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Chapter 13: Problem 50

Compounds \(A\) and \(B\) react to give a single product, \(C .\) Write the rate law for each of the following cases and determine the units of the rate constant by using the units \(M\) for concentration and s for time: a. The reaction is first order in \(A\) and second order in \(B\). b. The reaction is first order in \(A\) and second order overall. c. The reaction is independent of the concentration of \(\mathrm{A}\) and second order overall. d. The reaction is second order in both \(\mathrm{A}\) and \(\mathrm{B}\).

Short Answer

Expert verified
a. The reaction is first order in A and second order in B. b. The reaction is first order in A and second order overall. c. The reaction is independent of the concentration of A and second order overall. d. The reaction is second order in both A and B.

Step by step solution

01

Write the rate law

For first order in A and second order in B, the rate law is given by: Rate = k[A][B]^2
02

Determine the units of the rate constant

Let's find the units of k. The units of Rate are M/s, while the units of [A] and [B] are M. We have: [Unit of k] = (M/s) / (M)(M^2) [Unit of k] = M^{-1}s^{-1} b. The reaction is first order in A and second order overall.
03

Write the rate law

For first order in A and second order overall, the rate law is given by: Rate = k[A]
04

Determine the units of the rate constant

Let's find the units of k. The units of Rate are M/s, and the unit of [A] is M. [Unit of k] = (M/s) / (M) [Unit of k] = s^{-1} c. The reaction is independent of the concentration of A and second order overall.
05

Write the rate law

For independent of the concentration of A and second order overall, the rate law is given by: Rate = k[B]^2
06

Determine the units of the rate constant

Let's find the units of k. The units of Rate are M/s, and the unit of [B] is M. [Unit of k] = (M/s) / (M^2) [Unit of k] = M^{-1}s^{-1} d. The reaction is second order in both A and B.
07

Write the rate law

For the second order in both A and B, the rate law is given by: Rate = k[A]^2[B]^2
08

Determine the units of the rate constant

Let's find the units of k. The units of Rate are M/s, and the units of [A] and [B] are M. [Unit of k] = (M/s) / (M^2)(M^2) [Unit of k] = M^{-3}s^{-1}

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Most popular questions from this chapter

At room temperature in the gas phase, dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes to dinitrogen tetroxide and oxygen: $$2 \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow 2 \mathrm{N}_{2} \mathrm{O}_{4}(g)+\mathrm{O}_{2}(g)$$ Calculate the average rate of this reaction between consecutive measurements listed in the following table. $$\begin{array}{cc}\text { Time (s) } & {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right](\mathrm{M})} \\\0 & 0.200 \\\\\hline 300 & 0.180 \\\\\hline 600 & 0.161 \\\\\hline 900 & 0.144 \\\\\hline 1200 & 0.130 \\\\\hline\end{array}$$

On the basis of the frequency factors and activation energy values of the following two reactions, determine which one will have the larger rate constant at room temperature \((298 \mathrm{K})\). \(\mathrm{O}_{3}(g)+\mathrm{Cl}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) \(A=2.9 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{2}=2.16 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) \(A=2.0 \times 10^{-12} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{\mathrm{a}}=11.6 \mathrm{kJ} / \mathrm{mol}\)

The rate laws for the thermal and photochemical decomposition of \(\mathrm{NO}_{2}\) are different. Which of the following mechanisms are possible for the thermal decomposition of \(\mathrm{NO}_{2},\) and which are possible for the photochemical decomposition of \(\mathrm{NO}_{2} ?\) For the thermal decomposition, Rate \(=k\left[\mathrm{NO}_{2}\right]^{2},\) and for the photochemical decomposition, Rate \(=k\left[\mathrm{NO}_{2}\right]\). a. \(\quad \mathrm{NO}_{2}(g) \stackrel{\text { slow }}{\longrightarrow} \mathrm{NO}(g)+\mathrm{O}(g)\) \(\mathrm{O}(g)+\mathrm{NO}_{2}(g) \stackrel{\text { fast }}{\longrightarrow} \mathrm{NO}(g)+\mathrm{O}_{2}(g)\) b. \(\mathrm{NO}_{2}(g)+\mathrm{NO}_{2}(g) \stackrel{\text { fast }}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}_{4}(g)\) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \frac{\mathrm{slow}}{\mathrm{}_{\mathrm{fast}}} \mathrm{NO}(g)+\mathrm{NO}_{3}(g)\) \(\mathrm{NO}_{3}(g)+\mathrm{O}_{2}(g)\) c. \(\mathrm{NO}_{2}(g)+\mathrm{NO}_{2}(g) \stackrel{\text { slow }}{\longrightarrow} \mathrm{NO}(g)+\mathrm{NO}_{3}(g)\) \(\mathrm{NO}_{3}(g) \stackrel{\text { fast }}{\longrightarrow} \mathrm{NO}(g)+\mathrm{O}_{2}(g)\)

Activation Energy of Smog-Forming Reactions The initial step in the formation of smog is the reaction between nitrogen and oxygen. At the temperatures indicated, values of the rate constant of the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g)$$ are as follows: $$\begin{array}{cc}T(\mathrm{K}) & k\left(M^{-1 / 2} \mathrm{s}^{-1}\right) \\\2000 & 318 \\\\\hline 2100 & 782 \\\\\hline 2200 & 1770 \\\\\hline 2300 & 3733 \\\\\hline 2400 & 7396 \\\\\hline\end{array}$$ a. Calculate the activation energy of the reaction. b. Calculate the frequency factor for the reaction. c. Calculate the value of the rate constant at ambient temperature, \(T=300 \mathrm{K}\).

Why can't an elementary step in a mechanism have a rate law that is zero order in a reactant?
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