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Chapter 11: Problem 54

The rate law for the reaction $$2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ at some temperature is $$\text { Rate }=-\frac{\Delta[\mathrm{NOBr}]}{\Delta t}=k[\mathrm{NOBr}]^{2}$$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 \space M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to \(0.100 \space\mathrm{M} ?\)

Short Answer

Expert verified
The rate constant \(k\) for the reaction is 0.5556 M^-1 s^-1, and the time required for the concentration of NOBr to decrease to 0.100 M is 13.33 s.

Step by step solution

01

Derive the formula for the second-order reaction

For a second-order reaction of the type \[2 NOBr(g) \longrightarrow 2 NO(g) + Br_{2}(g)\] The rate law can be written as: \[Rate = -\frac{\Delta[\text{NOBr}]}{\Delta t} = k[\text{NOBr}]^{2}\] where: \(\Delta [\text{NOBr}]\) = change in concentration of NOBr \(\Delta t\) = change in time k = rate constant Half-life, t (given) = 2.00 s
02

Calculate the value of k for the reaction

First, we need to determine the initial concentration of NOBr: \[[\text{NOBr}]_{0} = 0.900\space M\] The half-life of a second-order reaction is given by the formula: \[t_{1/2} = \frac{1}{k[\text{NOBr}]_{0}}\] Now, substitute the given values in this formula and solve for k: \[2.00\space s =\frac{1}{k(0.900\space M)}\] \[k = \frac{1}{2.00\cdot 0.900}\] \[k = 0.5556\space M^{-1} s^{-1}\] The rate constant \(k\) for the reaction is 0.5556 M^-1 s^-1.
03

Determine the time required for the concentration of NOBr to decrease to 0.100 M

The integrated rate law for the second-order reaction is: \[\frac{1}{[\text{NOBr}]_{t}} = kt + \frac{1}{[\text{NOBr}]_{0}}\] Now, we need to calculate the time it takes for the concentration of NOBr to decrease to 0.100 M: \[[\text{NOBr}]_{t} = 0.100\space M\] Substitute the values of \([\text{NOBr}]_{0}\), \([\text{NOBr}]_{t}\), and k in the integrated rate law: \[\frac{1}{0.100\space M} = (0.5556\space M^{-1} s^{-1})t + \frac{1}{0.900\space M}\] Solve the equation for t: \[t = \frac{\frac{1}{0.100\space M} - \frac{1}{0.900\space M}}{0.5556\space M^{-1} s^{-1}}\] \[t = 13.33\space s\] The time required for the concentration of NOBr to decrease to 0.100 M is 13.33 s.

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Most popular questions from this chapter

You and a coworker have developed a molecule that has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law $$\text { Rate }=k[\mathrm{AV}]^{1}[\mathrm{V}]^{1}$$ You have been given the following data from your coworker: $$[\mathrm{V}]_{0}=0.20 \space\mathrm{M}$$ $$[\mathrm{AV}]_{0}=1.0 \times 10^{-4} \space\mathrm{M}$$A plot of \(\ln [\mathrm{AV}]\) versus \(t\) (s) gives a straight line with a slope of \(-0.32 \mathrm{s}^{-1} .\) What is the value of the rate constant \((k)\) for this reaction?

One mechanism for the destruction of ozone in the upper atmosphere is $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \text { Slow }$$ $$\frac{\mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g)}{\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)}\quad \text { Fast }$$ Overall reactiona. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction$$\mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2}(g)$$is \(14.0 \mathrm{kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is 11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factor \(A\) is the same for each reaction.

Make a graph of \([\mathrm{A}]\) versus time for zero-, first-, and second-order reactions. From these graphs, compare successive half-lives.

Table \(11-2\) illustrates how the average rate of a reaction decreases with time. Why does the average rate decrease with time? How does the instantaneous rate of a reaction depend on time? Why are initial rates used by convention?

The activation energy for the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is \(125 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)
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