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Chapter 12: Problem 42

A certain reaction has the following general form: $$ \mathrm{aA} \longrightarrow \mathrm{bB} $$ At a particular temperature and \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M,\) con- centration versus time data were collected for this reaction, and a plot of 1\(/[\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(+3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) . a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of A to decrease to $7.00 \times 10^{-4} M ?$

Short Answer

Expert verified
a. The reaction is second-order with rate law: \(\textrm{Rate} = k[\textrm{A}]^2\), integrated rate law: \(\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}]_{0}}\), and rate constant: \(k = 3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). b. The half-life of the reaction is: \(t_{1/2} = \frac{1}{(3.60 \times 10^{-2} \mathrm{L/mol \cdot s})(2.80 \times 10^{-3} \mathrm{M})}\) c. To find the time required for the concentration of A to decrease to \(7.00 \times 10^{-4} M\), use the equation: \(\frac{1}{7.00 \times 10^{-4} \mathrm{M}} = (3.60 \times 10^{-2} \mathrm{L/mol} \cdot \mathrm{s})\times t + \frac{1}{2.80 \times 10^{-3} \mathrm{M}}\). Solve for t.

Step by step solution

01

Determine the order of the reaction

Since the problem states that the plot of 1\(/[\mathrm{A}]\) versus time resulted in a straight line, this indicates that the reaction is of the second order. The general form of the second-order rate law is: $$\textrm{Rate} = k[\textrm{A}]^2$$
02

Find the integrated rate equation for the second order reaction

For a second-order reaction, the integrated rate equation is: $$\frac{1}{[\mathrm{A}]} = kt + \frac{1}{[\mathrm{A}]_{0}}$$
03

Calculate the rate constant

The problem statement provides the slope of the line of the graph of 1\(/[\mathrm{A}]\) versus time as \(+3.60 \times 10^{-2}\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). This slope corresponds to the rate constant k: $$k = 3.60 \times 10^{-2} \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}$$
04

Calculate the half-life for the second order reaction

The half-life for a second-order reaction is given by the equation: $$t_{1/2} = \frac{1}{k[\mathrm{A}]_{0}}$$ Using the initial concentration \([\mathrm{A}]_{0}=2.80 \times 10^{-3} M\) and the calculated value of k, the half-life can be determined: $$t_{1/2} = \frac{1}{(3.60 \times 10^{-2} \mathrm{L/mol \cdot s})(2.80 \times 10^{-3} \mathrm{M})}$$ Compute \(t_{1/2}\) to obtain the value for half-life.
05

Calculate the time required for the concentration of A to decrease to a certain value

We are given the final concentration of A, \([\mathrm{A}] = 7.00 \times 10^{-4} M\). We can use the integrated rate equation for the second-order reaction to calculate the time required: $$\frac{1}{7.00 \times 10^{-4} \mathrm{M}} = (3.60 \times 10^{-2} \mathrm{L/mol} \cdot \mathrm{s})\times t + \frac{1}{2.80 \times 10^{-3} \mathrm{M}}$$ Solve for time, t, to find how long it takes for the concentration of A to reach \(7.00 \times 10^{-4} M\).

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Most popular questions from this chapter

The mechanism for the gas-phase reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{array}{c}{\mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO}} \\ {\mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2}}\end{array} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?

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Sulfuryl chloride undergoes first-order decomposition at $320 .^{\circ} \mathrm{C}\( with a half-life of 8.75 \)\mathrm{h}$ . $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ What is the value of the rate constant, \(k,\) in \(\mathrm{s}^{-1}\) ? If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is 791 torr and the decomposition occurs in a \(1.25-\mathrm{L}\) container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after 12.5 \(\mathrm{h}\) ?

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