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:

For zero order reactions:

For first order reactions:

For second order reactions:

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*

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}

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})

1 1.512 1.496 at 30 s

2 2.584< 2.552 at 60 s

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

, which can only be true for m = 0.

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?

For a first order reaction, the integrated rate law is and the rate constant can be found from the half-life: .

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^{–kt} =
(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**

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 acidacetone

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

Hg^{2+}(*aq*) + Hg(*s*)
→
*k _{–1}*
Hg

Hg(*s*) + Tl^{3+}(*aq*)
→
*k _{2}*
Hg

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

so

[Hg] = *k _{1}*[Hg

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

Rate = *k _{2}*[Hg

or

Rate = (*k _{2}k_{1}*/

which exactly matches the observed rate law when *k* = *k _{2}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)

00

310.8

619.3

926.3

1232.4

1537.3

1841.3

2144.3

2446.5

2748.4

3050.4

∞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)

00 0/58.3 = 0.00 0.071(1 – 0) = 0.071

310.8 10.8/58.3 = 0.185 0.071(1 – 0.185) = 0.058

619.3 19.3/58.3 = 0.331 0.071(1 – 0.331) = 0.047

926.3 26.3/58.3 = 0.451 0.071(1 – 0.451) = 0.039

1232.4 32.4/58.3 = 0.556 0.071(1 – 0.556) = 0.032

1537.3 37.3/58.3 = 0.640 0.071(1 – 0.640) = 0.026

1841.3 41.3/58.3 = 0.708 0.071(1 – 0.708) = 0.021

2144.3 44.3/58.3 = 0.760 0.071(1 – 0.760) = 0.017

2446.5 46.5/58.3 = 0.798 0.071(1 – 0.798) = 0.014

2748.4 48.4/58.3 = 0.830 0.071(1 – 0.830) = 0.012

3050.4 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})

0.00400.0020
1.00
4.8×10^{–4}

0.00200.0040
1.00
5.0×10^{–4}

0.00200.0020
1.00
2.4×10^{–4}

0.00200.0020
0.50
4.6×10^{–4}

0.00200.0020
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})0.00400.0020 1.00 4.8×10

^{–4}60.0 s^{–1}0.00200.0040 1.00 5.0×10

^{–4}62.5 s^{–1}0.00200.0020 1.00 2.4×10

^{–4}60.0 s^{–1}0.00200.0020 0.50 4.6×10

^{–4}57.5 s^{–1}0.00200.0020 0.25 9.4×10

^{–4}58.75 s^{–1}

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

(c) First, add all the equations in the mechanism together to see if the stoichiometry of the mechanism is correct:

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.