14: Rates of Chemical Reactions

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Prelude

Textbook: Section 14.1

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Exercise \(\PageIndex{1a}\)

Solve the following equation for [A].

\[ln\left ( \frac{\left [ A \right ]}{\left [ A \right ]_{0}} \right )=-kt\]

Answer

\[ln\left ( \frac{\left [ A \right ]}{\left [ A \right ]_{0}} \right )=-kt\]

multiply by e

\[\frac{\left [ A \right ]}{\left [ A \right ]_{0}}=e^{-kt}\]

\[\left [ A \right ]=\left [ A \right ]_{0}e^{-kt}\]

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Exercise \(\PageIndex{1b}\)

Solve the following equation for [A]₀.

\[ln\left ( \frac{\left [ A \right ]}{\left [ A \right ]_{0}} \right )=-kt\]

Answer

\[ln\left ( \frac{\left [ A \right ]}{\left [ A \right ]_{0}} \right )=-kt\]

multiply by e

\[\frac{\left [ A \right ]}{\left [ A \right ]_{0}}=e^{-kt}\]

\[\left [ A \right ]_{0}=\frac{\left [ A \right ]}{e^{-kt}}\]

\[\left [ A \right ]_{0}=\left [ A \right ]e^{kt}\]

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Exercise \(\PageIndex{1c}\)

Solve the equation for k.

\[A=A_{0}e^{-kt}\]

 

Answer

\[A=A_{0}e^{-kt}\]

\[\frac{A}{A_{0}}=e^{-kt}\]

take the ln of both sides

\[ln\left ( \frac{A}{A_{0}}\right )=-kt\]

\[k=-\frac{ln\left ( \frac{A}{A_{0}} \right )}{t}\]

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Exercise \(\PageIndex{1d}\)

Solve the equation for E_a.

\[k=Ae^{-\frac{E_{a}}{RT}}\]

Answer

\[k=Ae^{\frac{-E_{a}}{RT}}\]

\[\frac{k}{A}=e^{\frac{-E_{a}}{RT}}\]

take the ln of both sides

\[ln\frac{k}{A}=\frac{-E_{a}}{RT}\]

\[E_{a}=-RTln\frac{k}{A}\]

\[E_{a}=RTln\frac{A}{k}\]

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Exercise \(\PageIndex{1e}\)

Solve the equation for T.

\[k=Ae^{-\frac{E_{a}}{RT}}\]

Answer

\[k=Ae^{-\frac{E_{a}}{RT}}\]

\[\frac{k}{A}=e^{\frac{-E_{a}}{RT}}\]

take the ln of both sides

\[ln\frac{k}{A}={\frac{-E_{a}}{RT}}\]

\[E_{a}=-RTln\frac{k}{A}\]

\[E_{a}=RTln\frac{A}{k}\]

\[T=\frac{E_{a}}{Rln\frac{A}{k}}\]

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Exercise \(\PageIndex{1f}\)

Solve the equation for t_1/2.

\[A=A_{0}\left ( \frac{1}{2} \right )^{\frac{t}{t_{1/2}}}\]

Answer

\[A=A_{0}\left ( \frac{1}{2} \right )^{\frac{t}{t_{1/2}}}\]

\[\frac{A}{A_{0}}=\left ( \frac{1}{2} \right )^{\frac{t}{t_{1/2}}}\]

take the log (or ln) of both sides

\[log\left ( \frac{A}{A_{0}}\right )={\frac{t}{t_{1/2}}}log\left ( \frac{1}{2} \right )\]

\[t_{1/2}*log\left ( \frac{A}{A_{0}}\right )=t*log\left ( \frac{1}{2} \right )\]

\[t_{1/2}=\frac{t*log\left ( \frac{1}{2} \right )}{log\left ( \frac{A}{A_{0}}\right )}\]

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Exercise \(\PageIndex{g}\)

Solve \(A=B(C)^n\), solve for n

Answer

\[A=B\left ( C \right )^n \\ \frac{A}{B}=\left ( C \right )^n \\ log\frac{A}{B}=log\left ( C \right )^n \\ log\frac{A}{B}=nlog\left( C \right ) \\ n= \frac{log\left ( \frac{A}{B} \right )}{logC}\]

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Exercise \(\PageIndex{h}\)

Show \(\frac{log_{10}(570)}{log_{10}(30)}= \frac{ln(570)}{ln(30)} \)

Answer

\[\frac{log570}{log30}=\frac{2.75587}{1.477}=1.86 \\ \; \\ \; \\  \frac{ln570}{ln30}=\frac{6.3456}{3.4012}=1.86\]

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Exercise \(\PageIndex{i}\)

Show \(\frac{log_{10}(570)}{log_{10}(30)}\neq log_{10}\frac{570}{30} \)

Answer

\[\frac{log570}{log30}=\frac{2.75587}{1.477}=1.86 \\ \; \\ \; \\  log_{10}\frac{570}{30}=log(19)=1.28\].

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Rates of Chemical Reactions

Textbook: Section 14.2

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Exercise \(\PageIndex{2a}\)

In the reaction 2NO₂ → 2NO + O₂ at 300⁰C, the concentration of NO₂ decreased from 0.0150 M to 0.0115M in 100s.  What is the rate of disappearance of NO₂?

Answer

\[rate=\frac{\Delta [ NO_{2}]}{\Delta t}=\frac{0.0115-0.0150}{100}=-3.5*10^{-5}M/s \]

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Exercise \(\PageIndex{2b}\)

In the reaction 2NO₂ → 2NO + O₂ at 300⁰C, the concentration of NO₂ decreased from 0.0150 M to 0.0115M in 100s.  what is the rate of appearance of O₂?

Answer

\[2{NO_{2}} \rightarrow 2{NO} + 1{O_{2}}\]

\[-\frac{1}{2}\frac{\Delta [{NO_{2}}]}{\Delta t} =\frac{1}{2}\frac{\Delta [{NO}]}{\Delta t}=\frac{\Delta [{O_{2}}]}{\Delta t}\]

\[\frac{\Delta [{O_{2}}]}{\Delta t}=-\frac{1}{2}\frac{\Delta [{NO_{2}}]}{\Delta t}=\frac{-1}{2}\frac{0.0115-0.0150}{100}=1.75*10^{-5}\,M/s\]

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Exercise \(\PageIndex{2c}\)

In the reaction 2NO₂ → 2NO + O₂ at 300⁰C, the disappearance of NO₂ (g) was monitored as the following

Time (s)	[NO2] (mol/L)
0.0	0.124
10.0	0.110
20.0	0.088
30.0	0.073
40.0	0.054

What is the average rate of disappearance of NO₂ between 10.0 and 30.0 seconds?

Answer

\[rate=\frac{[ NO_{2}]_{(2)}-[ NO_{2}]_{(1)}}{t_{(2)}-t_{(1)}}\]

\[rate=\frac{0.073-0.110}{30.0-10.0}=-0.00185M/s\]

Note, you need a negative sign in front of the rate of any reactant if you want to relate this rate to that of another species in the reaction

 

 

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Exercise \(\PageIndex{2d}\)

Use the data in Q 14.1.c  to determine the relative rate of the reaction for the appearance of O_{2} between 10.0 and 30.0 seconds.

Answer

\[rate=-\frac{1}{2} \frac{\Delta{[ NO_{2}]}}{\Delta t} =\frac{\Delta [{O_{2}}]}{\Delta t}\]

\[-\frac{1}{2}rate_{(NO_{2})} =rate_{(O_{2})} \]

\[rate_{(O_{2})}=\frac{1}{2} 0.00185 M/s=9.25*10^{-4}M/s\]

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Exercise \(\PageIndex{2e}\)

Which of the following has the fastest appearance or disappearance rate?

C₃H₈ (g) + 5O₂ (g) → 3CO₂ (g) +4H₂O(l)

Answer

Oxygen because it has the largest coefficient.

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Exercise \(\PageIndex{14.2f}\)

Consider the following reaction in aqueous solution,

5 Br^- (aq) + BrO₃^- (aq) + 6 H⁺ (aq) -> 3 Br₂ (aq) + 3 H₂O (l)

If the rate of disappearance of Br^- (aq) at a particular moment during the reaction is 3.5 x 10^-4 M s^-1, what is the rate of appearance of water at that moment?

Answer

note we are using derivative notation and d represents \(\Delta\)

\[
R \;=\; -\tfrac{1}{5}\tfrac{d[\mathrm{Br^-}]}{dt}
=\; -\tfrac{d[\mathrm{BrO_3^-}]}{dt}
=\; -\tfrac{1}{6}\tfrac{d[\mathrm{H^+}]}{dt}
=\; \tfrac{1}{3}\tfrac{d[\mathrm{Br_2}]}{dt}
=\; \tfrac{1}{3}\tfrac{d[\mathrm{H_2O}]}{dt}
\]

\[
\frac{d[\mathrm{H_2O}]}{dt} \;=\; -\frac{3}{5}\,\frac{d[\mathrm{Br^-}]}{dt}
= -\frac{3}{5}\,\left(-3.5\times 10^{-4}\right)
\]

\[
= 2.1 \times 10^{-4}\ \mathrm{M\,s^{-1}}
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Exercise \(\PageIndex{14.2g}\)

Consider the following reaction in aqueous solution,

5Br^-(aq) + BrO₃^-(aq) + 6H⁺(aq) --> 3Br₂(aq) + 3H₂O(l)

If the rate of appearance of Br₂ at a particular moment during the reaction is 0.125 M s^-1, what is the rate of disappearance (in M s^-1) of Br^- at that moment?

Answer

0.21

 

Reaction Conditions and Rate

Exercise \(\PageIndex{3a}\)

Name for factors that influence the rate of a reaction.

Answer

1. Concentration
2. Temperature
3. Catalysts
4. surface area on heterogenous systems.

 

Effect of Concentration on Reaction Rate

Textbook: Section 14.4

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Exercise \(\PageIndex{4a}\)

The reaction 2A+3B→products are second order in A and first order in B.  What is the rate law expression for this reaction?

Answer

\[rate=k[A]^{2}[B]\]

 

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Exercise \(\PageIndex{4b}\)

In a reaction of third order in A, increasing the concentration of A by a factor of 4 will cause what kind of change in reaction rate?

Answer

\[rate=k\left [ A \right ]^{3}\]

\[k\left [ 4A \right ]^{3}=64k\left [ A \right ]^{3}=64rate\]

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Exercise \(\PageIndex{4c}\)

For the reaction 2A+3B→ C + 2D

Experiment	[A] (M)	[B] (M)	initial rate (M/s)
1	0.012	0.035	0.10
2	0.024	0.070	0.80
3	0.024	0.035	0.10

What is the order of reaction with respect to A?

Answer

\[rate_{1}=k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}\]

\[rate_{3}=k\left [ A_{3} \right ]^{m}\left [ B_{3} \right ]^{n}\]

\[\frac{rate_{1}}{rate_{3}}=\frac{k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}}{k\left [ A_{3} \right ]^{m}\left [ B_{3} \right ]^{n}}=\left ( \frac{A_{1}}{A_{3}} \right )^{m}\left ( \frac{B_{1}}{B_{3}} \right )^{n}\]

select when \(\frac{B_{1}}{B_{3}}=1\) (data in the experiment 1 and 3)

\[\frac{rate_{1}}{rate_{3}}=\left ( \frac{A_{1}}{A_{3}} \right )^{m}=\left ( \frac{0.012}{0.024} \right )^{m}=\frac{0.10}{0.10}=1\]

therefore m=0

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Exercise \(\PageIndex{4d}\)

In Q 14.3.c, What is the order of reaction with respect to B?

Answer

\[\frac{rate_{2}}{rate_{3}}=\frac{k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}}{k\left [ A_{3} \right ]^{m}\left [ B_{3} \right ]^{n}}=\left ( \frac{A_{2}}{A_{3}} \right )^{m}\left ( \frac{B_{2}}{B_{3}} \right )^{n}\]

select when \(\frac{A_{2}}{A_{3}}=1\) (data in the experiment 2 and 3)

\[\frac{rate_{2}}{rate_{3}}=\left ( \frac{B_{2}}{B_{3}} \right )^{n}=\left ( \frac{0.070}{0.035} \right )^{n}=2^{n}=\frac{0.80}{0.10}=8\]

therefore n=3

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Exercise \(\PageIndex{4e}\)

In Q 14.3.c, What is the order of overall reaction?

Answer

The total reaction order is the sum of the exponents on each component. So 0+3=3

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Exercise \(\PageIndex{4f}\)

In Q 14.3.c, What is the rate constant for the reaction?

Answer

From the answer of Q 14.3.3 and Q 14.3.4,

we know that \(rate=k\left [ A \right ]^{0}\left [ B \right ]^{3}\),

exp 1: rate = 0.10 =k[0.035]³

k=2.3*10³ M^-2s^-1

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Exercise \(\PageIndex{4.g}\)

 Consider the following reaction in aqueous solution,

I^- (aq) + OCl^- (aq) -> IO^- (aq) + Cl^- (aq)

and the following initial concentration and initial rate data for this reaction:

[I^-], M	[OCl^-], M	initial rate, M s^-1
0.1000	0.0500	3.05 x 10^-4
0.2000	0.0500	6.10 x 10^-4
0.3000	0.0100	1.83 x 10^-4
0.3000	0.0200	3.66 x 10^-4

What is the Rate Law 

Answer

R = l[I^-][OCl^-]

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Exercise \(\PageIndex{4.h}\)

What is the overall order of reaction if a rate constant has units of M^-2s^-1 ?

Answer

Third

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Exercise \(\PageIndex{4.i}\)

Using the information below, the rate constant for the following reaction           is ____M^-1s^-1.

                                                A + B -->  P

                                                                                                            Initial

                        Experiment #               [A](M)             [B](M)             Rate (M/s)

                                    1                      0.273               0.763               2.83

                                    2                      0.273               1.526               2.83

                                    3                      0.819               0.763               25.47

.

Answer

\[
\textbf{Reaction:}\quad A + B \rightarrow P
\qquad
\textbf{Rate law:}\quad R = k[A]^m[B]^n
\]

\[
\begin{array}{c|ccc}
\text{Exp.} & [A]\ (\mathrm{M}) & [B]\ (\mathrm{M}) & R\ (\mathrm{M\,s^{-1}})\\ \hline
1 & 0.273 & 0.763 & 2.83\\
2 & 0.273 & 1.526 & 2.83\\
3 & 0.819 & 0.763 & 25.47
\end{array}
\]

\[
\textbf{Step 1: Solve for } n \text{ with } [A] \text{ constant (compare 1 and 2).}
\]
\[
\frac{R_2}{R_1} \;=\; \frac{k[A]^m[B_2]^n}{k[A]^m[B_1]^n}
\;=\;\left(\frac{B_2}{B_1}\right)^{\!n}
\quad\Longrightarrow\quad
\frac{2.83}{2.83} \;=\;\left(\frac{1.526}{0.763}\right)^{\!n}
\;=\; 2^{\,n}
\]
\[
1 \;=\; 2^{\,n} \;\;\Longrightarrow\;\; \boxed{\,n=0\,}
\]

\[
\textbf{Step 2: Solve for } m \text{ with } [B] \text{ constant (compare 1 and 3).}
\]
\[
\frac{R_3}{R_1} \;=\; \frac{k[A_3]^m[B]^n}{k[A_1]^m[B]^n}
\;=\;\left(\frac{A_3}{A_1}\right)^{\!m}
\quad\Longrightarrow\quad
\frac{25.47}{2.83} \;=\;\left(\frac{0.819}{0.273}\right)^{\!m}
\]
\[
9.000 \;=\; 3.000^{\,m}
\;\;\Longrightarrow\;\;
\boxed{\,m=2\,}
\]

\[
\textbf{Step 3: Determine } k \text{ using any experiment (use \#1). With } n=0,\, m=2:
\quad R = k[A]^2
\]
\[
k \;=\; \frac{R_1}{[A_1]^2}
\;=\; \frac{2.83\ \mathrm{M\,s^{-1}}}{(0.273\ \mathrm{M})^2}
\;=\; \frac{2.83}{0.273^2}\ \mathrm{M^{-1}\,s^{-1}}
\;=\; \frac{2.83}{0.074529}\ \mathrm{M^{-1}\,s^{-1}}
\;\approx\; 37.97\ \mathrm{M^{-1}\,s^{-1}}
\]

\[
\boxed{\,k \approx 3.80\times 10^{1}\ \mathrm{M^{-1}\,s^{-1}},\;\; m=2,\;\; n=0\,}
\]
 

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Integrated Rate Law

Textbook: Section 14.4

Chemical Kinetics & Half-Life

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Exercise \(\PageIndex{5.1a}\)

For the first-order reaction A + B → C, the initial concentration of A is 0.26M, how long it must take for the concentration of A to reduce to 0.176M?  k=0.33min^-1   

Answer

\[[A]_{0}e^{-kt}=[A]\]
\[0.26e^{-.33t}=0.176\]
\[e^{-.33t}=\frac{0.176}{0.26}\]
\[e^{-.33t}=0.677\]
\[-0.33t=ln(0.677)\]
\[-0.33t=-0.390\]
\[t=\frac{-0.390}{.-33}=1.18\,min\]

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Exercise \(\PageIndex{5.1b}\)

If the initial concentration of the reactant of a first order reaction is 0.135M, what is the concentration after 1.5s.  k=0.75s^-1

Answer

\[ln\frac{A}{A_{0}}=-kt\]

\[\frac{A}{A_{0}}=e^{-kt}\]

\[A=A_{0}e^{-kt}\]

\[A=0.135Me^{-0.75\left ( 1.5 \right )} \\ A=0.044M\]

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Exercise \(\PageIndex{5.1c}\)

The rate constant for a second-order reaction is 0.26M^-1s^-1, if the initial concentration of reactant is 0.26M, how long it must take for the concentration to drop to 0.13M?

Answer

\[\frac{1}{A}-\frac{1}{A_{0}}=kt\]

\[0.26t=\frac{1}{0.13}-\frac{1}{0.26}\]

\[t=14.79s\]

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Exercise \(\PageIndex{5.1d}\)

What is the rate constant of a first-order reaction that has a half-life of 144s?

Answer

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

\[k=\frac{0.693}{144.0}=4.8*10^{-3}\,s^{-1}\]

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Exercise \(\PageIndex{5.1e}\)

A reaction is in first-order of the reactant A.  A solution initially has 0.120M of A is found to have 0.015M after 54 min.  What is the half-life? 

Answer

\[ln\frac{0.015}{0.120}=-k*54\]

\[k=0.0385\,min^{-1}\]

\[t_{1/2}=\frac{0.693}{0.0385}=18.0\,min\]

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Exercise \(\PageIndex{5.1.f}\)

The gas phase decomposition of N₂O₃ is a first order process with a rate constant of 1.50x10^-3/sec at 55^oC.  The decomposition reaction is

 

            N₂O₃(g)   -->    2NO₂(g)+ 1/2O₂(g)

 

Ten (10.0) mol of N₂O₃ is placed in vessel 1 and five (5.0) mol of N₂O₃ is placed in vessel 2.  The vessels have the same volume and are at the same temperature.  You measure the time for the system to reach half the concentration in each vessel, which of the following statements are true?

a.  Vessel 1 requires twice as much time as vessel 2
b.  Vessel 2 requires twice as much time as vessel 1
c.  Vessel 1 requires three times as much time as vessel 2
d.  Vessel 1 requires four times as much time as vessel 2
e.  Vessel 1 requires the same amount of time as vessel 2

Answer

e.  Vessel 1 requires the same amount of time as vessel 2

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Exercise \(\PageIndex{5.1g}\)

5.  For the chemical reaction 2 NO₂(g) -> 2 NO (g) + O₂(g), a plot of [NO₂] vs. time gives a curved line, a plot of 1/[NO₂] gives a straight line with a positive slope, and a plot of ln[NO₂] vs. time gives a curved line. What is the order of reaction?

Answer

Second Order

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Zero-Order Integrated Rate Problems

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

Figure \(\PageIndex{1}\): Consider the zero-order reaction for the following problems

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Exercise \(\PageIndex{5.2a}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

The rate of a zero-order-reaction is dependent on which of the followings:

1. Reactant
2. Product
3. Both
4. Neither

Answer

d. Neither

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Exercise \(\PageIndex{5.2b}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

What is the rate constant if it takes 60 seconds for a 0.50 M of the ammonia gas to decompose to 0.25M?

Answer

\[A=A_{0}-kt\]

\[k=\frac{1}{t}(A_0-A)=\frac{1}{60 sec}(0.5-0.25) = 0.0042\,M/s\]

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Exercise \(\PageIndex{5.2c}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

How long does it take for 0.50M of the ammonia gas to decompose to 0.10M?

Answer

\[A=-kt+A_{0}=A_{0}-kt\]

\[0.1=0.5-0.0042t\]

\[t=96s\]

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Exercise \(\PageIndex{5.2d}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

How long does it take for 0.50M ammonia to decompose to 25% if the rate constant is 0.0042M/s

Answer

\[A=-kt+A_{0}=A_{0}-kt\]

\[t=\frac{1}{k}(A_0-A)=\frac{1}{0.0042}[0.50-0.25(0.50)] \]

\[t=89s\]

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Exercise \(\PageIndex{5.2e}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

How long does it take for 25% of the reactant in question 14.5.2d to be consumed?

Answer

\[A=A_{0}-kt\]

\[t=\frac{A_{0}-A}{k}=\frac{A_{0}-0.75A}{k}=\frac{A_{0}\left ( 1-0.75 \right )}{k}=\frac{0.25A_{0}}{k}\]

\[t=\frac{0.25\left ( 0.50 \right )}{0.0042}\]

\[t=30s\]

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Exercise \(\PageIndex{5.2f}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

What is [NH₃] of a 0.80M [NH₃] solution after 3 minutes?

Answer

\[A=A_{0}-kt=0.80-\left ( 0.0042*180 \right )=0.044M\]

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Exercise \(\PageIndex{5.2f}\)

\(2NH_{3}\,(g)\,\rightarrow N_{2}\,(g)\,+\,3H_{2}\,(g)\)

Calculate the initial [NH₃] if [NH₃] = 0.005M after 10 minutes?

Answer

\[A=A_{0}-kt\]

\[0.005=A_{0}-\left ( 0.0042*600\right )\]

\[A_{0}=2.5M\]

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Exercise \(\PageIndex{5.2g}\)

8. What is the rate constant of a zero-order reaction if it takes 87 seconds for a concentration of 0.32500M to decompose to 0.00387M?

Answer

Given

\[
[A]_0 = 0.32500\ \text{M}
\]

\[
[A]_t = 0.00387\ \text{M}
\]

\[
t = 87\ \text{s}
\]

---

Step 1: Use the integrated rate law for a zero-order reaction

\[
[A]_t = [A]_0 - kt
\]

---

Step 2: Solve for the rate constant \(k\)

\[
k = \frac{[A]_0 - [A]_t}{t}
\]

---

Step 3: Substitute values

\[
k =
\frac{0.32500 - 0.00387}{87}
\]

\[
k = 3.69 \times 10^{-3}\ \text{M s}^{-1}
\]

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First-Order Integrated Rate Problems

\(Br_{2}\rightarrow 2Br\)

Figure \(\PageIndex{2}\): Consider the first-order reaction for the following problems.

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Exercise \(\PageIndex{5.3a}\)

\(Br_{2}\rightarrow 2Br\)

What is the rate constant if it takes 60 seconds for a 0.50 M mixture to decompose to 0.25M?

Answer

\[A=A_{0}e^{-kt}\]

\[\frac{A}{A_{0}}=e^{-kt}\]

\[ln\frac{A}{A_{0}}=-kt\]

\[k=-\frac{1}{t}\left ( ln\,A-ln\,A_{0} \right )\]

\[k=-\frac{1}{60}\left ( ln\,0.25-ln\,0.5 \right )\]

\[k=0.012\, s^{-1}\]

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Exercise \(\PageIndex{5.3b}\)

\(Br_{2}\rightarrow 2Br\)

How long does it take for the reaction in Q 14.4.m to decompose to 0.10M?

Answer

\[A=A_{0}e^{-kt}\]

\[\frac{A}{A_{0}}=e^{-kt}\]

\[ln\frac{A}{A_{0}}=-kt\]

\[t=-\frac{1}{k}\left ( ln\,A-ln\,A_{0} \right )\]

\[t=-\frac{1}{0.012}\left ( ln\,0.1-ln\,0.5 \right )\]

\[t=134s\]

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Exercise \(\PageIndex{5.3c}\)

\(Br_{2}\rightarrow 2Br\)

How long does it take for the reactant in Q 14.4.m to decompose to 15%?

Answer

\[A=A_{0}e^{-kt}\]

\[\frac{A}{A_{0}}=e^{-kt}\]

\[ln\frac{A}{A_{0}}=-kt\]

\[t=-\frac{1}{k}\left ( ln\frac{A}{A_{0}} \right )\]

\[t=-\frac{1}{0.012}ln\,0.15\]

\[t=158s = 160 sec\]

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Exercise \(\PageIndex{5.3d}\)

\(Br_{2}\rightarrow 2Br\)

How long does it take for 15% of the reactant in Q 14.4.m to be consumed?

Answer

\[\frac{A}{A_{0}}=e^{-kt}\]

\[ln\frac{A}{A_{0}}=-kt\]

\[t=-\frac{1}{k}\left ( ln\frac{A}{A_{0}} \right )\]

\[t=-\frac{1}{0.012}\left ( ln\frac{\left (1-0.15 \right )A_{0}}{A_{0}} \right )=-\frac{1}{0.012}ln\,0.85\]

\[t=13.5s\]

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Exercise \(\PageIndex{5.3e}\)

\(Br_{2}\rightarrow 2Br\)

What is [Br₂] of a 0.80M [Br₂] solution after 3 minutes?

Answer

\[A=A_{0}e^{-kt}=0.80*e^{-0.012*180}=0.092M\]

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Exercise \(\PageIndex{5.3f}\)

\(Br_{2}\rightarrow 2Br\)

Calculate the initial [Br₂] if [Br₂] =0.005M after 10 minutes?

Answer

\[A=A_{0}e^{-kt}\]

\[\frac{A}{e^{-kt}}=A_{0}\]

\[A_{0}=\frac{0.005}{e^{-0.012*600}}=6.70M\]

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Exercise \(\PageIndex{5.3.g}\)

If k=0.012/sec for the first order reaction, how long does it take for the bromine concentration to decompose to 15% of the initial concentration?

Br₂→2Br

Answer

\[
\textbf{Given:}\quad \text{First-order decay of } \mathrm{Br_2}\ \ (\mathrm{Br_2}\rightarrow 2\mathrm{Br}),\quad
k = 0.012\ \mathrm{s^{-1}},\quad \frac{[\mathrm{Br_2}]}{[\mathrm{Br_2}]_0} = 0.15
\]

\[ [Br_2] = [Br_2]_o e^{-kt}\]

\[\frac{[Br_2]}{[Br_2]_o}=e^{-kt}\]

\[\text{take the natural log of both sides}\]

\[
\qquad
\ln\!\left(\frac{[\mathrm{Br_2}]}{[\mathrm{Br_2}]_0}\right) = -kt
\]

\[
\textbf{Solve algebraically for } t:\qquad
t \;=\; -\frac{1}{k}\,\ln\!\left(\frac{[\mathrm{Br_2}]}{[\mathrm{Br_2}]_0}\right)
\]

\[
\textbf{Substitute the fraction } \frac{[\mathrm{Br_2}]}{[\mathrm{Br_2}]_0} = 0.15 \text{ and } k=0.012\ \mathrm{s^{-1}}:
\qquad
t \;=\; -\frac{1}{0.012\ \mathrm{s^{-1}}}\,\ln(0.15)
\]

\[
\textbf{Evaluate:}\qquad
\ln(0.15) \approx -1.8971
\quad\Rightarrow\quad
t \approx \frac{1.8971}{0.012}\ \mathrm{s} \approx 1.58\times 10^{2}\ \mathrm{s}
\]

\[
\boxed{t \approx 1.6\times 10^{2}\ \mathrm{s} \;\;(\text{about }160\ \mathrm{s})}
\]
 

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Exercise \(\PageIndex{5.3h}\)

A certain first-order reaction has a rate constant, k, equal to 2.1 x 10^-5 s^-1 at 355 K. If the activation energy for this reaction is 135 kJ/mol, calculate the value of the rate constant (in s^-1) at 550 K.

Answer

Given

\[
k_1 = 2.1 \times 10^{-5}\ \mathrm{s^{-1}}
\]

\[
T_1 = 355\ \mathrm{K}
\]

\[
T_2 = 550\ \mathrm{K}
\]

\[
E_a = 135\ \mathrm{kJ\,mol^{-1}} = 1.35 \times 10^{5}\ \mathrm{J\,mol^{-1}}
\]

\[
R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}
\]

---

Step 1: Start with the exponential Arrhenius equation

\[
k = Ae^{-E_a/(RT)}
\]

Write the equation for both temperatures.

\[
k_1 = Ae^{-E_a/(RT_1)}
\]

\[
k_2 = Ae^{-E_a/(RT_2)}
\]

---

Step 2: Form a ratio to eliminate \(A\)

\[
\frac{k_2}{k_1} =
\frac{Ae^{-E_a/(RT_2)}}{Ae^{-E_a/(RT_1)}}
\]

\[
\frac{k_2}{k_1} =
e^{-E_a/R \left(\frac{1}{T_2}-\frac{1}{T_1}\right)}
\]

---

Step 3: Solve for \(k_2\)

\[
k_2 =
k_1 e^{-E_a/R \left(\frac{1}{T_2}-\frac{1}{T_1}\right)}
\]

---

Substitute values

\[
k_2 =
(2.1\times10^{-5})
e^{-\frac{1.35\times10^{5}}{8.314}
\left(\frac{1}{550}-\frac{1}{355}\right)}
\]

\[
k_2 \approx 2.3\times10^{2}\ \mathrm{s^{-1}}
\]

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Exercise \(\PageIndex{5.3i}\)

.  Radioactive phosphorus is used in the study of biochemical reaction mechanisms. The isotope phosphorus-33 decays by first-order kinetics with a half-life of 14.3 days. If a chemist initially has a 7.5 M solution of pure phosphorus-33, calculate the concentration (in M) of phosphorus-33 in the solution after 2.4 days.

Answer

Given

\[
t_{1/2} = 14.3\ \text{days}
\]

\[
[A]_0 = 7.5\ \text{M}
\]

\[
t = 2.4\ \text{days}
\]

---

Step 1: Start with the first-order integrated rate law

\[
[A]_t = [A]_0 e^{-kt}
\]

---

Step 2: Substitute the first-order half-life expression

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

\[
[A]_t = [A]_0 e^{-(0.693/t_{1/2})t}
\]

---

Step 3: Solve for the concentration after time \(t\)

\[
[A]_t =
(7.5)\,e^{-0.693(2.4/14.3)}
\]

\[
[A]_t \approx 6.7\ \text{M}
\]

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Second-Order Integrated Rate Problems

For the following questions use this equation

\[NOBr(g) \rightarrow 2NO(g) + Br_2(g) \nonumber\]

Exercise \(\PageIndex{5.4a}\)

What is the rate constant if it takes 75 seconds for a 0.90M mixture to decompose to 0.25M?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]

\[k=\frac{1}{t}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]

\[k=\frac{1}{75sec}\left ( \frac{1}{0.25M}-\frac{1}{0.90M}\right )\]

\[k=0.039\,M^{-1}s^{-1}\]

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Exercise \(\PageIndex{5.4b}\)

How long does it take for the reaction  to decompose to 0.10M?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]

\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]

\[t=\frac{1}{0.039\,M^{-1}s^{-1}}\left(\frac{1}{0.1M}-\frac{1}{0.90M}\right )\]

\[t=228sec\]

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Exercise \(\PageIndex{5.4c}\)

How long does it take  to decompose to 25%?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]

\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]

\[t=\frac{1}{0.039\,M^{-1}s^{-1}}\left ( \frac{1}{0.25\left ( 0.90M \right )}-\frac{1}{0.90M}\right )\]

\[t=85sec\]

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Exercise \(\PageIndex{5.4e}\)

How long does it take for 25% of the reactant to be consumed?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]

\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]

\[t=\frac{1}{0.039\,M^{-1}s^{-1}}\left ( \frac{1}{0.75\left ( 0.90M \right )}-\frac{1}{0.90M}\right )\]

\[t=9.5sec\]

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Exercise \(\PageIndex{5.4f}\)

What is [NOBr] of a .85M [NOBr] solution after 3 minutes?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]

\[A=\frac{1}{kt+\frac{1}{A_{0}}}\]

\[A=\frac{1}{0.039\,M^{-1}s^{-1}\left ( 180sec \right )+\frac{1}{0.85M}}\]

\[A=0.12M\]

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Exercise \(\PageIndex{5.4g}\)

Calculate the initial [NOBr] if [NOBr]=0.0080M after 15 minutes.

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}}\Rightarrow\frac{1}{A_{0}}=\frac{1}{A}-kt\]

\[A_{0}=\frac{1}{\frac{1}{A}-kt}\]

\[A_{0}=\frac{1}{\frac{1}{0.0080M}-0.039\,M^{-1}s^{-1}\left( 900sec \right )}\]

\[A_{0}=0.011M\]

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General Integrated Rate Problems 

Exercise \(\PageIndex{5.5a}\)

What is the rate constant of a zero-order reaction if it takes 87 seconds for a concentration of 0.32500M to decompose to 0.00387M?

Answer

\[A=-kt+A_{0}\]

\[0.00387M=-k\left ( 87s \right )+0.325M\]

\[k=-\frac{0.00387M-0.325M}{87s}\]

\[k=0.0037\,M/s\]

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Exercise \(\PageIndex{5.5b}\)

A compound of 0.0452M has a rate constant of 4.95 x 10^-5 M/s and decays to 0.0042M.  How long did this process take?

Answer

\[A=-kt+A_{0}\]

\[0.0042M=-4.95*10^{-5}\,M/s\left ( t \right )+0.0452M\]

\[t=-\frac{0.0042M-0.0452M}{-4.95*10^{-5}\,M/s}\]

\[t=828\,s\]

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Exercise \(\PageIndex{5.5c}\)

After 35 seconds, only 0.029M of the mixture is left.  Find the concentration of the initial mixture at a rate of 0.68M^-1s^-1.

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}} \]

\[\frac{1}{0.029M}=\left ( 0.68M^{-1}s^{-1}*35s \right )+\frac{1}{A_{0}} \]

\[\frac{1}{0.029M}=23.8M^{-1}+\frac{1}{A_{0}} \]

\[\frac{1}{A_{0}}=\frac{1}{0.029M}-23.8M^{-1} \]

\[\frac{1}{A_{0}}=10.68M \]

\[A_{0}=0.094M\]

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Exercise \(\PageIndex{5.5d}\)

For a first-order reaction, in 4 minutes a solution went from 0.0800M to 0.0060M.  What is the rate constant for the reaction?

Answer

\[ln\,A=-kt+ln\,A_{0} \]

\[ln\left ( 0.006M \right )=\left ( -k*240s \right )+ln\left ( 0.08M \right ) \]

\[k=-\frac{ln\left ( 0.006M \right )-ln\left ( 0.08M \right )}{240s} \]

\[k=-\frac{-2.59}{240s} \]

\[k=0.0108s^{-1}\]

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Exercise \(\PageIndex{5.5e}\)

Beginning with 0.5046M of a substance that has a rate constant of 0.085s^-1.  How much for the substance is left after 25 seconds?

Answer

\[ln\,A=-kt+ln\,A_{0} \]

\[ln\,A-ln\,A_{0}=-kt \]

\[ln\frac{A}{A_{0}}=-kt \]

\[A=A_{0}e^{-kt} \]

\[A=0.5046e^{-0.085\left ( 25 \right )} \]

\[A=0.06M\]

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Exercise \(\PageIndex{5.5f}\)

With a rate constant of .019M^-1s^-1, how long does it take a chemical to go from 0.0400M to 0.0020M?

Answer

\[\frac{1}{A}=kt+\frac{1}{A_{0}} \]

\[\frac{1}{0.0020M}=\left ( t*0.019M^{-1}s^{-1} \right )+\frac{1}{0.0400M} \]

\[\frac{1}{0.0020M}-\frac{1}{0.0400M}=\left ( t*0.019M^{-1}s^{-1} \right ) \]

\[t=\frac{500M-25m}{0.019M^{-1}s^{-1}} \]

\[t=25000sec\]

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Microscopic View of Reaction Rates

Textbook: Section 14.5

Two-state Arrhenius Equation

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Figure \(\PageIndex{3}\): Consider the Two-state Arrhenius equation for the following problems.

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Exercise \(\PageIndex{6.1.a}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for k₁:

Answer

\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}}\]

\[k_{1}=k_{2}\frac{e^{\frac{-E_{a}}{RT_{1}}}}{e^{\frac{-E_{a}}{RT_{2}}}}=k_{2}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}=k_{2}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

\[k_{1}=k_{2}e^{\frac{E_{a}}{R}\left (\frac{T_{1}-T_{2}}{T_{2}T_{1}} \right )}\]

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Exercise \(\PageIndex{6.1.b}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for k₂:

Answer

\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}}\]

\[k_{2}=k_{1}\frac{e^{\frac{-E_{a}}{RT_{2}}}}{e^{\frac{-E_{a}}{RT_{1}}}}=k_{1}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}=k_{1}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}\]

\[k_{2}=k_{1}e^{\frac{E_{a}}{R}\left (\frac{T_{2}-T_{1}}{T_{2}T_{1}} \right )}\]

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Exercise \(\PageIndex{6.1.c}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for E_a:

Answer

\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]

\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]

\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]

\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]

\[\left [ ln\left ( \dfrac{k_{1}}{k_{2}} \right )\right ]*R=\left [ \dfrac{-E_{a}}{\cancel{\textcolor{red}{R}}}\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right ]*\cancel{\textcolor{red}{R}}\]

\[ \dfrac{\left[  ln\left(  \dfrac{k_{1}}{k_{2}} \right) \right]*R}{\left(  \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}=\dfrac{-E_{a}*\cancel{\textcolor{red}{\left(  \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}}}{\cancel{\textcolor{red}{\left(  \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}}}\]

\[E_{a}=\frac{{}\left [ ln\left ( \frac{k_{1}}{k_{2}} \right )\right ]*R}{\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}\]

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Exercise \(\PageIndex{6.1.d}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for R:

Answer

\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]

\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]

\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]

\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]

\[\left [ ln\left ( \dfrac{k_{1}}{k_{2}} \right )\right ]*R=\left [ \dfrac{-E_{a}}{\cancel{\textcolor{red}{R}}}\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right ]*\cancel{\textcolor{red}{R}}\]

\[\dfrac{\cancel{\textcolor{red}{ln\left(\dfrac{k_{1}}{k_{2}}\right)}}*R}{\cancel{\textcolor{red}{ln\left(\dfrac{k_{1}}{k_{2}}\right)}}}=\dfrac{\left[-E_{a}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)\right]}{\left[ln\left(\dfrac{k_{1}}{k_{2}}\right)\right]}\]

\[ R=\dfrac{\left[-E_{a}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)\right]}{\left[ln\left(\dfrac{k_{1}}{k_{2}}\right)\right]} \]

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Exercise \(\PageIndex{6.1.e}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for T₁:

Answer

\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]

\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]

\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]

\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]

\[ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} = \left[ \cancel{\textcolor{red}{\dfrac{-E_{a}}{R}}} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right] * \cancel{\textcolor{red}{ \dfrac{R}{-E_{a}}}} \]

\[ \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} = \left(  \dfrac{1}{T_{1}}-\cancel{\textcolor{red}{\dfrac{1}{T_{2}}}} \right) + \cancel{\textcolor{red}{\dfrac{1}{T_{2}}}}\]

\[\dfrac{1}{T_{1}} = \left ( \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} \right) \]

\[T_{1} =\dfrac{1}{ \left ( \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} \right)} \]

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Exercise \(\PageIndex{6.1.f}\)

\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)

Solve for T₂:

Answer

\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]

\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]

\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]

\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]

\[ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} = \left[ \cancel{\textcolor{red}{\dfrac{-E_{a}}{R}}} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right] * \cancel{\textcolor{red}{ \dfrac{R}{-E_{a}}}} \]

\[ \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} = \left(  \dfrac{1}{T_{1}}-\cancel{\textcolor{red}{\dfrac{1}{T_{2}}}} \right) + \cancel{\textcolor{red}{\dfrac{1}{T_{2}}}}\]

\[ \left(\cancel{\textcolor{red}{\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}}} +\dfrac{1}{T_{2}} \right) - \cancel{\textcolor{red}{\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}}} = \dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\]

\[ \dfrac{1}{T_{2}}=\left(\dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\right) \]

\[ T_{2}=\dfrac{1}{\left(\dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\right)} \]

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Exercise \(\PageIndex{6.1.g}\)

The rate constant for a reaction is 1.3 /M^.s at 700 K and 23/M^.s at 800 K.
What is the activation energy?

Answer

\[
\textbf{Given:}\quad
k_1 = 1.3\ \mathrm{M^{-1}s^{-1}} \;\; \text{at } T_1 = 700\ \mathrm{K}, 
\qquad
k_2 = 23\ \mathrm{M^{-1}s^{-1}} \;\; \text{at } T_2 = 800\ \mathrm{K}
\]

\[\frac{k_{2}}{k_{1}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}\]

\[\text{Since we are after the activation energy which is in the exponent we will use the ln form of the above eq}\]

\[
\textbf{Arrhenius equation (two-point form):}\qquad
\ln\!\left(\frac{k_2}{k_1}\right) \;=\; \frac{E_a}{R}\,\left(\frac{1}{T_1} - \frac{1}{T_2}\right)
\]

\[
\textbf{Solve algebraically for } E_a:\qquad
E_a \;=\; \frac{R \,\ln(k_2/k_1)}{\tfrac{1}{T_1} - \tfrac{1}{T_2}}
\]

\[
\textbf{Substitute values:}\qquad
E_a \;=\; \frac{8.314\ \mathrm{J\,mol^{-1}K^{-1}} \times \ln\!\left(\tfrac{23}{1.3}\right)}
{\tfrac{1}{700\ \mathrm{K}} - \tfrac{1}{800\ \mathrm{K}}}
\]

\[
\ln\!\left(\tfrac{23}{1.3}\right) \approx \ln(17.692) \approx 2.871
\]

\[
\frac{1}{700} - \frac{1}{800} \;=\; 0.0014286 - 0.0012500 \;=\; 0.0001786
\]

\[
\text{Numerator: } 8.314 \times 2.871 \;\approx\; 23.87
\]

\[
E_a \;=\; \frac{23.87}{0.0001786}\ \mathrm{J\,mol^{-1}}
\;\approx\; 1.34\times 10^{5}\ \mathrm{J\,mol^{-1}}
\]

\[
\boxed{E_a \;\approx\; 134\ \mathrm{kJ\,mol^{-1}}}
\]
 

Arrhenius Equation Problems

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Exercise \(\PageIndex{6.2.a}\)

A first-order reaction has a rate constant of 4.40x10^-3s^-1 at 350K and a rate constant of 9.80x10^-2s^-1 at 580K.  What is the activation energy?

Answer

\[k_{1}=Ae^{-E_{a}/RT_{1}}\]

\[k_{2}=Ae^{-E_{a}/RT_{2}}\]

\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

\[ln\frac{k_{2}}{k_{1}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]

\[E_{a}=-R*\frac{ln\frac{k_{2}}{k_{1}}}{\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

R=8.314510*10^-3 kJ/K*mol

\[E_{a}=-\left (8.314510*10^{-3}\right )\frac{ln\frac{9.80*10^{-2}}{4.40*10^{-3}}}{\left ( \frac{1}{580}-\frac{1}{350} \right )}=22.8\,kJ/mol\]

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Exercise \(\PageIndex{6.2.b}\)

A reaction with activation energy of 120kJ/mol has a rate constant of 0.25s^-1 at 310K.  What will be the temperature for the rate constant to be doubled?

Answer

\[k_{1}=Ae^{-E_{a}/RT_{1}}\]

\[k_{2}=Ae^{-E_{a}/RT_{2}}\]

\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

\[ln\frac{k_{2}}{k_{1}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]

\[\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}=\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )\]

 

\[\frac{1}{T_2}= \frac{1}{T_1} -\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}\]

\[T_2 = \frac{1}{\frac{1}{T_1} -\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}}\]

\[k_{2}=2k_{1} \;\; so \; \; ln\frac{k_{2}}{k_{1}} = ln2   \]

\[T_2 = \frac{1}{\frac{1}{310K} -\frac{0.008314\frac{kJ}{mol*K}}{120\frac{kJ}{mol}}*ln2}\]

\[T_{2}=315K\]

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Exercise \(\PageIndex{6.2.c}\)

The rate constant of a reaction increases by a factor of 10.0 when the temperature increases from 300K to 330K.  What is the activation energy of the reaction?

Answer

\[k_{1}=Ae^{-E_{a}/RT_{1}}\]

\[k_{2}=Ae^{-E_{a}/RT_{2}}\]

\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

\[ln\frac{k_{2}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]

\[\frac{k_{2}}{k_{1}}=10 \]

\[ln\,10=-\frac{E_{a}}{8.314510*10^{-3}}*\left (\frac{1}{330}-\frac{1}{300}\right ) \]

\[E_{a}=63.2kJ\]

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Exercise \(\PageIndex{6.2.d}\)

The rate constant for a reaction is 1.5x10^-4M^-1s^-1 at 100⁰C, and 1.2 x10^-3M^-1s^-1at 150⁰C.  What is the energy of activation?

Answer

\[k_{1}=Ae^{-E_{a}/RT_{1}}\]

\[k_{2}=Ae^{-E_{a}/RT_{2}}\]

\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]

\[ln\frac{k_{2}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]

\[ln\frac{1.2*10^{-3}}{1.5*10^{-4}}=-\frac{E_{a}}{8.314510*10^{-3}}*\left (\frac{1}{423}-\frac{1}{373}\right ) \]

\[E_{a}=54.6kJ\]

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Exercise \(\PageIndex{6.2.e\)

For the reaction in Q 14.5.j, what would the rate constant (in M^-1s^-1) be at 200⁰C?

Answer

From the previous question, we know E_a=54.6kJ

\[k_{1}=Ae^{-E_{a}/RT_{1}}\]

\[k_{3}=Ae^{-E_{a}/RT_{3}}\]

\[\frac{k_{3}}{k_{1}}=\frac{e^{-E_{a}/RT_{3}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{3} \right )+E_{a}/RT_{3}}=e^{-E_{a}/R*\left ( \frac{1}{T_{3}}-\frac{1}{T_{3}} \right )}\]

\[ln\frac{k_{3}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{3}}-\frac{1}{T_{1}} \right )\]

\[ln\frac{k_{3}}{1.5*10^{-4}}=-\frac{54.6}{8.314510*10^{-3}}*\left (\frac{1}{473}-\frac{1}{373}\right ) \]

\[k_{3}=0.0064M^{-1}s^{-1}\]

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Exercise \(\PageIndex{6.2.f}\)

A certain first-order reaction has a rate constant, k, equal to 2.1 x 10-5 s-1 at 355 K. If the activation energy for this reaction is 135 kJ/mol, calculate the value of the rate constant (in s-1) at 550 K.

Answer

\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}} \nonumber\]

\[k_{1}=k_{2}\frac{e^{\frac{-E_{a}}{RT_{1}}}}{e^{\frac{-E_{a}}{RT_{2}}}}=k_{2}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}=k_{2}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )} \nonumber\]

\[k_{1}=2.1x10^{-5}s^{-1}e^{\frac{135kJ/mol}{0.008314J/mol \cdot K}\left ( \frac{1}{355}-\frac{1}{550} \right )} =230s^{-1}\nonumber\]

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Exercise \(\PageIndex{6.2g}\)

Consider an Arrhenius plot for a noncatalyzed reaction.  The Energy of reaction for the forward reaction is 26 kJ/mol and the energy of reaction for the reverse reaction is 46 kJ/mol.   Give a numerical value for the enthalpy of reaction

Answer

Given

\[
E_a^{\text{forward}} = 26\ \text{kJ mol}^{-1}
\]

\[
E_a^{\text{reverse}} = 46\ \text{kJ mol}^{-1}
\]

---

Step 1: Relate activation energies to the reaction enthalpy

\[
\Delta H_{\text{rxn}} = E_a^{\text{forward}} - E_a^{\text{reverse}}
\]

---

Step 2: Substitute values

\[
\Delta H_{\text{rxn}} =
26 - 46
\]

\[
\Delta H_{\text{rxn}} = -20\ \text{kJ mol}^{-1}
\]

---

\[
\boxed{\Delta H_{\text{rxn}} = -20\ \text{kJ mol}^{-1}}
\]

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Catalysts 

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Exercise \(\PageIndex{6.3.a}\)

Which statement is correct

a) increasing the surface area of a homogenous catalyst will increase the reaction rate.
b) decreasing the surface area of a homogenous catalyst will increase the reaction rate.
c) increasing the surface area of a heterogenous catalyst will increase the reaction rate.
d) decreasing the surface area of a heterogeneous catalyst will increase the reaction rate.
e) none of the above are correct.

Answer

c)  Increasing the surface area of a heterogenous catalyst will increase the reaction rate.

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Exercise \(\PageIndex{6.3.b}\)

The activation energy of a reaction can be decreased by

a. raising the temperature                              b. lowering the temperature
c. adding more reactants                                d. removing some products   
e. adding a catalyst

Answer

e. adding a catalyst

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Exercise \(\PageIndex{6.3.c}\)

Which of the following statements best describes how a catalyst works?

a. A catalyst changes the potential energies of the reactants and products.
b. A catalyst decreases the temperature of the reaction which leads to a faster rate.
c. A catalyst lowers the activation energy for the reaction by providing a different reaction mechanism.
d. A catalyst destroys some of the reactants, which lowers the concentration of the reactants.
e. A catalyst raises the activation energy for the reaction which produces a faster rate.

Answer

c. A catalyst lowers the activation energy for the reaction by providing a different reaction mechanism.

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Reaction Mechanisms

Textbook: Section 14.6

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Exercise \(\PageIndex{7a}\)

What is the intermediate reactant in the mechanism for the decomposition of ozone?

Step 1: \(O_{3}\,(g)\rightarrow O_{2}\,(g)+O\,(g)\)

Step 2: \(O_{3}\,(g)+O\,(g)\rightarrow 2O_{2}\,(g)\)

---

Total:   \(2O_{3}\,(g)\rightarrow 3O_{2}\,(g)\)

Answer

O, gaseous oxygen

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Exercise \(\PageIndex{7b}\)

What is the molecularity for each step in the reaction from Question 14.7.a?

Answer

unimolecular, bimolecular

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Exercise \(\PageIndex{7c}\)

What is the rate equation for step 2 in the reaction from Question 14.7.a?

Answer

k[O₃][O]

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Exercise \(\PageIndex{7d}\)

What is the rate determining step in the following mechanism?

Step 1: (fast) \(2NO\,(g)\underset{k_{2}}{\overset{k_1}{\rightleftharpoons}} N_{2}O_{2}(g)\)

Step 2: (slow) \(N_{2}O_{2}(g)\; + \; O_2(g)\overset{k_{2}}{\rightarrow}2NO_{2}(g)\)

---

Overall: \(2NO\,(g)+O_{2}(g)\rightarrow2NO_{2}(g)\)

Answer

Step 2

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Exercise \(\PageIndex{7e}\)

What is the rate equation for question 14.7.d?

Answer

k[O₂][NO]²

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Exercise \(\PageIndex{7f}\)

What is the rate determining step in the following mechanism?

Step 1: (slow) \(H_{2}\,+\,ICl\rightarrow HI\,+\,HCl\)

Step 2: (fast) \(HI\,+\,ICl\rightarrow I_{2}\,+\,HCl\)

---

Overall: \(H_{2}\,+\,2ICl\rightarrow I_{2}\,+\,2HCl\)

Answer

Step 1

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Exercise \(\PageIndex{7g}\)

What is the rate equation for question 14.7.f?

Answer

k[H₂][ICl]

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Exercise \(\PageIndex{7.h}\)

A possible mechanism for the stoichiometric reaction

Br₂ + 2NO  =>  2NOBr     is shown below.
 

2NO  <= =>   N₂O₂                   (slow)     has rate constants k₁, (forward)  k_-1 (back)

N₂O₂ =>  Br₂ à 2 NOBr          (fast ) has rate constant k₂ (forward)

 

a.  k_obs [NO]^1/2                        b. k_obs [Br₂]^1/2                         c. k_obs [NO]²[Br₂]      

d.  k_obs [NO]²                          e. k_obs [NO][Br₂]²      

Answer

The slow step is the rate determining step and so the observed rate law is defined by the molecularity of the first step.

d.  k_obs [NO]² 

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General Questions

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Exercise \(\PageIndex{a}\)

Which of the following is not a valid unit for a reaction rate?

1. g/s
2. M/s
3. mol/hr
4. mol/L
5. mol/L-hr

Answer

d. mol/L

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Exercise \(\PageIndex{b}\)

In the following reaction, [NO₂] drops from 0.0150 to 0.00975 M in 150s at 300°C. What is the rate of disappearance of NO₂ for this period?

\(2NO_{2}\rightarrow 2NO+O_{2}\)

Answer

\[Rate\,of\,reaction=\frac{1}{2}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{2}\frac{\Delta \left [NO\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]

\[Rate\,of\,disappearance=\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{0.0150-0.00975}{150}=3.5*10^{-5}\,M/s\]

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Exercise \(\PageIndex{c}\)

In the following reaction, [N₂O₅] drops from 0.0300 to 0.00195 M in 300s at 300°C. What is the rate of disappearance of O₂ for this period?

\(2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}\)

Answer

\[Rate\,of\,reaction=\frac{1}{2}\frac{\Delta \left [N_{2}O_{5}\right ]}{\Delta t}=\frac{1}{4}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]

\[\frac{1}{2}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]

\[\frac{\Delta \left [O_{2}\right ]}{\Delta t}=\frac{1}{2}\left (\frac{\Delta \left [NO_{2}\right ]}{\Delta t}\right )=\frac{1}{2}\left (\frac{0.0300-0.0195}{300}\right )=1.75*10^-5\,M/s\]

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Exercise \(\PageIndex{d}\)

What is the rate law of the following reaction if it is first order in A and second order in D?

\(2A+3D\rightarrow products\)

Answer

rate = [A][D]²

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Exercise \(\PageIndex{e}\)

For the following reaction, the reaction rate increased by a factor of 9 when the concentration of B was tripled. What is the order in B?

\(A+B\rightarrow P\)

Answer

\[rate=[A]^{x}\]

\[9=[3]^{x}\]

\[x=2\]

B is second order

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Exercise \(\PageIndex{f}\)

A reaction was found to be third order in A. By increasing the concentration of A by a factor of 3 will cause the reaction rate to _____.

1. decrease by a factor of the cube root of 3
2. increase by a factor of 27
3. increase by a factor of 9
4. remain constant
5. triple

Answer

\[rate=[A]^{x}\]

\[rate=[3]^{3}\]

\[rate=27\]

b. increase by a factor of 27

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Exercise \(\PageIndex{g}\)

Use the information below to answer the following questions:

1. Determine the order of the reaction in reactant A.
2. Determine the order of the reaction in reactant B.
3. Determine the overall order of the reaction.
4. Determine the rate constant for the reaction.

\(A+B\rightarrow P\)

Experiment Number	[A](M)	[B](M)	Initial Rate (M/s)
1	0.546	1.526	5.66
2	0.546	3.052	5.66
3	1.638	1.526	50.94

Answer for a.

\[rate_{1}=k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}\]

\[rate_{2}=k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}\]

\[\frac{rate_{1}}{rate_{2}}=\frac{k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}}{k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}}=\left ( \frac{A_{1}}{A_{2}} \right )^{m}\left ( \frac{B_{1}}{B_{2}} \right )^{n}\]

select when \(\frac{B_{1}}{B_{2}}=1\) (data in the experiment 1 and 3)

\[\frac{rate_{2}}{rate_{1}}=\left ( \frac{A_{2}}{A_{1}} \right )^{m}=\left ( \frac{1.638}{0.546} \right )^{m}=\frac{50.94}{5.66}=\]

\[3^{m}=9\]

therefore m=2

Second-order

Answer for b.

\[\frac{rate_{1}}{rate_{2}}=\frac{k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}}{k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}}=\left ( \frac{A_{1}}{A_{2}} \right )^{m}\left ( \frac{B_{1}}{B_{2}} \right )^{n}\]

select when \(\frac{A_{1}}{A_{2}}=1\) (data in the experiment 1 and 2)

\[\frac{rate_{1}}{rate_{2}}=\left ( \frac{B_{1}}{B_{2}} \right )^{n}=\left ( \frac{1.526}{3.052} \right )^{n}=2^{n}=\frac{5.66}{5.66}=1\]

\[\frac{1}{2}^{n}=1\]

therefore n=0

Zero-order

Answer for c.

The total reaction order is the sum of the exponents on each component. So 2+0=2

Second-order

Answer for d.

\[k=\frac{rate}{[A]^{2}[B]^{0}}=\frac{2.83}{[0.273]^{2}[0.763]^{0}}=38.0\,M^{-1}s^{-1}\]

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Exercise \(\PageIndex{h}\)

What is the rate law for the given data below for the following reaction?

\(yY+zZ\rightarrow P\)

Experiment Number	[Y](M)	[Z](M)	Initial Rate (M/s)
1	0.200	0.200	8.0 * 10^-5
2	0.400	0.200	3.2 * 10^-4
3	0.200	0.400	1.6 * 10^-4

Answer

rate = k[Y]²[Z]

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Exercise \(\PageIndex{i}\)

What is the rate law that corresponds to the following reaction?

\(2A+B\rightarrow C\)

Experiment Number	[A](M)	[B](M)	Initial Rate (M/s)
1	0.023	0.033	0.188
2	0.045	0.066	0.750
3	0.090	0.066	0.750
4	0.090	0.990	1.688

Answer

rate = k[B]²

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\[2ClO_{2}(aq)+2OH^{-}(aq)\rightarrow ClO_{3}^{-}(aq)+ClO_{2}^{-}(aq)+H_{2}O(l)\]

Experiment Number	[ClO2](M)	[OH^-](M)	Initial Rate (M/s)
1	0.060	0.030	2.48 * 10^-2
2	0.020	0.030	2.76 * 10^-3
3	0.020	0.090	8.28 * 10^-3

Figure \(\PageIndex{4}\): Use the following table of experimental data to answer the questions below.

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Exercise \(\PageIndex{j}\)

What is the order of the reaction with respect to ClO₂?

Answer

Second-order

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Exercise \(\PageIndex{k}\)

What is the order of the reaction with respect to OH^-?

Answer

First-order

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Exercise \(\PageIndex{l}\)

What is the order of the reaction overall?

Answer

Third-order

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Exercise \(\PageIndex{m}\)

What is the value of the rate constant for the reaction?

Answer

\[k=\frac{rate}{\left [ClO_{2}\right ]^{2}\left [OH^{-}\right ]^{1}}=\frac{2.48*10^{-2}}{\left [0.060\right ]^{2}\left [0.030\right ]^{1}}=230\]

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Exercise \(\PageIndex{n}\)

What are the units of the rate constant for the reaction?

Answer

M^-2s^-1

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Exercise \(\PageIndex{o}\)

What is the overall order of a reaction, if the rate constant is 1.3*10^-4 M^-1s^-1 at 100°C and 1.4*10^-3 M^-1s^-1 at 150°C?

Answer

\[M^{1-n}s^{-1}\]

\[1-n=-1\]

\[n=1+1\]

\[n=2\]

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Exercise \(\PageIndex{p}\)

For a zero-order reaction, a plot of _____ versus _____ is linear.

1. \(\frac{1}{\left [A\right ]},\,t\)
2. \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
3. \(ln\left [A\right ]_{t},\,t\)
4. \(\left [A\right ],\,t\)
5. \(t,\,\frac{1}{\left [A\right ]}\)

Answer

d. \(\left [A\right ],\,t\)

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Exercise \(\PageIndex{q}\)

For a first-order reaction, a plot of _____ versus _____ is linear.

1. \(\frac{1}{\left [A\right ]},\,t\)
2. \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
3. \(ln\left [A\right ]_{t},\,t\)
4. \(\left [A\right ],\,t\)
5. \(t,\,\frac{1}{\left [A\right ]}\)

Answer

c. \(ln\left [A\right ]_{t},\,t\)

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Exercise \(\PageIndex{r}\)

For a second-order reaction, a plot of _____ versus _____ is linear.

1. \(\frac{1}{\left [A\right ]},\,t\)
2. \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
3. \(ln\left [A\right ]_{t},\,t\)
4. \(\left [A\right ],\,t\)
5. \(t,\,\frac{1}{\left [A\right ]}\)

Answer

a. \(\frac{1}{\left [A\right ]},\,t\)

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Exercise \(\PageIndex{s}\)

Determine the rate constant (M^-1s^-1) by using integrated rate law for the following reaction. The reaction follows second-order kinetics, and [NO₂] drops from 0.0200 to 0.0130 M in 200s at 300°C.

\(2NO_{2}\rightarrow 2NO+O_{2}\)

Answer

\[\frac{1}{\left [A\right] _{t}}=kt+\frac{1}{\left [A\right] _{0}}\]

\[\frac{1}{0.0130}=k\left ( 200 \right )+\frac{1}{0.0200}\]

\[76.92=k\left ( 200 \right )+50\]

\[76.92-50=k\left ( 200 \right )\]

\[k=\frac{26.92}{200}=0.13\,M^{-1}s^{-1}\]

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Exercise \(\PageIndex{t}\)

The reaction below is a first-order reaction. At 230.3°C, k = 6.29*10^-4 s^-1. If [CH₃-N≡C]₀ is 0.00100 M, what is [CH₃-N≡C] in M after 1.000*10³ s?

\(CH_{3}-N\equiv C\rightarrow CH_{3}-C\equiv N\)

Answer

\[\left [ A \right ]=\left [ A \right ]_{0}e^{-kt}\]
\[\left [ A \right ]=\left ( 0.00100 \right )e^{-\left ( 6.29*10^{-4} \right )\left ( 1000 \right )}\]
\[\left [ A \right ]=\left ( 0.00100 \right )e^{-0.629}\]
\[\left [ A \right ]=0.00100*0.533\]
\[\left [ A \right ]=5.33*10^{-4}\]

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Exercise \(\PageIndex{u}\)

If a first-order reaction has a rate constant of 0.33 min^-1, what is the time for the reactant concentration to decrease from 0.13 M to 0.088M?

Answer

\[\left [ A \right ]e^{-kt}=\left [ A \right ]_{0}\]
\[0.13e^{-0.33*t}=0.088\]
\[e^{-0.33*t}=\frac{0.088}{0.13}\]
\[-0.33*t=ln\left ( 0.677 \right )\]
\[t=\frac{-0.390}{-0.33}\]
\[t=1.2\,min\]

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Exercise \(\PageIndex{v}\)

Why does the rate of reaction increase, when the temperature of the reaction increases?

1. activation energy is lowered
2. reactant molecules collide less frequently
3. reactant molecules collide less frequently and with greater energy per collision
4. reactant molecules collide with greater energy per collision
5. reactant molecules collide more frequently with less energy per collision

Answer

d. reactant molecules collide with greater energy per collision

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Exercise \(\PageIndex{w}\)

What do reaction rates depend on?

1. collision energy
2. collision frequency
3. collision orientation
4. all of these
5. none of these

Answer

d. all of these

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Exercise \(\PageIndex{x}\)

What is the species that exists at the maximum on the potential energy profile of a reaction?

1. activated complex
2. activation energy
3. atomic state
4. enthalpy of reaction
5. product

Answer

a. activated complex

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Exercise \(\PageIndex{y}\)

What is the activation energy (kJ/mol) of a reaction whose rate constant increases by a factor of 10.0 when the temperature is increased from 303K to 333K?

Answer

\[ln\left (\frac{k_{1}}{k_{2}}\right )=\frac{E_{a}}{R}\left (\frac{1}{T_{1}}-\frac{1}{T_{2}}\right )\]
\[ln\left (\frac{10}{1}\right )=\frac{E_{a}}{8.314}\left (\frac{1}{303}-\frac{1}{333}\right )\]
\[ln\left (10\right )=\frac{E_{a}}{8.314}\left (2.97*10^{-4}\right )\]
\[\frac{2.30}{2.97*10^{-4}}=\frac{E_{a}}{8.314}\]
\[7.74*10^{3}=\frac{E_{a}}{8.314}\]
\[E_{a}=6.44*10^{4}\,J/mol\]
\[E_{a}=64.4\,kJ/mol\]

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Exercise \(\PageIndex{z}\)

The rate constant for a reaction is 1.3*10^-4 M^-1s^-1 at 100°C, and 1.1*10^-3 at 150°C. What is the energy of activation (in kJ/mol) for this reaction?

Answer

\[ln\left (\frac{k_{1}}{k_{2}}\right )=\frac{E_{a}}{R}\left (\frac{1}{T_{1}}-\frac{1}{T_{2}}\right )\]
\[ln\left (\frac{1.1*10^{-3}}{1.3*10^{-4}}\right )=\frac{E_{a}}{8.314}\left (\frac{1}{273.15+100}-\frac{1}{273.15+150}\right )\]
\[ln\left (8.46\right )=\frac{E_{a}}{8.314}\left (3.17*10^{-4}\right )\]
\[\frac{2.14}{3.17*10^{-4}}=\frac{E_{a}}{8.314}\]
\[6.74*10^{3}=\frac{E_{a}}{8.314}\]
\[E_{a}=5.6*10^{4}\,J/mol\]
\[E_{a}=5.6\,kJ/mol\]

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Exercise \(\PageIndex{aa}\)

What is a plausible mechanism for the following reaction?

\(A + 2B \rightarrow C +D\,\,\,\,\,rate\,=\,k\left[A\right]^{2}\)

a. step 1: A + A → E + D           (slow)

    step 2: E + 2B → C + A          (fast)

b. step 1: A + B → E + C            (slow)

    step 2: E + B → D                  (fast)

c.  step 1: A + A → E + D           (fast)

     step 2: E + 2B → C + A          (slow)

d.  step 1: A + A + B → E + C     (slow)

     step 2: E + C → D + F            (fast)   

     step 3: F → C                           (fast)

e.       None of these mechanisms is plausible for this reaction

Answer

a. step 1: A + A → E + D           (slow)

    step 2: E + 2B → C + A          (fast)

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Exercise \(\PageIndex{ab}\)

What is the intermediate reactant in the following reaction mechanism for the formation of product X?

\(A\,+\,B \rightarrow C\,+\,D\)

\(B\,+\,D \rightarrow X\)

Answer

D

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Exercise \(\PageIndex{ac}\)

What is the molecularly and the rate law for the following reaction?

\(NO_{3}\,+\,CO\rightarrow NO_{2}\,+\,CO_{2}\)

Answer

2, k[NO₃][CO]

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Exercise \(\PageIndex{ad}\)

Which of the following will lower the activation energy for a reaction?

1. adding a suitable catalyst
2. increasing the concentration of reactants
3. raising the temperature of the reaction
4. all the above
5. none of the above

Answer

a. adding a suitable catalyst

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Exercise \(\PageIndex{ae}\)

What type of catalyst is used in automotive catalytic converters?

1. enzymes
2. heterogenous
3. homogenous
4. noble gas
5. nonmetal oxides

Answer

b. heterogenous

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Textbook: 14:Rates of Chemical Reactions

 

Robert E. Belford (University of Arkansas Little Rock; Department of Chemistry). The breadth, depth and veracity of this work is the responsibility of Robert E. Belford, [e-mail]. You should contact him if you have any concerns. This material has both original contributions, and content built upon prior contributions of the LibreTexts Community and other resources, including but not limited to:

• Emily Choate
• Liliane Poirot