The rate constant for the second-order reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ is \(0.80 / M \cdot \mathrm{s}\) at \(10^{\circ} \mathrm{C}\). (a) Starting with a concentration of \(0.086 M,\) calculate the concentration of NOBr after 22 s. (b) Calculate the half-lives when \([\mathrm{NOBr}]_{0}=0.072 M\) and \([\mathrm{NOBr}]_{0}=0.054 \mathrm{M}\)

Short Answer

Expert verified
The concentration of NOBr after 22 seconds is 0.0342 M. The half-life when [NOBr]_0 = 0.072 M is 1.7 s and when [NOBr]_0 = 0.054 M, the half-life is 2.3 s.

Step by step solution

01

Calculate the concentration of NOBr after 22 s

We use the second order reaction equation \(\frac{1}{{[NOBr]}} = kt + \frac{1}{{[NOBr]_0}}\). Let \(x = [NOBr]\), then we have to solve the equation for \(x\) in: \(\frac{1}{x} = (0.80/M.s)(22 s) + \frac{1}{0.086 M}\)
02

Solve the equation

Combining the terms on the right-hand side gives \(\frac{1}{x} = 17.6 M^{-1} + 11.6 M^{-1} = 29.2 M^{-1}\), or \(x = [NOBr] = 0.0342 M\)
03

Calculate the half-lives

Using the second order half-life equation \(\(\(t_{1/2} = \frac{1}{{k[A]_0}}\), we can calculate the half-lives for an initial concentration of 0.072 M and 0.054 M, respectively.
04

Half-life for [NOBr]_0 = 0.072 M

\(\(\(t_{1/2} = \frac{1}{{(0.80/M.s)(0.072 M)}} = 1.7 s\)
05

Half-life for [NOBr]_0 = 0.072 M

\(\(\(t_{1/2} = \frac{1}{{(0.80/M.s)(0.054 M)}} = 2.3 s\)

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Most popular questions from this chapter

Chlorine oxide (ClO), which plays an important role in the depletion of ozone (see Problem 13.101 ), decays rapidly at room temperature according to the equation $$ 2 \mathrm{ClO}(g) \longrightarrow \mathrm{Cl}_{2}(g)+\mathrm{O}_{2}(g) $$ From the following data, determine the reaction order and calculate the rate constant of the reaction. $$ \begin{array}{ll} \hline \text { Time (s) } & {[\mathrm{ClO}](M)} \\ \hline 0.12 \times 10^{-3} & 8.49 \times 10^{-6} \\ 0.96 \times 10^{-3} & 7.10 \times 10^{-6} \\ 2.24 \times 10^{-3} & 5.79 \times 10^{-6} \\ 3.20 \times 10^{-3} & 5.20 \times 10^{-6} \\ 4.00 \times 10^{-3} & 4.77 \times 10^{-6} \\ \hline \end{array} $$

The rate law for the reaction \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{NO}]^{2} .\) Which of the following mechanisms can be ruled out on the basis of the observed rate expression? Mechanism I $$ \begin{array}{c} \mathrm{H}_{2}+\mathrm{NO} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N} \\ \mathrm{N}+\mathrm{NO} \longrightarrow \mathrm{N}_{2}+\mathrm{O} \\ \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \end{array} $$ $$ \begin{aligned} &\text { Mechanism II }\\\ &\begin{array}{l} \mathrm{H}_{2}+2 \mathrm{NO} \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{array} \end{aligned} $$ Mechanism III $$ \begin{aligned} 2 \mathrm{NO} & \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} \\ \mathrm{~N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} & \longrightarrow \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O} \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow & \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \end{aligned} $$

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