanced technique is “hidden pairs.” Look for two squares within a row or column that only have two possible numbers left. If those two possible numbers exist in both squares, then those two squares must contain those specific numbers.

## Benefits of Solving Sudoku Puzzles

Not only is solving Sudoku puzzles fun and challenging, but it also has many benefits for your brain health. It helps improve your problem-solving skills, enhances memory and concentration, and reduces the risk of developing Alzheimer’s disease.

In conclusion, Sudoku is a great way to improve your problem-solving skills while also providing entertainment. With these starting and advanced strategies, you’ll be able to solve even the toughest Sudoku puzzles. So grab a pencil and paper and start sharpening those brain muscles.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.

## Practice Problems Chemical Kinetics: Rates and Mechanisms of Chemical Reactions

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.

The rate of a reaction is defined as the change in concentration as a function of time. Thus, the two quantities that must be measured are the molarity of either a reactant or product and the time.

The factors that affect a reaction rate include the temperature, the concentration of reactants, the surface area (if solids are involved in the reaction, and the presence or absence of a catalyst.

2. Explain why the rate of disappearance of NO and the rate of formation of N 2 are not the same in the reaction, 2CO( g ) + 2NO( g ) → 2CO 2 ( g ) + N 2 ( g ).

Because of the 2:1 stoichiometric ratio between NO and N 2 , the NO must use 2 moles for each mole of N 2 produced. This means that the rate of consumption of NO is twice as fast as the rate of production of N 2 .

3. What plot of experimental data can be used to evaluate the activation energy, E a , of a reaction? How is E a related to this plot?

The experimental data required to evaluate the activation energy are rate constants as a function of absolute temperature. If ln k is plotted against 1/T, a straight line should result and the slope of the line is equal to –E a /R, where R is the ideal gas constant in energy units.

4. What are the chief requirements that must be met by a plausible reaction mechanism? Why do we say "plausible" mechanism rather than "correct" mechanism?

A reaction mechanism must meet two criteria. 1) The sum of all of the steps in the mechanism must match the observed reaction, i.e., the stoichiometry of the reaction must be satisfied. 2) The reaction mechanism must account for the experimentally observed rate law.

Reaction mechanisms are considered "plausible" rather than "correct" because different sequences of elementary reactions may meet the two requirements.

5. In a reaction mechanism, (a) what is the difference between an activated complex and an intermediate ? (b) What is meant by the rate-determining step? Which elementary reaction in a reaction mechanism is often the rate-determining step?

(a) An activated complex is the structure along the reaction pathway of the highest energy, which determines the activation energy for the reaction. An intermediate can be any structure found in the reaction path.

(b) The rate-determining step is the elementary reaction that controls the mathematical form of the overall rate law. The rate-determining step is usually the slowest elementary reaction.

6. In the reaction H 2 O 2 ( aq ) → H 2 O( l ) + ½ O 2 ( g ), the initial concentration of H 2 O 2 is 0.2546 M, and the initial rate of reaction is 9.32×10 –4 M s –1 . What will be [H 2 O 2 ] at t = 35 s?

Use the definition of rate (change in concentration over change in time), recognizing that the initial rate for the loss of hydrogen peroxide is negative and at t = 0 s, [H 2 O 2 ] = 0.2546 M.

[H 2 O 2 ] = 0.2220 M

7. For the reaction, A → products, a graph of [A] versus time is a curve. What can be concluded about the order of this reaction?

The integrated rate laws, written as linear equations, are:

Since [A] vs. t is a curve, the reaction can not be zero order.

8. Following are two statements pertaining to the reaction 2A + B → 2C, for which the rate law is rate = k [A][B]. Identify which statement is true and which is false, and explain your reasoning.

(a) The value of k is independent of the initial concentrations [A] 0 and [B] 0 .

(b) The unit of the rate constant for this reaction can be expressed either as s –1 or min –1 .

The rate law is second order overall. The rate constant, by definition, is independent of concentrations, initial or otherwise, so ( a ) must be true. The units of the rate constant for a second order reaction are M –1 s –1 or M –1 min –1 (rate = Ms –1 = k[M][M]) so ( b ) is false.

9. The rate of the following reaction in aqueous solution is monitored by measuring the number of moles of Hg 2 Cl 2 that precipitate per liter per minute. The data obtained are listed in the table.

2 HgCl 2 ( aq ) + C 2 O 4 2– ( aq ) → 2 Cl – ( aq ) + 2 CO 2 ( g ) + Hg 2 Cl 2 ( s )

Experiment [HgCl 2 ] (M) [C 2 O 4 2– ] (M) Initial rate (mol L –1 min –1 )

1 0.105 0.15 1.8×10 –5

2 0.105 0.15 1.8×10 –5

3 0.052 0.30 7.1×10 –5

4 0.052 0.15 8.9×10 –6

(a) Determine the order of reaction with respect to HgCl 2 , with respect to C 2 O 4 2– and overall.

(b) What is the value of the rate constant k ?

(c) What would be the initial rate of reaction if [HgCl 2 ] = 0.094 M and [C 2 O 4 2– ] = 0.19 M?

(d) Are all four experiments necessary to answer parts (a) - (c)? Explain.

Use the method of initial rates to find the orders of reaction in each component. This will allow evaluation of the rate constant and the initial rate of reaction at any other condition.

(a) Rate = k [HgCl 2 ] m [C 2 O 4 2– ] n

Compare the rates in experiments 1 and 2 (or 3 and 4) to find the order in oxalate ion:

Compare the rates in experiments 2 and 3 (or 1 and 4) to find the order in mercury(II) chloride:

Therefore, the reaction is first order with respect to mercury(II) chloride and second order with respect to oxalate. The overall order is the sum of these, 2 + 1 = 3, third order.

(b) To find the rate constant, use the rate equation and solve for k :

Experiment [HgCl 2 ] (M) [C 2 O 4 2– ] (M) Initial rate (mol L –1 min –1 ) k

The average k = 7.6×10 –3 M –2 min –1

(c) Rate = k [HgCl 2 ][C 2 O 4 2– ] 2

Rate = (7.6×10 –3 M –2 min –1 )[0.094 M][0.19 M] 2 = 2.6×10 –5 M min –1

(d) Since there are only two reactants, three experiments are the minimum required to find the rate equation and rate constant. Experiments 1 - 3 would have sufficed to answer the questions posed.

10. In the reaction A → products, we find that when [A] has fallen to half of its initial value, the reaction proceeds at the same rate as its initial rate. Is the reaction zero order, first order, or second order? Explain.

Since the rate is not changing as a function of time, the reaction must be zero order.

11. In the reaction, A → products, with the initial concentration [A] 0 = 1.512 M, [A] is found to be 1.496 M at t = 30 s. With the initial concentration [A] 0 = 2.584 M, [A] is found to be 2.552 M at t = 1 min. What is the order of this reaction?

First, find the rate of reaction in each experiment, being sure to be consistent in units. Then use the Method of Initial rates to find the order of reaction.

Experiment [A] 0 (M) [A] (M) at t (s) Rate (M s –1 )

Since the rates are the same for both experiments, the reaction must be zero order. Mathematically:

12. A first order reaction, A → products, has a rate of reaction of 0.00250 M s –1 when [A] = 0.484 M. (a) What is the rate constant, k , for this reaction? (b) Does t 3/4 depend on the initial concentration? Does t 4/5 ? Explain.

(a) For a first order reaction, Rate = k [A]. Since the rate and concentration are known, solving the equation for k gives the required answer.

k = Rate/[A] = 0.00250 M s –1 /0.0484 M = 5.17×10 –3 s –1 .

(b) Neither t 3/4 nor t 4/5 depend upon intial concentration because this is a first order reaction, which means that the time associated reaction is simply related to the rate constant.

13. In the first-order decomposition of dinitrogen pentoxide at 335 K,

N 2 O 5 ( g ) → 2 NO 2 ( g ) + ½ O 2 ( g )

if we start with a 2.50-g sample of N 2 O 5 at 335 K and have 1.50 g remaining after 109 s, (a) What is the value of the rate constant k ? (b) What is the half-life of the reaction? (c) What mass of N 2 O 5 will remain after 5.0 min?

Since the concentration units cancel out, we can work directly in grams for this problem.

(a) [N 2 O 5 ] 0 = 2.50 g, [N 2 O 5 ] t = 1.50 g, t = 109 s, so

so k = 4.69×10 –3 s –1

(b) t ½ = 0.693/ k = 0.693/4.69×10 –3 = 148 s.

(c) t = 5.0 min = 300. s, so

14. The smog constituent peroxyacetyl nitrate (PAN) dissociates into peroxyacetyl radicals and NO 2 ( g ) in a first order reaction with a half-life of 32 min.

If the initial concentration of PAN in an air sample is 2.7×10 15 molecules/L, what will be the concentration 2.24 h later?

Thus, k = 0.693/t ½ = 0.693/32 = 0.022 min –1 . For an initial concentration [PAN] 0 = 2.7 ×10 15 molecules/L and at t = 2.24 h = 134 min:

[PAN] = [PAN] o e – k t = (2.7×10 15 )e –(0.022)(134) = 1.4×10 14 molecules/L.

15. The following data were obtained in two separate experiments in the reaction: A → products. Determine the rate law for this reaction, including the value of k .

Experiment 1 Experiment 2

[A] (M) time (s) [A] (M) time (s)

0.800 0 0.400 0

0.775 40 0.390 64

0.750 83 0.380 132

0.725 129 0.370 203

0.700 179 0.360 278

The general rate law is Rate = – k [A] m ; the method of initial rates can be used to establish the order of reaction and give an estimate of the rate constant.

The initial rate are found from the first two data points in each experiment:

Using the initial concentrations with the initial rates:

So this is a second order reaction. The rate constant can estimated from k = –Rate/[A] 2 for each experiment: k (exp. 1) = –(–6.25×10 –4 )/[0.800] 2 = 9.8×10 –4 M –1 s –1 and k (exp. 2) = –(–1.56×10 –4 )/[0.400] 2 = 9.8×10 –4 M –1 s –1 .

k = 9.8×10 –4 M –1 s –1

To confirm this and get a better value for the rate constant, a plot of [A] –1 vs. t should be linear with a slope equal to the rate constant:

The slopes of these plots show that a better value for the rate constant is k = 1.0×10 –3 M –1 s –1 .

16. Listed below are initial rates, expressed in terms of the rate of decrease of partial pressure of a reactant for the following reaction at 826 °C. Determine the rate law for this reaction, including the value for k .

NO( g ) + H 2 ( g ) → ½N 2 ( g ) + H 2 O( g )

With initial P H2 = 400 mmHg With initial P NO = 400 mmHg

Initial P NO (mmHg) Rate (mmHg/s) Initial P H2 (mmHg) Rate (mmHg/s)

359 0.750 289 0.800

300 0.515 205 0.550

152 0.125 147 0.395

The unknown rate law is given by Rate = k [NO] m [H 2 ] n . Using the Method of Initial rates will give the rate law and the value of the rate constant. Since the units cancel in the Method of Initial rates, we do not need to convert to molarity.

To find the order in NO, use the first set of data where the pressure of H 2 is kept constant. Experiments 2 and 3 are about a factor of 2 difference in initial pressure of NO, so will give easier numbers to work with:

Within experimental error, m = 2, or second order in NO.

To find the order in H 2 , use experiments 1 and 3 for the easiest numbers:

Within experimental error, n = 1, or first order in H 2 .

The rate constant can now be found from the rate law: k = Rate/[NO] 2 [H 2 ] for each experiment:

Initial P H2 (mmHg) Initial P NO (mmHg) Initial Rate (mmHg/s) k (mmHg –2 s –1 )

400 359 0.750 (0.750)/(400)(359) 2 = 1.45×10 –8

400 300 0.515 (0.515)/(400)(300) 2 = 1.43×10 –8

400 152 0.125 (0.125)/(400)(152) 2 = 1.35×10 –8

289 400 0.800 (0.800)/(289)(400) 2 = 1.73×10 –8

205 400 0.550 (0.550)/(205)(400) 2 = 1.68×10 –8

147 400 0.395 (0.395)/(147)(400) 2 = 1.68×10 –8

Averaging all of the k values gives k = 1.55×10 –8 mmHg –2 s –1

17. Rate constants for the first-order decomposition of acetonedicarboxylic acid

CO(CH 2 COOH) 2 ( aq ) → CO(CH 3 ) 2 ( aq ) + 2 CO 2 ( g )

acetonedicarboxylic acid acetone

are k = 4.75 ×10 –4 s –1 at 293 K and k = 1.63 ×10 –3 at 303 K. What is the activation energy, E a , for this reaction?

Use the two-point form of the Arrhenius equation to answer this question:

where k 2 = 1.63 ×10 –3 at T 2 = 303 K and k 1 = 4.75 ×10 –4 at T 1 = 293 K.

18. The following is proposed as a plausible reaction mechanism:

A + B → I (slow)

I + B → C + D (fast)

What is (a) the net reaction described by this mechanism and (b) a plausible rate law for the reaction?

(a) To find the net reaction, sum up all the reactions in the mechanism and eliminate common species on each side of the equation:

A + B + I + B → I + C + D

net: A + 2B → C + D

(b) The rate law is determined by the slow step in the reaction mechanism. Since a reaction mechanism is composed of elementary reactions, the rate law for the slow step can be written by inspection, with the orders of reaction being the stoichiometric coefficients:

Rate = k[A][B]

19. The following reaction exhibits the rate law: Rate = k [NO] 2 [Cl 2 ].

2 NO( g ) + Cl 2 ( g ) → 2 NOCl( g )

Explain why the following mechanism is not plausible for this reaction.

NO( g ) + Cl 2 → ← NOCl( g ) + Cl( g ) fast

NO( g ) + Cl( g ) → NOCl( g ) slow

The rate law is determined by the slow step: Rate = k slow [NO][Cl]

However, this rate law includes and intermediate that cannot be part of the rate law for the overall reaction - only NO or Cl 2 can be part of the rate law. The [Cl] term can be eliminated using the fast, reversible step where the rate of the forward reaction is the same as the rate of the reverse reaction:

Rate forward = Rate reverse

Rate forward = k forward [NO][Cl 2 ]

Rate reverse = k reverse [NOCl][Cl]

so k forward [NO][Cl 2 ] = k reverse [NOCl][Cl]

substituting for [Cl] in the slow step rate law gives:

This rate does not match the experimental rate law so this cannot be a plausible mechanism.

20. Show that the proposed mechanism is consistent with the rate law for the following reaction in aqueous solution,

Hg 2 2+ ( aq ) + Tl 3+ ( aq ) → 2 Hg 2+ ( aq ) + Tl + ( aq )

for which the observed rate law is

Proposed Mechanism:

Hg 2 2+ ( aq ) → ← Hg 2+ ( aq ) + Hg( s ) fast

Hg( s ) + Tl 3+ ( aq ) → Hg 2+ ( aq ) + Tl + ( aq ) slow

First, check to see that the proposed mechanism matches the experimental stoichiometry.

Add the two reactions together to give:

Hg 2 2+ ( aq ) + Hg( s ) + Tl 3+ ( aq ) → Hg 2+ ( aq ) + Hg( s ) + Hg 2+ ( aq ) + Tl + ( aq )

The Hg( s ) common to both sides of the equation cancel out and the two moles of Hg 2+ ( aq ) can be collected to give the observed reaction, so the stoichiometry matches.

The proposed mechanism can be rewritten as

Hg 2 2+ ( aq ) → k 1 Hg 2+ ( aq ) + Hg( s ) Rate = k 1 [Hg 2 2+ ] (fast)

Hg 2+ ( aq ) + Hg( s ) → k –1 Hg 2 2+ ( aq ) Rate = k –1 [Hg 2+ ][Hg] (fast)

Hg( s ) + Tl 3+ ( aq ) → k 2 Hg 2+ ( aq ) + Tl + ( aq ) Rate = k 2 [Hg 2+ ][Hg] (slow)

The overall rate is determined by the slow step, but this has an intermediate ([Hg}) that must be eliminated in order to evaluate the mechanism. This can be done using the two fast steps, which must have approximately the same rate, or:

k 1 [Hg 2 2+ ] = k –1 [Hg 2+ ][Hg]

[Hg] = k 1 [Hg 2 2+ ]/ k –1 [Hg 2+ ]

Using this expression in the rate law for the slow step gives

Rate = k 2 [Hg 2+ ] k 1 [Hg 2 2+ ]/ k –1 [Hg 2+ ]

Rate = ( k 2 k 1 / k –1 )[Hg 2+ [Hg 2 2+ ]/[Hg 2+ ]

which exactly matches the observed rate law when k = k 2 k 1 / k –1

21. Benzenediazonium chloride decomposes in water yielding N 2 ( g ).

C 6 H 5 N 2 Cl( aq ) → C 6 H 5 Cl( aq ) + N 2 ( g )

The data tabulated below were obtained for the decomposition of a 0.071 M solution at 50 °C ( t = ∞ corresponds to the completed reaction). To obtain [C 6 H 5 N 2 Cl] as a function of time, note that during the first 3 min, the volume of N 2 ( g ) produced was 10.8 mL of a total of 58.3 mL, corresponding to this fraction of the total reaction: 10.8 mL/58.3 mL = 0.185. An equal fraction of the available C 6 H 5 N 2 Cl was consumed during the same time.

time (min) N 2 ( g ) (mL)

∞ 58.3

(a) Plot graphs showing the disappearance of C 6 H 5 N 2 Cl and the formation of N 2 ( g ) as a function of time.

(b) What is the initial rate of formation of N 2 ( g )?

(c) What is the rate of disappearance of C 6 H 5 N 2 Cl at t = 20 min?

(d) What is the half-life, t ½ , of the reaction?

(e) Write the rate law for this reaction, including a value for k .

(a) First, find the concentrations of C 6 H 5 N 2 Cl remaining at each time:

time (min) N 2 ( g ) (mL) fraction N 2 = mL/58.3 mL C 6 H 5 N 2 Cl (M) = 0.071×(1 – fraction)

0 0 0/58.3 = 0.00 0.071(1 – 0) = 0.071

3 10.8 10.8/58.3 = 0.185 0.071(1 – 0.185) = 0.058

6 19.3 19.3/58.3 = 0.331 0.071(1 – 0.331) = 0.047

9 26.3 26.3/58.3 = 0.451 0.071(1 – 0.451) = 0.039

12 32.4 32.4/58.3 = 0.556 0.071(1 – 0.556) = 0.032

15 37.3 37.3/58.3 = 0.640 0.071(1 – 0.640) = 0.026

18 41.3 41.3/58.3 = 0.708 0.071(1 – 0.708) = 0.021

21 44.3 44.3/58.3 = 0.760 0.071(1 – 0.760) = 0.017

24 46.5 46.5/58.3 = 0.798 0.071(1 – 0.798) = 0.014

27 48.4 48.4/58.3 = 0.830 0.071(1 – 0.830) = 0.012

30 50.4 50.4/58.3 = 0.864 0.071(1 – 0.864) = 0.0097

∞ 58.3 58.3/58.3 = 1.00 0.071(1 – 1.00) = 0.00

The plots are:

(b) From the definition of rate,

(c) Again, using the definition of rate and approximating t = 20 min using the t = 18 min and t = 21 min points,

(d) The initial concentration of benzenediazonium chloride is 0.071 M, so the first half-life is reached when the concentration has reached 0.0355 M. Using the graph from part ( a ) gives t ½ ~ 10 min.

(e) The order of the reaction can be determined by looking at the time of the second half-life, when the benzenediazonium chloride concentration has been reduced to ½ of its initial value, 0.018 M. This occurs at ~21 min, just twice the time for the first half-life. Since t ½ is the same for the two different time intervals, this means that the reaction is first order.

Rate = k [C 6 H 5 N 2 Cl]

Plotting ln[C 6 H 5 N 2 Cl] vs. t confirms this and the slope of the plot gives the rate constant, k .

The slope of the plot = –0.0664, so k = 0.0664 min –1 . (This allows a better estimate of the half-life to be t ½ = 0.693/ k = 0.693/0.0664 = 10.4 min.)

22. Hydroxide ion is involved in the mechanism but not consumed in this reaction in aqueous solution.

OCl – ( aq ) + I – ( aq ) → OH – OI – ( aq ) + Cl – ( aq )

(a) From the data in the table, determine the order of reaction with respect to OCl – , I – , and OH – , and the overall order.

[OCl – ] (M) [I – ] (M) [OH – ] (M) Rate of formation of OI – (mol L –1 s –1 )

0.0040 0.0020 1.00 4.8×10 –4

0.0020 0.0040 1.00 5.0×10 –4

0.0020 0.0020 1.00 2.4×10 –4

0.0020 0.0020 0.50 4.6×10 –4

0.0020 0.0020 0.25 9.4×10 –4

(b) Write the rate law, and determine the value of the rate constant, k .

(c) Show that the following mechanism is consistent with the net equation and with the rate law. Which is the rate–determining step?

OCl – ( aq ) + H 2 O( l ) → ← HOCl( aq ) + OH – ( aq ) I – ( aq ) + HOCl( aq ) → HOI( aq ) + Cl – ( aq ) HOI( aq ) + OH – ( aq ) → H 2 O( l ) + OI – ( aq )

(d) Is it appropriate to refer to OH – as a catalyst in this reaction? Explain.

(a) Use the method of initial rates for a rate law of the general form Rate = k [OCl – ] m [I – ] n [OH – ] p . Experiments 1 and 3 give the order in hypochlorite ion (iodide and hydroxide are constant): doubling the concentration of hypochlorite doubles the initial rate, meaning that the reaction is 1 st order in hypochlorite ion, m = 1. Experiments 2 and 3 can be used to find the order in iodide ion (hypochlorite and hydroxide are unchanged): doubling the iodide concentration doubles the rate so the reaction is also first order in iodide ion, n = 1. Experiments 3 and 4 can be used to find the order in hydroxide (hypochlorite and iodide are unchanged): doubling the concentration halves the rate, so the order in hydroxide must be –1, p = –1. Mathematically:

The overall order is the sum of each individual order = 1 + 1 + –1 = 1

(b) Based on the orders found above, the rate law is:

The rate constant can be found by using the rate law and the data:

[OCl – ] (M) [I – ] (M) [OH – ] (M) Rate of formation of OI – (mol L –1 s –1 ) 0.0040 0.0020 1.00 4.8×10 –4 60.0 s –1 0.0020 0.0040 1.00 5.0×10 –4 62.5 s –1 0.0020 0.0020 1.00 2.4×10 –4 60.0 s –1 0.0020 0.0020 0.50 4.6×10 –4 57.5 s –1 0.0020 0.0020 0.25 9.4×10 –4 58.75 s –1

The average rate constant (to the correct number of significant figures) = 60. s –1

OCl – ( aq ) + H 2 O( l ) + I – ( aq ) + HOCl( aq ) + HOI( aq ) + OH – ( aq ) → HOCl( aq ) + OH – ( aq ) + HOI( aq ) + Cl – ( aq ) + H 2 O( l ) + OI – ( aq )

After eliminating common species:

OCl – ( aq ) + I – ( aq ) → OI – ( aq ) + Cl – ( aq )

which is the correct stoichiometry.

To find the rate–determining step, consider the rate law for each elementary reaction:

Step 1, forward reaction: Rate = k forward [OCl – ][H 2 O]

Step 1, reverse reaction: Rate = k reverse [HOCl][ OH – ]

Step 2: Rate = k 2 [I – ][HOCl]

Step 3: Rate = k 3 [HOI][OH – ]

Step 1, forward does not have enough terms in the rate law. Step 1, reverse and Step 3 both have hydroxide in the numerator but the experimental rate law requires it to be in the denominator. This leaves Step 2 as the likely slow step, but this has an intermediate that must be eliminated by using Step 1, assuming it to be a fast step:

For Step 1, Rate forward = Rate reverse so k forward [OCl – ][H 2 O] = k reverse [HOCl][OH – ]

Solving for the intermediate, [HOCl] gives:

Substituting this into the rate law for Step 2, the proposed slow step, gives:

Since water is the solvent, its concentration does not change and should be grouped with the constants:

This now matches the experimental rate law.

(d) Since hydroxide is found in the denominator of the rate law, an increase in concentration slows the reaction down so OH – is an inhibitor, not a catalyst. Compare experiments 3, 4, and 5 to see how decreasing the hydroxide increases the reaction rate.

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## AP®︎/College Chemistry

Course: ap®︎/college chemistry   >   unit 5.

• Rate law and reaction order

## Worked example: Determining a rate law using initial rates data

• Units of the rate constant
• Introduction to rate law

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## Core Concepts

Some reactions will go fast, and some will go slow – the speed of the reaction is it’s reaction rate , which is dictated by a rate law . In this article, we will learn about reaction rates , rate laws, the rate constant, and the reaction order.

The rate of a chemical reaction is determined—and altered—by many factors, including the nature (of reactivity) of reactants, surface area, temperature, concentration, and catalysts. For each unique chemical reaction, rate laws can be written at a rate law equation to show how the concentrations of reactants affect the rate of the reaction. It is important to note that you can only determine rate law experimentally !

## Topics Covered in Other Articles

• Activation Energy
• Equilibrium Constant
• Chemical Reaction Types
• Influences on Reaction Rate
• Concentration Units: Normality and Parts Per Million

## What is a Rate Law? The Rate Law Equation

The reaction rate can depend on how concentrated our reactants are. A chemical reaction’s rate law is an equation that describes the relationship between the concentrations of reactants in the reaction and the reaction rate. In the standard form, the rate law equation is written as:

Let’s break down each of these components.

## Reaction Rate

Specific rate constant.

Furthermore, the units of the specific rate constant are dependent on the orders of the reaction. This will be discussed more in detail later.

## Molar Concentrations of Reactants

The rate law uses the molar concentrations of reactants to determine the reaction rate. Typically, increased concentrations of reactants increases the speed of the reaction, because there are more molecules colliding and reacting with each other.

## How to Determine the Rate Law

There are 2 main questions you’ll see when asked to determine the rate law. The first type asks you to find the rate law from elementary steps. The second type asks you to find the rate law from a table listing different experiments with different reactant concentrations and reaction rates.

Sometimes, you’ll have to find a rate law for a reaction with an intermediate. For that, you’ll need to find the rate determining step.

## From Elementary Steps

In many reactions, the chemical equation oversimplifies the reaction process. Usually, there are many intermediate reactions, or elementary steps, that occur to get from the reactants to the products.

The NO 3 and NO 2 cancel out on both sides of the equation, so you’re left with the original equation.

In these problems, you will usually be given the elementary steps and the rates of each of the steps. For example in the equation provided above, step 1 is the slow step, and step 2 is faster. The slower step is used as the rate-determining step —because the rate of reaction can only go as fast as the slowest step. You would use the rate-determining step to write the rate law by using its reactants.

The rate law does not include CO (the second reactant in the original chemical equation). This is because CO is not used in the slower, rate-determining step, so it does not affect the reaction rate.

## From a Table

To determine the rate law from a table, you must mathematically calculate how differences in molar concentrations of reactants affect the reaction rate to figure out the order of each reactant. Then, plug in values of the reaction rate and reactant concentrations to find the specific rate constant. Finally, rewrite the rate law by plugging in the specific rate constant and the orders for the reactants.

In the experiment, hydrogen iodide HI is the reactant, and H 2 and I 2 are the products. From the table, you can tell that 3 experiments of the same reaction were run, with varying concentrations of HI. In each experiment, there reaction rate was different, as a result of the different concentrations of HI.

## Determine Rate Law Using the Table

Finding the order of reactants.

Mathematically, you can use the same process to find reactant orders by plugging values into the following equation:

## Finding the Specific Rate Constant

Now that you know the order of reactant HI, you can start to write the rate law. First, plug in the order into the rate law equation.

## Units for the Specific Rate Constant

As mentioned earlier, the units for the specific rate constant depend on the order of the reaction. Keep in mind:

• The order of the reactant changes the units on the right side of the equation

As you can see, the order of each reactant affects the units of the specific rate constant.

## For More Help, Watch our Interactive Video on Rate Laws!

Example questions for determining rate law.

Write the rate law for the following reaction given the reaction mechanism elementary steps:

Explanation: Since step 1 is the slower step, it is the rate-determining step for this reaction. Write the rate law by plugging in the reactants into the rate law equation.

Explanation:

## Finding the Orders of Reactants

Let’s start by finding the order of Reactant A. As you can see in the table, between experiments 1 and 2, the concentration of B changed, but the concentration of A did not—this would not be useful in finding the order of A. However, between experiments 1 and 3, the concentration of A changed, while B did not—this is perfect for finding the order of A because A is the only thing that changed, and therefore is the only variable that could have affected the reaction rate.

Plugging in the values from the table, you get:

*As a side note, you could also do this by comparing values on the table, without using the equation. Between experiments 1 and 2, as B was halved, the reaction rate was also halved. Thus, you know that the concentration of B had a directly proportional effect on the reaction rate, and the order of B was 1.

Finally, find the value of k by plugging in the values from any of the experiments. If we choose to use experiment 1, we get:

## Our Latest Awesome Chemistry Video

Rate law practice problems.

In an overall second-order reaction involving two reactants (Reactant X and Reactant Y), you observe the reaction rate double when you double the concentration of X. In another experiment, you halve the concentration of Y. How would you expect the reaction rate to change?

What is the rate law for this reaction?

## Rate Law Practice Problem Solutions

1: The rate would halve.

A + 2B ---> C + 2D Exp. Initial A (mol/L) Initial B (mol/L) Init. Rate of Formation of C (M min -1 ) 1 0.10 0.10 3.0 x 10 -4 2 0.30 0.30 9.0 x 10 -4 3 0.10 0.30 3.0 x 10 -4 4 0.20 0.40 6.0 x 10 -4
1) compare exp. 1 and exp. 3. A remains constant and B is tripled. The rate from 1 to 3 remains constant. Conclusion: B is not in the rate law expression. 2) compare exp 1 to exp 2. The concentration of A triples (and we don't care what happens to B). The rate triples. Conclusion: first order in A. 3) we can also show first order in A by comparing exp 1 to exp 4. The concentration of A doubles and the rate doubles. Remember, B is not part of the rate law, so we don't pay any attention to it at all. rate = k[A]
Trial 1: [A] = 0.50 M; [B] = 1.50 M; Initial rate = 4.2 x 10 -3 M/min Trial 2: [A] = 1.50 M; [B] = 1.50 M; Initial rate = 1.3 x 10 -2 M/min Trial 3: [A] = 3.00 M; [B] = 3.00 M; Initial rate = 5.2 x 10 -2 M/min
Look at trial 1 and trial 2. B is held constant while A triples. The result is that the rate triples. Conclusion: A is first order.
Look at trials 2 and 3. The key to this is that we already know that the order for A is first order. Both concentrations were doubled from 2 to 3 and the rate goes up by a factor of 4. Since A is first order, we know that a doubling of the rate is due to the concentration of A being doubled. So, we look at the concentration change for B (a doubling) and the consequent rate change (another doubling - remember the overall increase was a factor of 4 - think of 4 as being a doubled doubling). Conclusion: the order for B is first order. The rate law is rate = k [A] [B]
Exp. Initial A (mmol/L) Initial B (mmol/L) Init. Rate of Formation of products (mM min -1 ) 1 4.0 6.0 1.60 2 2.0 6.0 0.80 3 4.0 3.0 0.40
1) Look at experiments 2 and 1. From 2 to 1, we see that A is doubled (while B is held constant). A doubling of the rate with a doubling of the concentration shows that the reaction is first order with respect to A. 2) Now compare experiments 1 and 3. The concentration of A is held constant while the concentration of B is cut in half. When B is cut in half, the overall rate is cut by a factor of 4 (which is the square of 2). This shows the reaction is second order in B. 3) The rate law is this: rate = k [A] [B] 2 4) Note that the comparison in (2) can be reversed. Consider that the concentration of B is doubled as you go from exp. 3 to exp. 1. When the concentration is doubled, the rate goes up by a factor of 4 (which is 2 2 ). 5) We can use any set of values to determine the rate constant: rate = k [A] [B] 2 1.60 mM min -1 = k (4.0 mM) (6.0 mM) 2 k = 0.011 mM -2 min -2 the units on k can be rendered in this manner: k = 0.067 L 2 mmol -2 min -1
Exp. Initial A (mol/L) Initial B (mol/L) Init. Rate of Formation of products (M s -1 ) 1 0.040 0.040 9.6 x 10 -6 2 0.080 0.040 1.92 x 10 -5 3 0.080 0.020 9.6 x 10 -6
1) Examine exps. 2 and 3. A remains constant while B is doubled in concentration from 3 to 2. The result of this change is that the rate of the reaction doubles. We conclude that the reaction is first order in B. 2) Now we look at exps 1 and 2. B remains constant while the concentration of A doubles. As a result of the doubled concentration, the rate also doubles. Conclusion: the reaction is first order in A. 3) The rate law for this reaction is: rate = k [A] [B]
rate = k [A] [B] 9.6 x 10 -6 M s -1 = k (0.040 M) (0.040 M) k = 0.0060 M -1 s -1
rate = (0.0060 M -1 s -1 ) (0.12 M) (0.015 M) rate = 1.08 x 10 -5 M s -1
2ClO 2 (aq) + 2OH¯(aq) ---> ClO 3 ¯(aq) + ClO 2 ¯(aq) + H 2 O(l)
Determination #1: [ClO 2 ] o = 1.25 x 10¯ 2 M; [OH¯] o = 1.30 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 2.33 x 10¯ 4 M s -1
Determination #2: [ClO 2 ] o = 2.50 x 10¯ 2 M; [OH¯] o = 1.30 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 9.34 x 10¯ 4 M s -1
Determination #3: [ClO 2 ] o = 2.50 x 10¯ 2 M; [OH¯] o = 2.60 x 10¯ 3 M Initial rate for formation of ClO 3 ¯ = 1.87 x 10¯ 3 M s -1
1) Compare #1 and #2. The concentration of ClO 2 doubles (hydroxide remains constant) and the rate goes up by a factor of four (think of it as two squared). This means the reaction is second order with respect to ClO 2 . 2) Compare #2 and #3. The concentration of hydroxide is doubled while the [ClO 2 ] remains constant. The rate doubles, showing that the reaction is first order in hydroxide. 3) The rate law is: rate = k [ClO 2 ] 2 [OH¯] 4) Calculation for the rate constant: 1.87 x 10¯ 3 M s -1 = k (2.50 x 10¯ 2 M) 2 (2.60 x 10¯ 3 M) k = 1.15 x 10 3 M¯ 2 s -1 Often the rate constant unit is rendered thusly: L 2 mol -2 s -1 . Note that the overall order of the rate law is third order and that this is reflected in the unit associated with the rate constant.
rate = (1.15 x 10 3 M¯ 2 s -1 ) (8.25 x 10¯ 3 M) 2 (5.35 x 10¯ 2 M) rate = 4.19 x 10¯ 3 M s -1
Exp. [A] [B] Initial rate of formation of C 1 0.6 0.15 6.3 x 10 -3 2 0.2 0.6 2.8 x 10 -3 3 0.2 0.15 7.0 x 10 -4
rate 1 / rate 3 = k 1 [A 1 ] x [B 1 ] y / k 3 [A 3 ] x [B 3 ] y k's will cancel and [B] will cancel rate 1 / rate 3 = [A 1 ] x / [A 3 ] x 0.0063 / 0.0007 = (0.6) x / (0.2) x 9 = 0.6 x / 0.2 x 9 = 3 x rate = k[A] 2
Exp. [CH 3 CHO] [CO] Rate (M s -1) 1 0.30 0.20 0.60 2 0.10 0.30 0.067 3 0.10 0.20 0.067
1) Note experiments 2 and 3. The [CH 3 CHO] remains constant while the [CO] changes. However, no change of the reaction rate is observed. From this, we conclude that CO is not part of the rate law. 2) Examine experiment 3 compared to experiment 1. from 3 to 1, the [CH 3 CHO] triples and the rate goes up by a factor of 9 (which is 3 2 ). Conclusion: the reaction is second order in CH 3 CHO. 3) The rate expression is this: rate = k [CH 3 CHO] 2
Q + X --> products Trial [Q] [X] Rate 1 0.12 M 0.10 M 1.5 x 10 -3 M/min 2 0.24 M 0.10 M 3.0 x 10 -3 M/min 3 0.12 M 0.20 M 1.2 x 10 -2 M/min
1) Examine trials 1 and 2. [X] is held constant while [Q] is doubled (from 1 to 2). As a result, the rate doubles. We conclude that the reaction is first order in Q. 2) Examine trials 3 and 1. The concentration of Q does not change from 3 to 1 but the concentration for X is doubled. When this happens, we observe an eight-fold increase in the rate of the reaction. We conclude that the reaction is third order in X. This arises from the fact that 2 3 = 8. In more detail: rate 3 / rate 1 = k 3 [Q 3 ] x [X 3 ] y / k 1 [Q 1 ] x [X 1 ] y k's will cancel and [Q] will cancel rate 3 / rate 1 = [X 3 ] x / [X 1 ] x 0.012 / 0.0015 = (0.20) x / (0.10) x 8 = 0.20 x / 0.10 x 8 = 2 x x = 3
rate = k [Q] [X] 3 0.012 M min -1 = k (0.12 M) (0.20 M) 3 k = 12.5 L 3 mol -3 min -1
Trial [NO] o [O 2 ] o Initial reaction rate, M/s 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.040 0.020 0.227
1) Doubling O 2 doubles reaction rate (trials 2 and 1). 2) Doubling NO increases reaction rate 8x (trials 2 and 3). 3) Conclusion: first order O 2 , third order NO.
A + B ---> 2C
(a) rate = k [A] 2 [B] (b) rate = k [A] [B] (c) rate = k [A] [B] 2 (d) rate = k [A] 1/2 [B]
R = k[1] 0.5 = 1 R = k[2] 0.5 = 1.4 R = k[4] 0.5 = 2 R = k[8] 0.5 = 2.8 R = k[16] 0.5 = 4
The reaction rate is not doubled when you double the concentration. Notice that when the concentration changed from 2 to 8, the R went from 1.4 to 2.8. So when the concentration is quadrupled, the R doubles.
When the concentration doubles, the rate goes up by a factor which is the square root of two.

Chapter 17. Kinetics

Jessie A. Key

Learning Objectives

• To gain an understanding of rate laws and determine rate laws from initial rates.
• To gain an understanding of and the ability to determine reaction orders (including units).

The mathematical relationship of reaction rate with reactant concentrations is known as the rate law . This relationship may rely more heavily on the concentration of one particular reactant, and the resulting rate law may include some, all, or none of the reactant species involved in the reaction.

For the following hypothetical reaction

the rate law can be expressed as:

The proportionality constant, k , is known as the rate constant and is specific for the reaction shown at a particular temperature. The rate constant changes with temperature, and its units depend on the sum of the concentration term exponents in the rate law. The exponents ( y and z ) must be experimentally determined and do not necessarily correspond to the coefficients in the balanced chemical equation.

## Reaction Order

The sum of the concentration term exponents in a rate law equation is known as its reaction order . We can also refer to the relationship for each reactant in terms of its exponent as an order.

For the following reaction between nitrogen dioxide and carbon monoxide:

NO 2 (g) + CO(g) → NO(g) + CO 2 (g)

The rate law is experimentally determined to be: rate = k [NO 2 ] 2

Therefore, we would say that the overall reaction order for this reaction is second-order (the sum of all exponents in the rate law is 2), but zero-order for [CO] and second-order for [NO 2 ].

The reaction order is most often a whole number such as 0, 1, or 2; however, there are instances where the reaction order may be a fraction or even a negative value.

Earlier, it was mentioned that the units of the rate constant depend on the order of the reaction. Let’s quickly examine why this occurs. A simplified rate law can be expressed generically in the following way:

Therefore, the units of the rate constant should be:

## Determining Rate Laws from Initial Rates

The rate law can be determined experimentally using the method of initial rates, where the instantaneous reaction rate is measured immediately on mixing the reactants. The process is repeated over several runs or trials, varying the concentration one reactant at a time. These runs can then be compared to elucidate how changing the concentration of each reactant affects the initial rate.

Example 17.3

The initial rate of reaction for the reaction E + F → G was measured at three different initial concentrations of reactants as shown in the table.

• Determine the rate law of the reaction.
• Determine the rate constant.

Therefore, the reaction is first order with respect to [E].

Comparing trials 1 and 3, [F] is doubled, while [E] and the rate constant are held constant. This comparison will allow us to determine the order of reactant F:

Therefore, the reaction is zero order with respect to [F].

The rate law can now be written as:

Using Trial 1:

Key Takeaways

• The rate law is a mathematical relationship obtained by comparing reaction rates with reactant concentrations.
• The reaction order is the sum of the concentration term exponents in a rate law equation.
• A reaction’s rate law may be determined by the initial rates method.

• Determining n, m, and p from reaction orders
• Determining n, m, and p from initial rate data
• Determining the rate constant

#### IMAGES

1. Integrated Rate Law Problems 1

2. How to Solve rate problems in algebra « Math :: WonderHowTo

3. Solving a Rate Law Using the Initial Rates Method

4. Rate Law Problem

5. Example of a Rate Law problem

6. Rate law practice

#### VIDEO

1. PS 6.3- Solving Rate Problems- Lesson Video

2. 3.9

3. Initial Rate Method

4. Experiment: Determination of rate law

5. How to solve MORE Integrated Rate Law Problems in AP Chemistry

6. Desmos 6.3.07 Solving Rate Problems

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