# 8.E: Chemical Kinetics (Exercises)

- Page ID
- 207018

## 9.1: Reaction Rates

### Q9.1

Write the rate of reaction in terms of the rate of disappearance of reactant and the rate of formation of products:

- \(NO_{(g)} + O_{3 (g)} \rightarrow NO_{2(g)} + O_{2(g)}\)
- \(2C_2H_{6 (g)} + 7O_{2(g)} \rightarrow 4 CO_{2(g)} + 6 H_2O_{(aq)}\)
- \(H_{2 (g)} + I_{2 (g)} \rightarrow 2HI_{(g)} \)
- \(4OH_{(g)} + H_2S_{(g)} \rightarrow SO_{2(g)} + 2H_2O_{(aq)} + H_{2(g)}\)

### S9.1

- \(\text{rate of reaction} = \dfrac{-∆[NO]}{∆t} = \dfrac{-∆[O_3]}{∆t} = \dfrac{∆[NO_2]}{∆t} = \dfrac{∆[O_2]}{∆t} \)
- \(\text{rate of reaction} = \dfrac{-∆[C_2H_6]}{2∆t} = \dfrac{-∆[O_2]}{ 7∆t} =\dfrac{∆[CO_2]}{4∆t} = \dfrac{∆[H_2O]}{6∆t} \)
- \(\text{rate of reaction} = \dfrac{-∆[H_2]}{ ∆t} = \dfrac{-∆[I_2]}{∆t} = \dfrac{∆[HI]}{2∆t} \)
- \(\text{rate of reaction} = \dfrac{-∆[OH]}{4∆t} = \dfrac{-∆[H_2S]}{∆t} = \dfrac{∆[SO_2]}{∆t} = \dfrac{∆[H_2O]}{∆t} = \dfrac{∆[H_2]}{∆t} \)

## 9.2: Reaction Order

### Q9.2

Determine the value of the rate constant for the elementary reaction:

\[I_{2(g)} + H_{2 (g)} \rightarrow 2HI_{(aq)}\]

given that when [Br_{2}] is 0.15 M and [H_{2}] is 0.2M, the rate of reaction is 0.005 M s^{-1} at 298 K.

### S9.2

rate of reaction =** **k[Br_{2}][H_{2}]

0.005 Ms^{-1 }=k ( 0.15M)^{2}(0.2M)

k= 1.11 M^{-1}s^{-1}

### Q9.3

Given the rate of the third order reaction:

\[A + B + C \rightarrow P \]

is 0.05 Ms^{-1}

If the [A] = 0.05 M, [B] = 0.01M, and [C] = 0.25M. What is the third order rate constant?

### S9.3

rate of reaction = k[A][B][C]

0.05Ms^{-1 }=k (0.05M)(0,01M)(0.25M)

k= 0.05Ms^{-1}/( 0.05) (0.01)( 0.25) M^{3}

k= 400M^{-2}s^{-1}

### Q9.3

What are the units of the rate constant for a second-order reaction?

### S9.3

The reaction rate:^{-1}min

^{-1}or M

^{-1}s

^{-1}

### Q9.4

Calculate the fraction of the starting quantity of A that will be used up after 60 s. Given the reaction below which is found to be the first order in A and \(t_{1/2} = 40\;s\)

\[ A \rightarrow B + C\]

### S9.4

with \(t_{1/2} = 40 \; s\]

The remain of fraction A after 60 s:

The fraction will be used up after 60s:

### Q9.5

Given the first order reaction is completed 90% in 30 mins at 298 K. Calculate the rate constant.

### Q9.6

Assume the half life of the first order decay of radioactive isotope takes about 1 year (365 days). How long will it take the radioactivity of that isotope to decay by 60%?

### Q9.7a

The decomposition of dinitrogen peroxide (\(N_2O_5\)) is a first-order reaction with a rate constant of 0.045 min^{-1} at 300 K.

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

If there were initially 0.040 mol of \(N_2O_5\), calculate the moles of \(N_2O_5\) remaining after 5 minutes.### S9.7a

The integrated rate equation of a first-order reaction is:

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

Substituting concentration for moles of reactant and plugging in the known values:

\[n_{A} = 0.040_{0}e^{-0.045*5}\]

\[n_{A} = 0.032\]

### Q9.7b

### S9.7b

First, you need to figure the half life of your compound. We do this by solving for** k**:

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

\[k=\dfrac{0.693}{t\dfrac{1}{2}}\]

\[k=\dfrac{0.693}{2 hrs} =0.3465\]

with this half life, we can find the time it will take by solving for **t:**

\[ln\dfrac{[A]}{[A]^{_{o}}}=-kt\]

We do not have the initial and final concentration, but that is okay.

Assuming the initial concentration is 100 g, we can assume that the final concentration is 10 g because that would be reduced 90% as stated in the problem.

\[ln\dfrac{[10]}{[100]}=-kt\]

\[\dfrac{ln[0.1]}{-0.3465}=t=6.65\; hours\]

So 5 hours will be definitely not enough time to reduce the compound

### Q9.8a

^{-4}/min, but your employer wants to the half-life of the reactant?

### S9.8a

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

\[t_{1/2}=\dfrac{0.693}{1.5x10^{^{-4}}}\]

\[=4.62 \times 10^3\; mins\]

### Q9.8b

The half-life of a second order reaction \[2A\rightarrow P\] is given by:

\[t_{1/2}=\dfrac{1}{k[A]_{o}}\]

Calculate the half-life of a reaction with initial reactant concentration [A] = 0.90 M and a rate constant of 0.20 M^{-1}min

^{-1}.

### S9.8b

\[t_{1/2} = \dfrac{1}{0.20\times 0.90} = 0.18 min^{-1}\]### Q9.9a

### S9.9a

A reaction whose half-life changes when the reactant concentration is changed is a second-order reaction. \[t_{1/2} = \dfrac{1}{k[A]_{o}}\] \[k = \dfrac{1}{[A]_{0}t_{1/2}}\] \[k = \dfrac{1}{(0.4M)(5min)} = 0.5 M^{-1}min^{-1}\]### Q9.9b

### S9.9b

### Q9.10a

Calculate the order of the reaction and the rate constant of Cytobutane decompose to ethylene based on equation

\[C_{_4}H_{8 g)} \rightarrow 2C__{2}H_{_{4}(g)}\]

^{0}C, constant volume, the pressure 200, 158, 124, 98, 77.5, 61 mmHg.

### Q9.10b

Given the following data of concentration [A] over a period of time, decide if the data represents first order or second order. Solve for K. Show graphs.

### S9.10b

The data best fits a Second order graph.

The equation for a second order reaction is:

1/[A]= kt+ 1/[A_{0}]

When solving for K, the equation is rearranged to:

__1/[A]- 1/[A _{0}]__ =k

t

Plug in the numbers given:

__1/[.3]- 1/[.5]__= k

(0-54)

.025=k

One, can also obtain the slop of the graph and k=.025

### Q9.11a

If a compound’s ½ life is 15.6 days. What is the value of k? How long will it take to decompose to 10%. Use first order reaction.

### S9.11a__ __

Equation for half-life of a First-Order Reaction is:

\[ t_½ = \dfrac{\ln (2)}{k}\]

If we plug in the information given:

\[15.6 days= \dfrac{\ln(2)}{k}\]

\[k=4.4 \times 10^{-2}\, day^{-1}\]

The second equation needed is the first order reaction, which is:

ln([A]/[A_{0}])= -kt_{1/2 }

so, ln ([A]/[A_{0}])=.1

t=-1/k x ln([A]/[A_{0}])

t= -1/(4.4 x 10^{-2} day^{-1}) x (.1)

t=52 days

12) In a second order reaction 2A--> products, the final concentration is .28M. What is the initial concentration if k=.32M^{-1}s^{-1} and the times is 5 seconds.

__Solution:__

Second order reaction equation is:

1/[A]= kt+ 1/[A_{0}]

Filling the information given:

1/[.28M]=.32M^{-1}s^{-1}(5)+ 1/[A_{0}]

1/[A_{0}]= 1.97

[A_{0}]= .50M

### Q9.11b

Calculate the half life of a compound if 90% of a given sample of the compound decomposed in 30min### S9.11b

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

[A]_{0} = 90%

[A] = 10%

t = 30min * 60s/1min = 1800s

[10%]/[90%]=e^ -k*1800

ln[10%/90%]/1800 = k = 3.95E-4

t1/2= ln(2)/k

t1/2 = .693/3.95E-4= 1754s

1754s*1min/60s = 29 min

### Q9.12

Given rate constant for second order reaction

\[2 NO_{2(g)} \rightarrow 2NO_{(g)} + O_{2(g)}\]

is 1.08 M^{-1}s^{-1 }at 600 ^{0 }C. Find the time that would take for the concentration of NO_{2} from 1.24 M to 0.56 M?

### S9.12

1/ [A] – 1/[A]_{0} = kt

t= 1/k ( 1/[A] – 1/[A]_{0})

= (1/ 1.08 M^{-1}s^{-1 }) (1/0.56 – 1/1.24) = 7.2 s

### Q9.13a

Is which order of reaction half-life is independent of initial concentration?

### S9.13a

First order because the half-life equation for first order is

t_{1/2= }ln(2)/k, it does not have [A_{0}]

### Q9.13b

The decomposition of N2) is the first order. At 365^{0}C, t1/2 is 1.79 x 10^{3 }min. given intial pressure of 1.05 atm.. Calculate total pressure.

### S9.13b

\[\begin{align} P &= P_{N_2O} + P_{N_2} + P_{O_2} \\ &= 0.525 + 0.525 +0.2625 = 1.31 \;atm \end{align}\]

### Q9.14a

The integrated rate law for the zero-order reaction A → B is [A]_{t} = [A]_{0 }- *kt *

a) skektch the following plots:

(i) rate vs. [A]

(ii) [A] vs. t

### S9.14a

rate = k

rate is independent of [A]

(ii) [A] vs. t

[A]_{t} = [A]_{0 }- *kt *

b) Derive an expression for the half-life of the reaction.

At t = t_{1/2} , [A] = [A]_{0}/2 so, [A]_{0}/2 = [A]_{0 }- *kt _{1/2}*

t_{1/2 }= 1/2*k *[A}_{0}

c) Calculate the time in half-lives when the integrated rate law is no longer valid (that is, when [A] = 0)

[A]_{t} = 0 = [A]_{0 }- *kt *

t_{1/2 }= 1/2*k *[A}_{0} ⇒ k_{ }= 1/2t_{1/2 }[A}_{0}

Therefore, to consume all of the reactants it takes

t_{ }= [A}_{0}/k = [A]_{0}/2 so, [A]_{0 }/ (1/(2t_{1/2})) [A]_{0 }= 2t_{1/2}

integrated rate law is no longer valid after 2 half-lives

### Q9.14b

Jack, Jill, and you are in a physical chemistry class. The professor writes the following equations on the black board.

\[A\rightarrow B\]

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

- The professor assigns you to derive the first-order reaction in front of the class.
- Jack was assigned to find the rate constant if the reaction half-life is 10 hours. Since you are a good friend of Jack, you decide to help him double check his answer by solving for the rate constant.

### S9.14b

a)

\[\int ^{[A]}_{[A]_{0}} \dfrac{d[A]}{[A]}=\int^{t}_{0} -kdt\]

\[\ln[A] \mid ^{[A]}_{[A]_{0}} = -kt\]

\[\ln[A] -\ln[A]_{0}=-kt\]

\[\ln\dfrac{[A]}{[A]_{0}} =-kt\]

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

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

b)

\[t_{1/2}=\dfrac{\ln 2}{k}\]

\[10 hours = \dfrac{\ln2}{k}\]

\[k=\dfrac{\ln2}{10 hours}\times\dfrac{1hour}{60min}\times\dfrac{1min}{60secs}\]

\[k=\]

### Q9.15a

In the nuclear industry, workers use a rule of thumb that the readioactivity from any sample will be relatively harmless after 10 half-lives. Calculate the fraction of a radioactive sample that remains after this time period (hint: Radioactive decay obeys first-order kinetics)

### S9.15a

[A] = [A]_{o}e^{-kt}

t_{1/2} = ln2 / k

[A] = (100)e^{-(1)(10) }=.00454

% remaining after 10 half-lives

### Q9.15b

- Radioactive decay occurs in what order?
- The gas used in the simulation is harmless because the fraction of the original remaining is 0.004 and the time that passed 30 years. What is the rate constant?

### S9.15b

a) Radioactive decay occurs in first order.

b)

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

\[0.004=e^{-k\,30years}\]

\[\ln[0.004]=-k\,30years\]

\[k=\dfrac{-30}{\ln[0.004]}\]

### Q9.16a

Many reactions involving heterogeneous catalysis are zero order; that is, rate = k. An example is the decomposition of phosphine (\(\ce{PH3}\)) over tungsten (W):

\[\ce{4PH3(g) →P4(g) + 6H2(g)}\]

The rate for this reaction is independent of [PH_{3}] as long as phosphine's pressure is sufficiently high (>= 1 atm). Explain.

### S9.16a

With sufficient PH_{3}, all of the catalytic sites on the tungsten surface are occupied. Further increases in the amount of phosphine cannot affect the reaction, and the rate is independent of [PH_{3}].

### Q9.16b

Jill and you just went to an enzyme kinetic workshop. However, Jill is confused about why at high concentrations of substrates the reaction order is zero-order. Please explain to Jill in a way that anyone can understand.### S9.16b

## 9.3: Molecularity of a Reaction

### Q9.41a

A mixture of compounds M and N whose half-lives are 40 minutes and 17 minutes, respectively. They decompose by first-order kinetics. If their concentrations are equal initially, how long does concentration of N to be half that of M?

### S9.41a

[I] is the initial concentration of M and N

-> t = 30.1 min

### Q9.41b

Compounds A and B both decay by first-order kinetics. The half-life of A is 20 minutes and the half-life of B is 48 minutes. If a container initially contains equal concentrations of compounds A and B, after how long will the concentration of B be twice that of A?

### S9.41b

1. Write, in mathematical terms, the information given in the problem and what the problem is asking for.

\[ t_{1/2, A} = 20.0 min \]

\[ t_{1/2, B} = 48.0 min \]

\[ [A]_0 = [B]_0 \]

A and B decay by first-order, so

\[ -\dfrac{d[A]}{dt} = k_A [A] \]

\[ -\dfrac{d[B]}{dt} = k_B [B] \]

(Note: the rate constants for A and B are not equal, so indicate which is which with subscripts.)

Want find t at which the following is true:

\[ [B] = 2[A] \]

2. Substitute the integrated rate equations for [A] and [B]

\[ [B]_0 e^{-k_B t} = 2 [A]_0 e^{-k_A t} \]

3. Write expressions for the rate constants in terms of half-lives, and substitute into the equation.

\[ t_{1/2, A} = \dfrac{ln2}{k_A} \Rightarrow k_A = \dfrac{ln2}{t_{1/2, A}} \]

\[ t_{1/2, B} = \dfrac{ln2}{k_B} \Rightarrow k_B = \dfrac{ln2}{t_{1/2, B}} \]

\[ [B]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 [A]_0 e^{-\dfrac{ln2}{t_{1/2, B}} t} \]

4. Solve for t

Since initial concentrations of A and B are equal:

\[ e^{-\dfrac{ln2}{t_{1/2, B}} t} = 2 e^{-\dfrac{ln2}{t_{1/2, B}} t} \]

Take natural log of both sides:

\[ -\dfrac{t}{t_{1/2, B}} ln2 = \left ( 1 - \dfrac{t}{t_{1/2, A}} \right ) ln2 \]

\[ t = \dfrac{1}{\dfrac{1}{t_{1/2, A}} - \dfrac{1}{t_{1/2, B}}} \]

5. Plug in values for half-lives

\[ t = \dfrac{1}{\dfrac{1}{20.0 min} - \dfrac{1}{48.0 min}} = 34.2\ min \]

Answer: 34.2 minutes

### Q9.42a

In Q3 thermodynamics and in Q9 chemical kinetics, the term “reversible “ is used. How do you understand this term? (it has a same meaning in these two chapters)

### S9.42a

Actually, this word is used to describe a “reversible” reaction in which both forward and backward reactions can happen in kinetics. In thermodynamics, “reversible” is used to describe a process that is in equilibrium along the pathway from the initial to final states.

### Q9.42b

If a reaction has come to thermodynamic equilibrium, can we say anything in particular about the system's kinetics?

### S9.42b

Equilibrium occurs when all reactants and products are being consumed at the same rate that they are created. Take the simple example:

\[ A\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}}B \]

When the system is at equilibrium,

\[ \dfrac{d[A]}{dt} = \dfrac{d[B]}{dt} = 0 \]

\[ k_{-1} [A] = k_1 [B] \]

\[ \dfrac{k_{-1}}{k_1} = \dfrac{[B]}{[A]} \]

\[ K_{eq} = \dfrac{k_{-1}}{k_1} \]

### Q9.43a

The recombination of bromine atoms in an organic solvent, like carbon tetrafloride, is considered as a diffusion-controlled process

\[Br + Br \rightarrow Br_2\]

we have the viscosity of CF_{4} is 9.80 x 10^{-4} Nsm^{-2 }at 30^{o}C, what is the rate of recombination at 30^{o}C?

### S9.43a

### Q9.43b

Calculate the rate constant of the diffusion-controlled reaction

\[ 2 I \rightarrow I_2 \]

in dichloromethane at 15°C, which has a viscosity of 0.449 mPa·s at 15°C.

### S9.43b

1. Use equation 9.50 to calculate the rate constant.

\[ k_D = \dfrac{8}{3}\dfrac{RT}{\eta} \]

\[ k_D = \dfrac{8}{3} \dfrac{8.314 \dfrac{J}{mol \ K} \times 288\ K \times \dfrac{N\ m}{J}}{0.449\ mPa \ s \times \dfrac{Pa}{1000\ mPa} \times\dfrac{N/m^2}{Pa}} \times \dfrac{1000\ L}{m^3} \]

Answer:

\[ k_D = 1.42 \times 10^{10} M^{-1} s^{-1} \]

### Q9.52

Japanese survivors have been exposed to the risk of radiation after the atomic bomb. One man was measured to have iodine-131 levels at 9.7 mC. Calculate the number of atoms of I-131 to which this radioactivity corresponds.### S9.52

First, convert the rate mCi s^{-1}

1 mCi=1.10X10^{-3 }Ci

1 Ci=3.7 X 10^{10 }s^{-1}

The rate (\(r\) can be derived as such

\[r= (9.7\; Ci) \left( \dfrac{1\;Ci}{1000\; mCi} \right) \left( \dfrac{3.7 \times 10^{10}\; s^{-1}}{1\; Ci} \right) = 3.59 \times 10^8 s^{-1}\]

The accepted value for the half life of I-131 is 8.02 days. Using this information, the number of I-131 atoms can be calculated using the nuclear decay equation. Use the

Nuclear Decay Equation:

\[ \lambda N = \dfrac{\ln 2}{t_{1/2}} N\]

or

\[N = \dfrac{\ln 2}{t_{1/2}} r\]

The half-life for the radioactive beta decay of iodine is 8.02 days

\[\ce{^{131}_{53}I} \rightarrow \ce{^{131}_{54}Xe} + \beta + \bar{\nu_e} \]

therefore, the rate is

\[r= 3.59 \times 10^8 s^{-1}\]

Plug into the equation and convert days to seconds.^{-8}) = 2.49 X 10

^{14}I-131 atoms.

### Q9.54

Calculate the rate law for the following acid-catalyzed reaction:

\[CH_3COCH_3 + Br_2 \overset{H^+}{\longrightarrow} CH_3COCH_2Br + H^+ + Br^-\]

Expt. # | \([CH_3COCH_3]_o\)/ M | \([Br_2]_o\) /M | \([H^+]_o\) /M | Rate of Disappearance Br_{2}/ M*s^{-1} |
---|---|---|---|---|

1 |
0.3 |
0.05 | 0.05 | \(5.7 \times 10^{-5}\) |

2 |
0.3 | 0.1 | 0.05 | \(5.7 \times 10^{-5}\) |

3 |
0.3 | 0.05 | 0.05 | \(1.2 \times 10^{-4}\) |

4 |
0.4 | 0.05 | 0.2 | \(3.1 \times 10^{-4}\) |

5 |
0.4 | 0.05 | 0.05 | \(7.6 \times 10^{-5}\) |

### S9.54

Find the rate law.

rate = k[CH3COCH3]^x [Br2]^y [H+]^z

[exp 1] / [exp 5]:

5.7e-5 / 7.6e-5 = (0.30/0.40)^x (0.050/0.050)^y (0.050/0.050)^z

3/4 = (3/4)^x

x = 1

[exp 1] / [exp 2]:

5.7e-5 / 5.7e-5 = (0.30/0.30) (0.050/0.10)^y (0.050/0.050)^z

1 = (1/2)^y

y = 0

[exp 1] / [exp 3]:

5.7e-5 / 1.2e-4 = (0.30/0.30) (0.050/0.10)^z

19/40 = (1/2)^z

z ≈ 1

Use [exp 1] to find rate constant.

5.7e-5 M/s = k(0.30 M)(0.050 M)

k = 3.8e-3 M^(-1)-s^(-1)

"calculate the rate of disappearance of bromine if the initial concentration are .600mol/L, 0.200 mol/L, and .10 mol/L for propanone, bromine and H+."

rate = (3.8e-3 M^(-1)-s^(-1))(0.600 M)(0.10 M)

rate = 2.28e-4 M/s

### Q9.55

Determine the rate law for the following reaction:

\[N_2O_2 + H_2 \rightarrow H_2O + N_2O\]

In addition, determine which of the following actions would alter the value of \(k\)?

- Increase in pressure of \(N_2O_2\)
- Increase in volume size of container
- Increase in temperature
- Addition of catalyst to the container.
- None of the above ;a rate constant is always constant.

### Q9.56

Consider the mechanism for the association of iodine atoms to create molecular iodine.

\[2I_{(g)} \rightleftharpoons I_{2(g)}^*\]

\[I_2^*(g) + M(g) \rightarrow I_{2(g)} + M_{(g)}\]

With the respect of the first step is at equilibrium, determine the expected rate law (d/dt)[I2(g)] in terms of k1, k-1, k2, [I], and [M].

### Q9.57

Consider the following reaction:

\[C_3H_{8(g)} + 5O_{2(g)} -> 3CO_{2(g)} + 4H_2O_{(g)}\]

If propane (C3H8) is burning at a rate of 0.15 M/s^-1, calculate the rate of formation of CO2.

## 9.4: More Complex Reactions

### S9.57

First, express the reaction with the differential rate equation for the reactants and products involved.

-(d/dt)[C3H8] = (1/3)(d/dt)[CO2]

Then, use the given burning rate of propane and plug it into the differential equation.

(d/dt)[CO2] = 3(0.15M/s^-1)

## 9.5: The Effect of Temperature on Reaction Rates

### Q9.26

If kept in a refrigerator, fresh fish will last for 3 days. If kept in a freezer, it will last for 6 months. Assuming that the temperature in the refrigerator is 5°C, and the temperature in the freezer is -10°C, calculate the activation energy for the bacterial spoiling of fish. Assume that the spoiling time is the 1/e lifetime instead of the half-life.

### Q9.27

Find the activation energy of a reaction whose rate constant is multiplied by 6.50 when T is increased from 300.0 K to 310.0 K. For a reaction with E_{a} = 19 kJ/mol, by what factor is k multiplied when T increases from 300.0 K to 310.0 K?

### Q9.28

The kinetics of the browning of juice from Golden Delicious apples was studied; at 20°C k=7.87×10^{-3}/week, and at 37°C k=0.139/week. What is the activation energy for the browning of Golden Delicious apple juice?