3.12 Practice Problems

Attempt these problems as if they were real exam questions in an exam environment.

Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.

Consider the following reaction.

\[5\mathrm{Br^-}(aq) + \mathrm{BrO_3}(aq) + 6\mathrm{H^+}(aq) \longrightarrow \mathrm{3Br_2}(aq) + \mathrm{3H_2O}(l)\]

If the rate of disappearance of Br^–(aq) at a particular moment during the reaction is 3.7 × 10^–4, what is the rate of appearance (in M s^–1) of Br₂(aq) at that moment?
1. 2.2 × 10^–4
2. 4.4 × 10^–3
3. 6.6 × 10^–2
4. 0.88
5. 1.03
Solution

Answer: A

Concept: Average Rate

Write out the relevant terms of the average rate of reaction expression for the balanced reaction and solve for the rate of appearance of Br₂.

\[\begin{align*} \dfrac{1}{5}\dfrac{-\Delta[\mathrm{Br^-}]}{\Delta t} &= \dfrac{1}{3}\dfrac{\Delta[\mathrm{Br_2}]}{\Delta t} \longrightarrow\\[1.5ex] \dfrac{\Delta[\mathrm{Br_2}]}{\Delta t} &= \dfrac{3}{5} \left (\dfrac{-\Delta[\mathrm{Br^-}]}{\Delta t} \right) \\[1.5ex] &= \dfrac{3}{5} \left ( 3.7\times 10^{-4}~M~\mathrm{s}^{-1} \right ) \\[1.5ex] &= 2.22\times 10^{-4}~M~\mathrm{s}^{-1} \end{align*}\]
In the first 10.0 s of the following reaction, [I^–] drops from 1.000 M to 0.868 M. What is the average rate of reaction (in M s^–1) in this time interval?

\[\mathrm{H_2O_2}(aq) + 3\mathrm{I^-}(aq) + 2\mathrm{H^+}(aq) \longrightarrow \mathrm{I_3^-}(aq) + \mathrm{2H_2O}(l)\]
1. –8.23 × 10^–6
2. –4.40 × 10^–3
3. 3.14 × 10^–15
4. 4.40 × 10^–3
5. 1.44
Solution

Answer: D

Concept: Average Rate

Find Rate of Disappearance of I^–

\[\begin{align*} \dfrac{-\Delta[\mathrm{I^-}]}{\Delta t} &= \dfrac{-(M_{\mathrm{final}} - M_{\mathrm{initial}})} {t_{\mathrm{final}} - t_{\mathrm{initial}}} \\[1.5ex] &= -\dfrac{-(0.868~M - 1.000~M)}{10~\mathrm{s} - 0~\mathrm{s}}\\[1.5ex] &= 1.32\times 10^{-2}~M~\mathrm{s}^{-1} \end{align*}\]

Find Average Rate of Reaction

\[\begin{align*} \mathrm{average~rate} &= \dfrac{1}{3} \dfrac{-\Delta[\mathrm{I^-}]}{\Delta t} \\[1.5ex] &= \dfrac{1}{3} \left ( 1.32\times 10^{-2}~M~\mathrm{s}^{-1} \right ) \\[1.5ex] &= 4.40\times 10^{-3}~M~\mathrm{s}^{-1} \end{align*}\]
The balanced reaction below had an initial concentration of A of 4.09 M. After 40.0 s, the concentration of A was found to be 1.93 M. What is the average rate of reaction (in M s^–1)?

\[\mathrm{A}\longrightarrow \mathrm{B}\]
1. 5.23 × 10^–4
2. 4.23 × 10^–2
3. 5.40 × 10^–2
4. 1.023
5. 300.4
Solution

Answer: C

Concept: Average Rate

Find Rate of Disappearance of A

\[\begin{align*} \dfrac{-\Delta[\mathrm{A}]}{\Delta t} &= \dfrac{-(M_{\mathrm{final}} - M_{\mathrm{initial}})} {t_{\mathrm{final}} - t_{\mathrm{initial}}} \\[1.5ex] &= -\dfrac{-(1.93~M - 4.09~M)}{40~\mathrm{s} - 0~\mathrm{s}}\\[1.5ex] &= 5.4\times 10^{-2}~M~\mathrm{s}^{-1} \end{align*}\]

Find Average Rate of Reaction

\[\begin{align*} \mathrm{average~rate} &= \dfrac{-\Delta[\mathrm{I^-}]}{\Delta t} \\[1.5ex] &= \left ( 5.4\times 10^{-2}~M~\mathrm{s}^{-1} \right ) \\[1.5ex] &= 5.4\times 10^{-2}~M~\mathrm{s}^{-1} \end{align*}\]
A reaction is second-order and has a single reactant, A. What is the rate constant (in M^–1 s^–1) if the reaction rate at 450 °C is 1.13 × 10^–1 mol L^–1 s^–1 when A is 0.206 mol L^–1?
1. 0.001
2. 0.11
3. 2.66
4. 1.33
5. 11.34
Solution

Answer: C

Concept: Rate Law

The reaction is second-order with a single reactant. Therefore, the rate law is

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

We are given the rate of reaction as well as the concentration of reactant A. Solve for the rate constant, k.

\[\begin{align*} \mathrm{rate} &= k[\mathrm{A}]^2 \longrightarrow \\[1.5ex] k &= \dfrac{\mathrm{rate}}{[\mathrm{A}]^2} \\[1.5ex] &= \dfrac{1.13\times 10^{-1}~\mathrm{mol~L^{-1}~s^{-1}}}{(0.206~\mathrm{mol~L^{-1}})^2} \\[1.5ex] &= 2.663~M^{-1}~\mathrm{s^{-1}} \end{align*}\]
The rate constant for a first-order decomposition of cyclobutane, C₄H₈, at 500 °C is 9.2 × 10^–3 s^–1. How long (in s) will it take for 20.0% of the reactant to decompose?
1. 24
2. 42
3. 87
4. 175
5. 430
Solution

Answer: A

Concept: Integrated Rate Law

This problem involves the integrated rate law since time is involved. The first-order integrated rate law is

\[\ln\mathrm{[A]}_t = -kt + \ln\mathrm{[A]_0}\]

The initial concentration, A₀ can be any non-zero amount. Here I choose to begin with a 1 M concentration. The problem wants us to determine how long it takes for 20% of the reactant to decompose. Therefore, the final concentration of reactant is 0.8 (i.e. 80% of 1).

Solve for t.

\[\begin{align*} \ln\mathrm{[A]}_t &= -kt + \ln\mathrm{[A]_0} \longrightarrow \\[1.5ex] t &= \dfrac{-(\ln\mathrm{[A]}_t - \ln\mathrm{[A]}_0)}{k} \\[1.5ex] &= \dfrac{-[\ln (0.8~M) - \ln (1~M)]}{9.2\times 10^{-3}~\mathrm{s^{-1}}} \\[1.5ex] &= 24.25~\mathrm{s} \end{align*}\]
Molecular iodine dissociates at 625 K with a rate constant of 0.271 M s^–1. What is the half-life (in s) of this reaction when [I₂] = 1.05 M?
1. 1.94
2. 2.56
3. 6.42
4. 8.36
5. 9.34
Solution

Answer: A

Concept: Half-life

The reaction can be determined to be zeroth-order by looking at the units of the rate constant.

Solve for t_1/2.

\[\begin{align*} t_{1/2} &= \dfrac{[\mathrm{A}]_0}{2k} \\[1.5ex] &= \dfrac{1.05~M}{2(0.271~M~\mathrm{s^{-1}})} \\[1.5ex] &= 1.9373~\mathrm{s} \end{align*}\]
What is the half-life (in h) for the decomposition of O₃ when the concentration is 2.42 × 10^–6 M? The rate constant is 50.4 L mol^–1 h^–1.
1. 4.80 × 10^–8
2. 0.014
3. 623.15
4. 8,199
5. 2.86 × 10⁵
Solution

Answer: D

Concept: Half-life

The reaction can be determined to be second-order by looking at the units of the rate constant.

Solve for t_1/2.

\[\begin{align*} t_{1/2} &= \dfrac{1}{k[\mathrm{A}]_0} \\[1.5ex] &= \dfrac{1}{50.4~M\mathrm{^{-1}~h^{-1}} \left ( 2.42\times 10^{-6}~M\right )} \\[1.5ex] &= 8198.87~\mathrm{h} \end{align*}\]
A decomposition reaction of molecule X is performed and three plots are generated with the x-axis as time. You notice that the plot which gave a straight line of data points had 1/X on the y-axis. What is the order of the decomposition reaction?
1. zeroth-order
2. first-order
3. second-order
4. third-order
5. none of these
Solution

Answer: C

Concept: Integrated Rate Law

Recall that rate laws can be expressed as integrated rate laws. We can obtain linear correlations of concentration vs. time depending on how the these quantities are expressed. For a second-order reaction, this relation is 1/X vs. time according to the second-order integrated rate law given in the form of a line (y=mx+b).

\[\dfrac{1}{[\mathrm{A}]_t} = kt + \dfrac{1}{[\mathrm{A}]_0}\]
What is the rate constant (in s^–1) for a decomposition of molecule X if 95% of the reactant decomposed in 5.43 minutes?
1. 9.2 × 10^–3
2. 0.55
3. 1.2
4. 68.2
5. 174.9
Solution

Answer: A

Concept: Integrated Rate Law

This reaction is first-order since the rate constant is given as the inverse of time (i.e. 1/t). We can assume any initial concentration. I will conveniently choose 1 M. Recognize that the final concentration is 5% of the initial concentration since 95% has decomposed. Therefore, the final concentration is 0.05~M. Solve for the rate constant, k.

\[\begin{align*} \ln\mathrm{[A]}_t &= -kt + \ln\mathrm{[A]_0} \longrightarrow \\[1.5ex] k &= \dfrac{-(\ln\mathrm{[A]}_t - \ln\mathrm{[A]}_0)}{t} \\[1.5ex] &= \dfrac{-[\ln (0.05~M) - \ln (1~M)]}{5.43~\mathrm{min}\left ( \dfrac{60~\mathrm{s}}{\mathrm{min}}\right )} \\[1.5ex] &= 9.195\times 10^{-3}~\mathrm{s^{-1}} \end{align*}\]
What is the activation energy (in kJ) for a reaction that has a rate constant of 0.253 s^–1 and a frequency factor of 1.06 × 10¹¹ s^–1 at 44.0 °C?
1. –7.06 × 10⁴
2. 7.21
3. 17.3
4. 70.6
5. 700.69
Solution

Answer: D

Concept: Arrhenius Equation

Use the Arrhenius equation to solve for the activation energy. Put temperature in Kelvin (44.0 °C → 317.15 K).

\[\begin{align*} k &= Ae^{\frac{-E_{\mathrm{a}}}{RT}} \longrightarrow \\[1.5ex] E_{\mathrm{a}} &= -RT \ln\dfrac{k}{A} \\[1.5ex] &= -\left (8.314~\mathrm{J~mol^{-1}~K^{-1}} \right ) \left (317.15~\mathrm{K} \right ) \ln\left (\dfrac{0.253~\mathrm{s^{-1}}}{1.06\times 10^{11}~\mathrm{s^{-1}}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 70.563~\mathrm{kJ} \end{align*}\]
For the following reaction, the rate constant at 701 K is measured as 2.57 M^–1 s^–1 and at 895 K is measured as 567 M^–1 s^–1. What is the activation energy (in kJ mol^–1 to one decimal place)?
1. –1.4 × 10^–2
2. 4.9 × 10^–3
3. 1.5 × 10²
4. 8.3 × 10²
5. 2.6 × 10³
Solution
Answer: C

Concept: Arrhenius Equation

Organizing the given information
- k₁ = 2.57 M^–1 s^–1
- T₁ = 701 K
- k₂ = 567 M^–1 s^–1
- T₂ = 895 K
- E_a = ?
leads us to conclude that the Arrhenius equation two-point form is needed to solve for the activation energy.

\[\begin{align*} \ln\dfrac{k_2}{k_1} &= \dfrac{E_{\mathrm{a}}}{R} \left ( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right ) \longrightarrow \\[1.5ex] E_{\mathrm{a}} &= R \left ( \dfrac{\ln \left ( \dfrac{k_2}{k_1} \right )}{\dfrac{1}{T_1} - \dfrac{1}{T_2}} \right ) \\[1.5ex] &= (8.314~\mathrm{J~mol^{-1}~K^{-1}}) \left ( \dfrac{\ln \left ( \dfrac{567~M^{-1}~\mathrm{s^{-1}}}{2.57~M^{-1}~\mathrm{s^{-1}}} \right )} {\dfrac{1}{701~\mathrm{K}} - \dfrac{1}{895~\mathrm{K}}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[1.5ex] &= 145.097~\mathrm{kJ~mol^{-1}} \end{align*}\]

Math String: 8.314 * ln(567/2.57) / (1/701 - 1/895) / 1000
What is the molecularity for the following elementary step?

\[2\mathrm{N_2O}(g) + \mathrm{Cl_2}(g) \longrightarrow 2\mathrm{N_2}(g) + 2\mathrm{ClO}(g)\]
1. Unimolecular
2. Bimolecular
3. Termolecular
Solution

Answer: C

Concept: Molecularity

Recall that elementary steps are reaction steps that cannot be simplified any further. Simply count the number of reactant particles involved in the balanced reaction. Here we have three particles: two N₂O particles and one Cl₂ particle.

The molecularity for this step is termolecular.
Which rate law for an elementary step corresponds to a molecularity that is bimolecular?
1. rate = k[A][B]²
2. rate = k[A]
3. rate = k[A]²[B]²
4. rate = k[A]²
5. none of these
Solution

Answer: D

Concept: Elemenary Steps

Recall that we can write out a rate law with defined exponents for an elementary step. Given that the step is bimolecular, we know that the sum of the exponents must equal two. The only rate law in the given choices that would fulfill this criteria is

\[\mathrm{rate} = k[\mathrm{A}]^2\]
What is the rate law for the following elementary step?

\[\mathrm{NO}(g) + \mathrm{O_2}(g) \longrightarrow \mathrm{NO_2}(g)\]
1. rate = k[NO][O₂]³
2. rate = k[NO]²[O₂]
3. rate = k[NO][O₂]
4. rate = [NO₂]/[NO][O₂]
5. none of these
Solution

Answer: B

Concept: Elemenary Steps

Recall that we can write out a rate law with defined exponents for an elementary step. First, we must balance the reaction.

\[2\mathrm{NO}(g) + \mathrm{O_2}(g) \longrightarrow 2\mathrm{NO_2}(g)\]

Next, write out the rate law.

\[\mathrm{rate} = k[\mathrm{NO}]^2[\mathrm{O_2}]\]
Consider the following two-step reaction mechanism.

\[\begin{align*} \mathrm{Cl}(g) + \mathrm{O_3}(g) ~ &\longrightarrow ~ \mathrm{ClO}(g) + \mathrm{O_2}(g) \\ \mathrm{ClO}(g) + \mathrm{O}(g) ~ &\longrightarrow ~ \mathrm{Cl}(g) + \mathrm{O_2}(g) \end{align*}\]

Identify the intermediate of reaction.
1. O₂
2. ClO
3. O
4. Cl
5. none of these
Solution

Answer: B

Concept: Intermediates

An intermediate is any species that is produced in a multi-step reaction and is later subsequently consumed. Here, ClO is produced in the first step and consumed in the second step.
A single-reactant is found to have an decreasing half-life as the initial concentration of the reactant increases. What is the order of the reaction?
1. zeroth
2. first
3. second
Solution

Answer: C

Concept: Half-life

The second-order half-life equation

\[t_{1/2} = \dfrac{1}{k[\mathrm{A}]_0}\]

shows that as the reactant concentration, A, increases, t_1/2 gets smaller.
The rate constant for a decomposition reaction is 4.5 × 10^–3 s^–1. How long (in min) will it take for 75% of the reactant to decompose?
1. 122.41
2. 5.13
3. 308.07
4. 1.07
5. 63.93
Solution

Answer: B

Concept: Integrated Rate Law

The reaction is first-order and can be determined by analyzing the units of the rate constant. Use the first-order integrated rate law to solve for time. We can assume any initial concentration we wish. Here I will assume a 1 M concentration. The final concentration is therefore 0.25 M since 75% of the reactants has decomposed.

Solve for t.

\[\begin{align*} \ln\mathrm{[A]}_t &= -kt + \ln\mathrm{[A]_0} \longrightarrow \\[1.5ex] t &= \dfrac{-(\ln\mathrm{[A]}_t - \ln\mathrm{[A]}_0)}{k} \\[1.5ex] &= \dfrac{-[\ln (0.25~M) - \ln (1~M)]}{4.5\times 10^{-3}~\mathrm{s^{-1}}} \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}}\right ) \\[1.5ex] &= 5.134~\mathrm{min} \end{align*}\]
If a 5.0 M solution of substance A decomposes for 50 min and the remaining is found to be 0.5 M, what is the half life of this reaction (in h) if it follows first-order kinetics?
1. 0.25
2. 0.83
3. 1.0
4. 15.1
5. 32.4
Solution

Answer: A

Concept: Half-life

The first-order half-life equation is

t_1/2 = ln(2)/k

To solve for the half-life we need the rate constant for the reaction. We can use the first-order integrated rate law.

Find k

\[\begin{align*} \ln\mathrm{[A]}_t &= -kt + \ln\mathrm{[A]_0} \longrightarrow \\[1.5ex] k &= \dfrac{-(\ln\mathrm{[A]}_t - \ln\mathrm{[A]}_0)}{t} \\[1.5ex] &= \dfrac{-[\ln (0.5~M) - \ln (5~M)]}{50~\mathrm{min}\left ( \dfrac{\mathrm{h}}{60~\mathrm{min}}\right )} \\[1.5ex] &= 2.763~\mathrm{h^{-1}} \end{align*}\]

Find t_1/2

\[\begin{align*} t_{1/2} &= \dfrac{\ln 2}{k} \\[1.5ex] &= \dfrac{\ln 2}{2.763~\mathrm{h^{-1}}} \\[1.5ex] &= 0.2509~\mathrm{h} \end{align*}\]
Choose the false statement regarding the Arrhenius equation and reaction rates.
1. Increasing T results in a faster rate of reaction.
2. Increasing the pre-exponential factor results in a larger k.
3. Decreasing k results in a slower rate of reaction.
4. Increasing E_a results in a faster rate of reaction.
5. none of these
Solution

Answer: D

Concept: Arrhenius Equation

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

As E_a increases, k decreases leading to a slower rate of reaction according to any given rate law which begins as

\[\mathrm{rate} = k\cdots\]

As k gets smaller, rate gets smaller. Therefore, D is the false statement.
A two-step reaction mechanism describing an aqueous phase reaction has the following rate constants for each step: k₁ = 1.843 and k₂ = 0.0325. Each step is unimolecular and equimolar. A catalyst is used in the overall reaction. Which of the following statements is true?
1. The catalyst is consumed in the second step of the reaction
2. The catalyst lowers the activation energy for the first step of the reaction
3. The catalyst lowers the activation energy for the second step of the reaction
4. The catalyst shifts the equilibrium towards the products
5. The catalyst is an intermediate of reaction
Solution

Answer: C

Concept: Catalysis

A catalyst speeds up a reaction. The only way for a catalyst to do this in a multi-step reaction is to lower the activation energy of the slowest step. Here, the second step is the slowest because its rate constant, k, is the smallest. Therefore, C is the true statement.

The mechanism for a chemical reaction is given below.

\[\begin{alignat*}{3} \mathrm{(CH_3)_3CCl} ~ &\longrightarrow ~ \mathrm{(CH_3)_3C^+} + \mathrm{Cl^-} \quad &&k_1 \quad &&(E_{\mathrm{a}}~\text{is large}) \\ \mathrm{(CH_3)_3C^+} + \mathrm{OH^-} ~ &\longrightarrow ~ \mathrm{(CH_3)_3COH} \quad &&k_2 \quad &&(E_{\mathrm{a}}~\text{is small}) \end{alignat*}\]

Which of the following statements is/are correct?

Number	Statement
1	The overall balanced reaction is (CH₃)₃CCl + OH^– → (CH₃)₃COH + Cl^–
2	The chloride ion is a reaction intermediate
3	The following experimental rate law is consistent with the mechanism: rate = k[(CH₃)₃CCl]

1 and 3
1
3
1, 2, and 3
1 and 2

Solution

What is the predicted rate law for the following reaction mechanism?

\[\begin{align*} &\text{Step 1 (fast):} ~\mathrm{2NO}(g) \longrightarrow \mathrm{N_2O_2}(g)\\[1.5ex] &\text{Step 2 (slow):} ~\mathrm{H_2}(g) + \mathrm{N_2O_2}(g) \longrightarrow \mathrm{H_2O}(g) + \mathrm{N_2O}(g)\\[1.5ex] &\text{Step 3 (fast):} ~\mathrm{N_2O}(g) + \mathrm{H_2}(g) \longrightarrow \mathrm{N_2}(g) + \mathrm{H_2O}(g) \end{align*}\]
1. rate = k[NO]²
2. rate = k[N₂O][H₂]
3. rate = k[H₂][N₂O₂]
4. rate = k[H₂]²
5. none of these
Solution

Answer: C

Concept: Reaction Mechanisms

See the given example on mechanisms with a slow second-step.