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Chapter 13: Problem 43

The rate constants of some reactions double with every 10 -degree rise in temperature. Assume that a reaction takes place at \(295 \mathrm{~K}\) and \(305 \mathrm{~K}\). What must the activation energy be for the rate constant to double as described?

Short Answer

Expert verified
The activation energy for the rate constant to double when the temperature increases by 10 K is \( 10400 \) J/mol.

Step by step solution

01

Write the Arrhenius equation for the two temperatures

The Arrhenius equation can be written for the two temperatures \( T_1 \) and \( T_2 \) as follows:\[ k_1 = Ae^{-\frac{E_a}{RT_1}} \] \[ k_2 = Ae^{-\frac{E_a}{RT_2}} \]
02

Form the ratio \( k_2/k_1 \)

Dividing \( k_2 \) by \( k_1 \) cancels out the pre-exponential factor \( A \) and results in the equation: \[ \frac{k_2}{k_1} = \frac{e^{-\frac{E_a}{RT_2}}}{e^{-\frac{E_a}{RT_1}}} = e^{\frac{E_a}{RT_1}-\frac{E_a}{RT_2}} \]
03

Insert given values

Substituting \( k_2 = 2k_1 \) (the rate constant doubles), \( T_1 = 295 \) K, \( T_2 = 305 \) K (temperature increases by 10 K), and \( R = 8.314 \) J/(mol·K) (the gas constant in SI units) into the equation gives: \[ 2 = e^{\frac{E_a}{8.314*295}-\frac{E_a}{8.314*305} } = e^{\frac{E_a}{8.314}\left(\frac{1}{295}-\frac{1}{305}\right)} \]
04

Solve for \( E_a \)

To isolate the activation energy \( E_a \), first take the natural logarithm of both sides, then multiply the equation by \( 8.314 \) and divide it by the term in parentheses on the right-hand side, which gives: \[ E_a = 8.314*\left(\frac{1}{295}-\frac{1}{305}\right)^{-1} * ln(2) \] Evaluating this expression with a calculator yields the activation energy \( E_a = 10400 \) J/mol.

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Most popular questions from this chapter

Consider the following elementary steps for a consecutive reaction: $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(A\).

For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life \(\left(t_{1}\right)\) and in terms of the rate constant \(k\).

For the reaction \(\mathrm{X}_{2}+\mathrm{Y}+\mathrm{Z} \longrightarrow \mathrm{XY}+\mathrm{XZ}\) it is found that doubling the concentration of \(\mathrm{X}_{2}\) doubles the reaction rate, tripling the concentration of Y triples the rate, and doubling the concentration of \(\mathrm{Z}\) has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of \(Z\) has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

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 constant for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is \(1.64 \times 10^{-6} / M \cdot\) s." What is incomplete about this statement?
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