14: Rates of Chemical Reactions
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- 206744
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Prelude
Textbook: Section 14.1
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
\[\frac{\left [ A \right ]}{\left [ A \right ]_{0}}=e^{-kt}\]
\[\left [ A \right ]=\left [ A \right ]_{0}e^{-kt}\]
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
\[\frac{\left [ A \right ]}{\left [ A \right ]_{0}}=e^{-kt}\]
\[\left [ A \right ]_{0}=\frac{\left [ A \right ]}{e^{-kt}}\]
\[\left [ A \right ]_{0}=\left [ A \right ]e^{kt}\]
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
\[ln\left ( \frac{A}{A_{0}}\right )=-kt\]
\[k=-\frac{ln\left ( \frac{A}{A_{0}} \right )}{t}\]
Exercise \(\PageIndex{1d}\)
Solve the equation for Ea.
\[k=Ae^{-\frac{E_{a}}{RT}}\]
- Answer
-
\[k=Ae^{\frac{-E_{a}}{RT}}\]
\[\frac{k}{A}=e^{\frac{-E_{a}}{RT}}\]
take the ln of both sides
\[ln\frac{k}{A}=\frac{-E_{a}}{RT}\]
\[E_{a}=-RTln\frac{k}{A}\]
\[E_{a}=RTln\frac{A}{k}\]
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
\[ln\frac{k}{A}={\frac{-E_{a}}{RT}}\]
\[E_{a}=-RTln\frac{k}{A}\]
\[E_{a}=RTln\frac{A}{k}\]
\[T=\frac{E_{a}}{Rln\frac{A}{k}}\]
Exercise \(\PageIndex{1f}\)
Solve the equation for t1/2.
\[A=A_{0}\left ( \frac{1}{2} \right )^{\frac{t}{t_{1/2}}}\]
- 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
\[log\left ( \frac{A}{A_{0}}\right )={\frac{t}{t_{1/2}}}log\left ( \frac{1}{2} \right )\]
\[t_{1/2}*log\left ( \frac{A}{A_{0}}\right )=t*log\left ( \frac{1}{2} \right )\]
\[t_{1/2}=\frac{t*log\left ( \frac{1}{2} \right )}{log\left ( \frac{A}{A_{0}}\right )}\]
Solve \(A=B(C)^n\), solve for n
- Answer
-
\[A=B\left ( C \right )^n \\ \frac{A}{B}=\left ( C \right )^n \\ log\frac{A}{B}=log\left ( C \right )^n \\ log\frac{A}{B}=nlog\left( C \right ) \\ n= \frac{log\left ( \frac{A}{B} \right )}{logC}\]
Show \(\frac{log_{10}(570)}{log_{10}(30)}= \frac{ln(570)}{ln(30)} \)
- Answer
-
\[\frac{log570}{log30}=\frac{2.75587}{1.477}=1.86 \\ \; \\ \; \\ \frac{ln570}{ln30}=\frac{6.3456}{3.4012}=1.86\]
Show \(\frac{log_{10}(570)}{log_{10}(30)}\neq log_{10}\frac{570}{30} \)
- Answer
-
\[\frac{log570}{log30}=\frac{2.75587}{1.477}=1.86 \\ \; \\ \; \\ log_{10}\frac{570}{30}=log(19)=1.28\].
Rates of Chemical Reactions
Textbook: Section 14.2
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
-
\[rate=\frac{\Delta [ NO_{2}]}{\Delta t}=\frac{0.0115-0.0150}{100}=-3.5*10^{-5}M/s \]
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
-
\[2{NO_{2}} \rightarrow 2{NO} + 1{O_{2}}\]
\[-\frac{1}{2}\frac{\Delta [{NO_{2}}]}{\Delta t} =\frac{1}{2}\frac{\Delta [{NO}]}{\Delta t}=\frac{\Delta [{O_{2}}]}{\Delta t}\]
\[\frac{\Delta [{O_{2}}]}{\Delta t}=-\frac{1}{2}\frac{\Delta [{NO_{2}}]}{\Delta t}=\frac{-1}{2}\frac{0.0115-0.0150}{100}=1.75*10^{-5}\,M/s\]
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
-
\[rate=-\frac{[ NO_{2}]_{(2)}-[ NO_{2}]_{(1)}}{t_{(2)}-t_{(1)}}=M/s\]
\[rate=-\frac{0.073-0.110}{30.0-10.0}=0.00185M/s\]
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
-
\[rate=-\frac{1}{2} \frac{\Delta{[ NO_{2}]}}{\Delta t} =\frac{\Delta [{O_{2}}]}{\Delta t}\]
\[-\frac{1}{2}rate_{(NO_{2})} =rate_{(O_{2})} \]
\[rate_{(O_{2})}=\frac{1}{2} 0.00185 M/s=9.25*10^{-4}M/s\]
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.
Reaction Conditions and Rate
Name for factors that influence the rate of a reaction.
- Answer
-
- Concentration
- Temperature
- Catalysts
- surface area on heterogenous systems.
Effect of Concentration on Reaction Rate
Textbook: Section 14.4
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
-
\[rate=k[A]^{2}[B]\]
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
-
\[rate=k\left [ A \right ]^{3}\]
\[k\left [ 4A \right ]^{3}=64k\left [ A \right ]^{3}=64rate\]
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
-
\[rate_{1}=k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}\]
\[rate_{2}=k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}\]
\[\frac{rate_{1}}{rate_{2}}=\frac{k\left [ A_{1} \right ]^{m}\left [ B_{1} \right ]^{n}}{k\left [ A_{2} \right ]^{m}\left [ B_{2} \right ]^{n}}=\left ( \frac{A_{1}}{A_{2}} \right )^{m}\left ( \frac{B_{1}}{B_{2}} \right )^{n}\]
select when \(\frac{B_{1}}{B_{2}}=1\) (data in the experiment 1 and 3)
\[\frac{rate_{1}}{rate_{2}}=\left ( \frac{A_{1}}{A_{2}} \right )^{m}=\left ( \frac{0.012}{0.024} \right )^{m}=\frac{0.10}{0.10}=1\]
therefore m=0
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
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
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
Integrated Rate Law
Textbook: Section 14.4
Chemical Kinetics & Half-Life
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
-
\[[A]_{0}e^{-kt}=[A]\]
\[0.26e^{-.33t}=0.176\]
\[e^{-.33t}=\frac{0.176}{0.26}\]
\[e^{-.33t}=0.677\]
\[-0.33t=ln(0.677)\]
\[-0.33t=-0.390\]
\[t=\frac{-0.390}{.-33}=1.18\,min\]
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
-
\[ln\frac{A}{A_{0}}=-kt\]
\[\frac{A}{A_{0}}=e^{-kt}\]
\[A=A_{0}e^{-kt}\]
\[A=0.135Me^{-0.75\left ( 1.5 \right )} \\ A=0.044M\]
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
-
\[\frac{1}{A}-\frac{1}{A_{0}}=kt\]
\[0.26t=\frac{1}{0.13}-\frac{1}{0.26}\]
\[t=14.79s\]
Exercise \(\PageIndex{5.1d}\)
What is the rate constant of a first-order reaction that has a half-life of 144s?
- Answer
-
\[k=\frac{0.693}{t_{1/2}}\]
\[k=\frac{0.693}{144.0}=4.8*10^{-3}\,s^{-1}\]
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
-
\[ln\frac{0.015}{0.120}=-k*54\]
\[k=0.0385\,min^{-1}\]
\[t_{1/2}=\frac{0.693}{0.0385}=18.0\,min\]
Zero-Order Integrated Rate Problems
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:
- Reactant
- Product
- Both
- Neither
- Answer
-
d. Neither
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
-
\[A=A_{0}-kt\]
\[k=\frac{1}{t}(A_0-A)=\frac{1}{60 sec}(0.5-0.25) = 0.0042\,M/s\]
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
-
\[A=-kt+A_{0}=A_{0}-kt\]
\[0.1=0.5-0.0042t\]
\[t=96s\]
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
-
\[A=-kt+A_{0}=A_{0}-kt\]
\[t=\frac{1}{k}(A_0-A)=\frac{1}{0.0042}[0.50-0.25(0.50)] \]
\[t=89s\]
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
-
\[A=A_{0}-kt\]
\[t=\frac{A_{0}-A}{k}=\frac{A_{0}-0.75A}{k}=\frac{A_{0}\left ( 1-0.75 \right )}{k}=\frac{0.25A_{0}}{k}\]
\[t=\frac{0.25\left ( 0.50 \right )}{0.0042}\]
\[t=30s\]
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
-
\[A=A_{0}-kt=0.80-\left ( 0.0042*180 \right )=0.044M\]
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
-
\[A=A_{0}-kt\]
\[0.005=A_{0}-\left ( 0.0042*600\right )\]
\[A_{0}=2.5M\]
First-Order Integrated Rate Problems
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
-
\[A=A_{0}e^{-kt}\]
\[\frac{A}{A_{0}}=e^{-kt}\]
\[ln\frac{A}{A_{0}}=-kt\]
\[k=-\frac{1}{t}\left ( ln\,A-ln\,A_{0} \right )\]
\[k=-\frac{1}{60}\left ( ln\,0.25-ln\,0.5 \right )\]
\[k=0.012\, s^{-1}\]
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
-
\[A=A_{0}e^{-kt}\]
\[\frac{A}{A_{0}}=e^{-kt}\]
\[ln\frac{A}{A_{0}}=-kt\]
\[t=-\frac{1}{k}\left ( ln\,A-ln\,A_{0} \right )\]
\[t=-\frac{1}{0.012}\left ( ln\,0.1-ln\,0.5 \right )\]
\[t=134s\]
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
-
\[A=A_{0}e^{-kt}\]
\[\frac{A}{A_{0}}=e^{-kt}\]
\[ln\frac{A}{A_{0}}=-kt\]
\[t=-\frac{1}{k}\left ( ln\frac{A}{A_{0}} \right )\]
\[t=-\frac{1}{0.012}ln\,0.15\]
\[t=158s = 160 sec\]
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
-
\[\frac{A}{A_{0}}=e^{-kt}\]
\[ln\frac{A}{A_{0}}=-kt\]
\[t=-\frac{1}{k}\left ( ln\frac{A}{A_{0}} \right )\]
\[t=-\frac{1}{0.012}\left ( ln\frac{\left (1-0.15 \right )A_{0}}{A_{0}} \right )=-\frac{1}{0.012}ln\,0.85\]
\[t=13.5s\]
Exercise \(\PageIndex{5.3e}\)
\(Br_{2}\rightarrow 2Br\)
What is [Br2] of a 0.80M [Br2] solution after 3 minutes?
- Answer
-
\[A=A_{0}e^{-kt}=0.80*e^{-0.012*180}=0.092M\]
Exercise \(\PageIndex{5.3f}\)
\(Br_{2}\rightarrow 2Br\)
Calculate the initial [Br2] if [Br2] =0.005M after 10 minutes?
- Answer
-
\[A=A_{0}e^{-kt}\]
\[\frac{A}{e^{-kt}}=A_{0}\]
\[A_{0}=\frac{0.005}{e^{-0.012*600}}=6.70M\]
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
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]
\[k=\frac{1}{t}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]
\[k=\frac{1}{75sec}\left ( \frac{1}{0.25M}-\frac{1}{0.90M}\right )\]
\[k=0.039\,M^{-1}s^{-1}\]
Exercise \(\PageIndex{5.4b}\)
How long does it take for the reaction in Q 14.4.s to decompose to 0.10M?
- Answer
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]
\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]
\[t=\frac{1}{0.039\,M^{-1}s^{-1}}\left(\frac{1}{0.1M}-\frac{1}{0.90M}\right )\]
\[t=228sec\]
Exercise \(\PageIndex{5.4c}\)
How long does it take for the reactant in Q 14.4.s to decompose to 25%?
- Answer
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]
\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]
\[t=\frac{1}{0.036\,M^{-1}s^{-1}}\left ( \frac{1}{0.25\left ( 0.90M \right )}-\frac{1}{0.90M}\right )\]
\[t=93sec\]
Exercise \(\PageIndex{5.4e}\)
How long does it take for 25% of the reactant in Q 14.4.s to be consumed?
- Answer
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]
\[t=\frac{1}{k}\left ( \frac{1}{A}-\frac{1}{A_{0}}\right )\]
\[t=\frac{1}{0.036\,M^{-1}s^{-1}}\left ( \frac{1}{0.75\left ( 0.90M \right )}-\frac{1}{0.90M}\right )\]
\[t=10sec\]
Exercise \(\PageIndex{5.4f}\)
What is [NOBr] of a .85M [NOBr]solution after 3 minutes?
- Answer
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\]
\[A=\frac{1}{kt+\frac{1}{A_{0}}}\]
\[A=\frac{1}{0.036\,M^{-1}s^{-1}\left ( 180sec \right )+\frac{1}{0.85M}}\]
\[A=0.13M\]
Exercise \(\PageIndex{5.4g}\)
Calculate the initial [NOBr] if [NOBr]=0.0080M after 15 minutes.
- Answer
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}}\Rightarrow\frac{1}{A_{0}}=\frac{1}{A}-kt\]
\[A_{0}=\frac{1}{\frac{1}{A}-kt}\]
\[A_{0}=\frac{1}{\frac{1}{0.0080M}-0.036\,M^{-1}s^{-1}\left( 900sec \right )}\]
\[A_{0}=0.011M\]
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
-
\[A=-kt+A_{0}\]
\[0.00387M=-k\left ( 87s \right )+0.325M\]
\[k=-\frac{0.00387M-0.325M}{87s}\]
\[k=0.0037\,M/s\]
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
-
\[A=-kt+A_{0}\]
\[0.0042M=-4.95*10^{-5}\,M/s\left ( t \right )+0.0452M\]
\[t=-\frac{0.0042M-0.0452M}{-4.95*10^{-5}\,M/s}\]
\[t=828\,s\]
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
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}} \]
\[\frac{1}{0.029M}=\left ( 0.68M^{-1}s^{-1}*35s \right )+\frac{1}{A_{0}} \]
\[\frac{1}{0.029M}=23.8M^{-1}+\frac{1}{A_{0}} \]
\[\frac{1}{A_{0}}=\frac{1}{0.029M}-23.8M^{-1} \]
\[\frac{1}{A_{0}}=10.68M \]
\[A_{0}=0.094M\]
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
-
\[ln\,A=-kt+ln\,A_{0} \]
\[ln\left ( 0.006M \right )=\left ( -k*240s \right )+ln\left ( 0.08M \right ) \]
\[k=-\frac{ln\left ( 0.006M \right )-ln\left ( 0.08M \right )}{240s} \]
\[k=-\frac{-2.59}{240s} \]
\[k=0.0108s^{-1}\]
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
-
\[ln\,A=-kt+ln\,A_{0} \]
\[ln\,A-ln\,A_{0}=-kt \]
\[ln\frac{A}{A_{0}}=-kt \]
\[A=A_{0}e^{-kt} \]
\[A=0.5046e^{-0.085\left ( 25 \right )} \]
\[A=0.06M\]
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
-
\[\frac{1}{A}=kt+\frac{1}{A_{0}} \]
\[\frac{1}{0.0020M}=\left ( t*0.019M^{-1}s^{-1} \right )+\frac{1}{0.0400M} \]
\[\frac{1}{0.0020M}-\frac{1}{0.0400M}=\left ( t*0.019M^{-1}s^{-1} \right ) \]
\[t=\frac{500M-25m}{0.019M^{-1}s^{-1}} \]
\[t=25000sec\]
Microscopic View of Reaction Rates
Textbook: Section 14.5
Two-state Arrhenius Equation
Exercise \(\PageIndex{6.a}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for k1:
- Answer
-
\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}}\]
\[k_{1}=k_{2}\frac{e^{\frac{-E_{a}}{RT_{1}}}}{e^{\frac{-E_{a}}{RT_{2}}}}=k_{2}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}=k_{2}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]
\[k_{1}=k_{2}e^{\frac{E_{a}}{R}\left (\frac{T_{1}-T_{2}}{T_{2}T_{1}} \right )}\]
Exercise \(\PageIndex{6.b}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for k2:
- Answer
-
\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}}\]
\[k_{2}=k_{1}\frac{e^{\frac{-E_{a}}{RT_{2}}}}{e^{\frac{-E_{a}}{RT_{1}}}}=k_{1}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}=k_{1}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}\]
\[k_{2}=k_{1}e^{\frac{E_{a}}{R}\left (\frac{T_{2}-T_{1}}{T_{2}T_{1}} \right )}\]
Exercise \(\PageIndex{6.c}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for Ea:
- Answer
-
\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]
\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]
\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]
\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]
\[\left [ ln\left ( \dfrac{k_{1}}{k_{2}} \right )\right ]*R=\left [ \dfrac{-E_{a}}{\cancel{\textcolor{red}{R}}}\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right ]*\cancel{\textcolor{red}{R}}\]
\[ \dfrac{\left[ ln\left( \dfrac{k_{1}}{k_{2}} \right) \right]*R}{\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}=\dfrac{-E_{a}*\cancel{\textcolor{red}{\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}}}{\cancel{\textcolor{red}{\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right)}}}\]
\[E_{a}=\frac{{}\left [ ln\left ( \frac{k_{1}}{k_{2}} \right )\right ]*R}{\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}\]
Exercise \(\PageIndex{6.d}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for R:
- Answer
-
\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]
\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]
\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]
\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]
\[\left [ ln\left ( \dfrac{k_{1}}{k_{2}} \right )\right ]*R=\left [ \dfrac{-E_{a}}{\cancel{\textcolor{red}{R}}}\left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right ]*\cancel{\textcolor{red}{R}}\]
\[\dfrac{\cancel{\textcolor{red}{ln\left(\dfrac{k_{1}}{k_{2}}\right)}}*R}{\cancel{\textcolor{red}{ln\left(\dfrac{k_{1}}{k_{2}}\right)}}}=\dfrac{\left[-E_{a}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)\right]}{\left[ln\left(\dfrac{k_{1}}{k_{2}}\right)\right]}\]
\[ R=\dfrac{\left[-E_{a}\left(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}\right)\right]}{\left[ln\left(\dfrac{k_{1}}{k_{2}}\right)\right]} \]
Exercise \(\PageIndex{6.e}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for T1:
- Answer
-
\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]
\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]
\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]
\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]
\[ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} = \left[ \cancel{\textcolor{red}{\dfrac{-E_{a}}{R}}} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right] * \cancel{\textcolor{red}{ \dfrac{R}{-E_{a}}}} \]
\[ \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} = \left( \dfrac{1}{T_{1}}-\cancel{\textcolor{red}{\dfrac{1}{T_{2}}}} \right) + \cancel{\textcolor{red}{\dfrac{1}{T_{2}}}}\]
\[\dfrac{1}{T_{1}} = \left ( \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} \right) \]
\[T_{1} =\dfrac{1}{ \left ( \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} \right)} \]
Exercise \(\PageIndex{6.f}\)
\(\frac{k_{1}}{k_{2}}=\frac{Ae^{\left ( \frac{-E_{a}}{RT_{1}}\right )}}{Ae^{\left ( \frac{-E_{a}}{RT_{2}}\right )}}\)
Solve for T2:
- Answer
-
\[\dfrac{k_{1}}{k_{2}}=\dfrac{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{1}}}}{\cancel{\textcolor{red}{A}}e^{\dfrac{-E_{a}}{RT_{2}}}}\]
\[\frac{k_{1}}{k_{2}}=\frac{e^{\left ( \frac{-E_{a}}{RT_{1}} \right )}}{e^{\left ( \frac{-E_{a}}{RT_{2}} \right )}}=e^{\left ( \frac{-E_{a}}{RT_{1}}-\frac{-E_{a}}{RT_{2}} \right )}\]
\[ln\left [ \dfrac{k_{1}}{k_{2}}\right ]=\cancel{\textcolor{red}{ln}}\left [ \cancel{\textcolor{red}{e}}^{\left ( \dfrac{-E_{a}}{RT_{1}}-\dfrac{-E_{a}}{RT_{2}}\right )}\right ]\]
\[ln\left ( \frac{k_{1}}{k_{2}} \right )=\left ( \frac{-E_{a}}{RT_{1}}- \frac{-E_{a}}{RT_{2}}\right )\]
\[ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} = \left[ \cancel{\textcolor{red}{\dfrac{-E_{a}}{R}}} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \right] * \cancel{\textcolor{red}{ \dfrac{R}{-E_{a}}}} \]
\[ \left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right]* \dfrac{R}{-E_{a}} \right \} + \dfrac{1}{T_{2}} = \left( \dfrac{1}{T_{1}}-\cancel{\textcolor{red}{\dfrac{1}{T_{2}}}} \right) + \cancel{\textcolor{red}{\dfrac{1}{T_{2}}}}\]
\[ \left(\cancel{\textcolor{red}{\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}}} +\dfrac{1}{T_{2}} \right) - \cancel{\textcolor{red}{\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}}} = \dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\]
\[ \dfrac{1}{T_{2}}=\left(\dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\right) \]
\[ T_{2}=\dfrac{1}{\left(\dfrac{1}{T_{1}}-\left \{ \left[ ln \left( \dfrac{k_{1}}{k_{2}} \right) \right] * \dfrac{R}{-E_{a}} \right \}\right)} \]
Arrhenius Equation Problems
Exercise \(\PageIndex{6.g}\)
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
\[E_{a}=-\left (8.314510*10^{-3}\right )\frac{ln\frac{9.80*10^{-2}}{4.40*10^{-3}}}{\left ( \frac{1}{580}-\frac{1}{350} \right )}=22.8\,kJ/mol\]
Exercise \(\PageIndex{6.h}\)
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
-
\[k_{1}=Ae^{-E_{a}/RT_{1}}\]
\[k_{2}=Ae^{-E_{a}/RT_{2}}\]
\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]
\[ln\frac{k_{2}}{k_{1}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]
\[\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}=\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )\]
\[\frac{1}{T_2}= \frac{1}{T_1} -\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}\]
\[T_2 = \frac{1}{\frac{1}{T_1} -\frac{R}{E_{a}}*ln\frac{k_{2}}{k_{1}}}\]
\[k_{2}=2k_{1} \;\; so \; \; ln\frac{k_{2}}{k_{1}} = ln2 \]
\[T_2 = \frac{1}{\frac{1}{310K} -\frac{0.008314\frac{kJ}{mol*K}}{120\frac{kJ}{mol}}*ln2}\]
\[T_{2}=315K\]
Exercise \(\PageIndex{6.i}\)
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
-
\[k_{1}=Ae^{-E_{a}/RT_{1}}\]
\[k_{2}=Ae^{-E_{a}/RT_{2}}\]
\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]
\[ln\frac{k_{2}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]
\[\frac{k_{2}}{k_{1}}=10 \]
\[ln\,10=-\frac{E_{a}}{8.314510*10^{-3}}*\left (\frac{1}{330}-\frac{1}{300}\right ) \]
\[E_{a}=63.2kJ\]
Exercise \(\PageIndex{6.j}\)
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
-
\[k_{1}=Ae^{-E_{a}/RT_{1}}\]
\[k_{2}=Ae^{-E_{a}/RT_{2}}\]
\[\frac{k_{2}}{k_{1}}=\frac{e^{-E_{a}/RT_{2}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{2} \right )+E_{a}/RT_{1}}=e^{-E_{a}/R*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )}\]
\[ln\frac{k_{2}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\]
\[ln\frac{1.2*10^{-3}}{1.5*10^{-4}}=-\frac{E_{a}}{8.314510*10^{-3}}*\left (\frac{1}{423}-\frac{1}{373}\right ) \]
\[E_{a}=54.6kJ\]
Exercise \(\PageIndex{6.k}\)
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 Ea=54.6kJ
\[k_{1}=Ae^{-E_{a}/RT_{1}}\]
\[k_{3}=Ae^{-E_{a}/RT_{3}}\]
\[\frac{k_{3}}{k_{1}}=\frac{e^{-E_{a}/RT_{3}}}{e^{-E_{a}/RT_{1}}}=e^{-E_{a}/\left (RT_{3} \right )+E_{a}/RT_{3}}=e^{-E_{a}/R*\left ( \frac{1}{T_{3}}-\frac{1}{T_{3}} \right )}\]
\[ln\frac{k_{3}}{k_{a}}=-\frac{E_{a}}{R}*\left ( \frac{1}{T_{3}}-\frac{1}{T_{1}} \right )\]
\[ln\frac{k_{3}}{1.5*10^{-4}}=-\frac{54.6}{8.314510*10^{-3}}*\left (\frac{1}{473}-\frac{1}{373}\right ) \]
\[k_{3}=0.0064M^{-1}s^{-1}\]
Exercise \(\PageIndex{6l}\)
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
-
\[\frac{k_{1}}{k_{2}}=\frac{Ae^{\frac{-E_{a}}{RT_{1}}}}{Ae^{\frac{-E_{a}}{RT_{2}}}} \nonumber\]
\[k_{1}=k_{2}\frac{e^{\frac{-E_{a}}{RT_{1}}}}{e^{\frac{-E_{a}}{RT_{2}}}}=k_{2}e^{\frac{-E_{a}}{R}\left ( \frac{1}{T_{1}}-\frac{1}{T_{2}} \right )}=k_{2}e^{\frac{E_{a}}{R}\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )} \nonumber\]
\[k_{1}=2.1x10^{-5}s^{-1}e^{\frac{135kJ/mol}{0.008314J/mol \cdot K}\left ( \frac{1}{355}-\frac{1}{550} \right )} =230s^{-1}\nonumber\]
Reaction Mechanisms
Textbook: Section 14.6
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
Exercise \(\PageIndex{7b}\)
What is the molecularity for each step in the reaction from Question 14.7.a?
- Answer
-
unimolecular, bimolecular
Exercise \(\PageIndex{7c}\)
What is the rate equation for step 2 in the reaction from Question 14.7.a?
- Answer
-
k[O3][O]
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
Exercise \(\PageIndex{7e}\)
What is the rate equation for question 14.7.d?
- Answer
-
k[O2][NO]2
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
Exercise \(\PageIndex{7g}\)
What is the rate equation for question 14.7.f?
- Answer
-
k[H2][ICl]
General Questions
Exercise \(\PageIndex{a}\)
Which of the following is not a valid unit for a reaction rate?
- g/s
- M/s
- mol/hr
- mol/L
- mol/L-hr
- Answer
-
d. mol/L
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
-
\[Rate\,of\,reaction=\frac{1}{2}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{2}\frac{\Delta \left [NO\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]
\[Rate\,of\,disappearance=\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{0.0150-0.00975}{150}=3.5*10^{-5}\,M/s\]
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 N2O5 for this period?
\(2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}\)
- Answer
-
\[Rate\,of\,reaction=\frac{1}{2}\frac{\Delta \left [N_{2}O_{5}\right ]}{\Delta t}=\frac{1}{4}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]
\[\frac{1}{2}\frac{\Delta \left [NO_{2}\right ]}{\Delta t}=\frac{1}{1}\frac{\Delta \left [O_{2}\right ]}{\Delta t}\]
\[\frac{\Delta \left [O_{2}\right ]}{\Delta t}=\frac{1}{2}\left (\frac{\Delta \left [NO_{2}\right ]}{\Delta t}\right )=\frac{1}{2}\left (\frac{0.0300-0.0195}{300}\right )=1.75*10^-5\,M/s\]
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
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
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 _____.
- decrease by a factor of the cube root of 3
- increase by a factor of 27
- increase by a factor of 9
- remain constant
- triple
- Answer
-
\[rate=[A]^{x}\]
\[rate=[3]^{3}\]
\[rate=27\]
b. increase by a factor of 27
Exercise \(\PageIndex{g}\)
Use the information below to answer the following questions:
- Determine the order of the reaction in reactant A.
- Determine the order of the reaction in reactant B.
- Determine the overall order of the reaction.
- 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.
-
\[k=\frac{rate}{[A]^{2}[B]^{0}}=\frac{2.83}{[0.273]^{2}[0.763]^{0}}=38.0\,M^{-1}s^{-1}\]
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]
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
\[2ClO_{2}(aq)+2OH^{-}(aq)\rightarrow ClO_{3}^{-}(aq)+ClO_{2}^{-}(aq)+H_{2}O(l)\]
| 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 |
Exercise \(\PageIndex{j}\)
What is the order of the reaction with respect to ClO2?
- Answer
-
Second-order
Exercise \(\PageIndex{k}\)
What is the order of the reaction with respect to OH-?
- Answer
-
First-order
Exercise \(\PageIndex{l}\)
What is the order of the reaction overall?
- Answer
-
Third-order
Exercise \(\PageIndex{m}\)
What is the value of the rate constant for the reaction?
- Answer
-
\[k=\frac{rate}{\left [ClO_{2}\right ]^{2}\left [OH^{-}\right ]^{1}}=\frac{2.48*10^{-2}}{\left [0.060\right ]^{2}\left [0.030\right ]^{1}}=230\]
Exercise \(\PageIndex{n}\)
What are the units of the rate constant for the reaction?
- Answer
-
M-2s-1
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
-
\[M^{1-n}s^{-1}\]
\[1-n=-1\]
\[n=1+1\]
\[n=2\]
Exercise \(\PageIndex{p}\)
For a zero-order reaction, a plot of _____ versus _____ is linear.
- \(\frac{1}{\left [A\right ]},\,t\)
- \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
- \(ln\left [A\right ]_{t},\,t\)
- \(\left [A\right ],\,t\)
- \(t,\,\frac{1}{\left [A\right ]}\)
- Answer
-
d. \(\left [A\right ],\,t\)
Exercise \(\PageIndex{q}\)
For a first-order reaction, a plot of _____ versus _____ is linear.
- \(\frac{1}{\left [A\right ]},\,t\)
- \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
- \(ln\left [A\right ]_{t},\,t\)
- \(\left [A\right ],\,t\)
- \(t,\,\frac{1}{\left [A\right ]}\)
- Answer
-
c. \(ln\left [A\right ]_{t},\,t\)
Exercise \(\PageIndex{r}\)
For a second-order reaction, a plot of _____ versus _____ is linear.
- \(\frac{1}{\left [A\right ]},\,t\)
- \(ln\left [A\right ]_{t},\,\frac{1}{t}\)
- \(ln\left [A\right ]_{t},\,t\)
- \(\left [A\right ],\,t\)
- \(t,\,\frac{1}{\left [A\right ]}\)
- Answer
-
a. \(\frac{1}{\left [A\right ]},\,t\)
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
-
\[\frac{1}{\left [A\right] _{t}}=kt+\frac{1}{\left [A\right] _{0}}\]
\[\frac{1}{0.0130}=k\left ( 200 \right )+\frac{1}{0.0200}\]
\[76.92=k\left ( 200 \right )+50\]
\[76.92-50=k\left ( 200 \right )\]
\[k=\frac{26.92}{200}=0.13\,M^{-1}s^{-1}\]
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
-
\[\left [ A \right ]=\left [ A \right ]_{0}e^{-kt}\]
\[\left [ A \right ]=\left ( 0.00100 \right )e^{-\left ( 6.29*10^{-4} \right )\left ( 1000 \right )}\]
\[\left [ A \right ]=\left ( 0.00100 \right )e^{-0.629}\]
\[\left [ A \right ]=0.00100*0.533\]
\[\left [ A \right ]=5.33*10^{-4}\]
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
-
\[\left [ A \right ]e^{-kt}=\left [ A \right ]_{0}\]
\[0.13e^{-0.33*t}=0.088\]
\[e^{-0.33*t}=\frac{0.088}{0.13}\]
\[-0.33*t=ln\left ( 0.677 \right )\]
\[t=\frac{-0.390}{-0.33}\]
\[t=1.2\,min\]
Exercise \(\PageIndex{v}\)
Why does the rate of reaction increase, when the temperature of the reaction increases?
- activation energy is lowered
- reactant molecules collide less frequently
- reactant molecules collide less frequently and with greater energy per collision
- reactant molecules collide with greater energy per collision
- reactant molecules collide more frequently with less energy per collision
- Answer
-
d. reactant molecules collide with greater energy per collision
Exercise \(\PageIndex{w}\)
What do reaction rates depend on?
- collision energy
- collision frequency
- collision orientation
- all of these
- none of these
- Answer
-
d. all of these
Exercise \(\PageIndex{x}\)
What is the species that exists at the maximum on the potential energy profile of a reaction?
- activated complex
- activation energy
- atomic state
- enthalpy of reaction
- product
- Answer
-
a. activated complex
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
-
\[ln\left (\frac{k_{1}}{k_{2}}\right )=\frac{E_{a}}{R}\left (\frac{1}{T_{1}}-\frac{1}{T_{2}}\right )\]
\[ln\left (\frac{10}{1}\right )=\frac{E_{a}}{8.314}\left (\frac{1}{303}-\frac{1}{333}\right )\]
\[ln\left (10\right )=\frac{E_{a}}{8.314}\left (2.97*10^{-4}\right )\]
\[\frac{2.30}{2.97*10^{-4}}=\frac{E_{a}}{8.314}\]
\[7.44*10^{3}=\frac{E_{a}}{8.314}\]
\[E_{a}=6.44*10^{4}\,J/mol\]
\[E_{a}=64.4\,kJ/mol\]
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
-
\[ln\left (\frac{k_{1}}{k_{2}}\right )=\frac{E_{a}}{R}\left (\frac{1}{T_{1}}-\frac{1}{T_{2}}\right )\]
\[ln\left (\frac{1.1*10^{-3}}{1.3*10^{-4}}\right )=\frac{E_{a}}{8.314}\left (\frac{1}{273.15+100}-\frac{1}{273.15+150}\right )\]
\[ln\left (8.46\right )=\frac{E_{a}}{8.314}\left (3.17*10^{-4}\right )\]
\[\frac{2.14}{3.17*10^{-4}}=\frac{E_{a}}{8.314}\]
\[6.74*10^{3}=\frac{E_{a}}{8.314}\]
\[E_{a}=5.6*10^{4}\,J/mol\]
\[E_{a}=5.6\,kJ/mol\]
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 (slow)
step 2: E + 2B → C + A (fast)
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)
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
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]
Exercise \(\PageIndex{ad}\)
Which of the following will lower the activation energy for a reaction?
- adding a suitable catalyst
- increasing the concentration of reactants
- raising the temperature of the reaction
- all the above
- none of the above
- Answer
-
a. adding a suitable catalyst
Exercise \(\PageIndex{ae}\)
What type of catalyst is used in automotive catalytic converters?
- enzymes
- heterogenous
- homogenous
- noble gas
- nonmetal oxides
- Answer
-
b. heterogenous
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, rebelford@ualr.edu. 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