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Chapter 14: Problem 29

Use the Arrhenius equation to show why the rate constant of a reaction (a) decreases with increasing activation energy and (b) increases with increasing temperature.

Short Answer

Expert verified
According to the Arrhenius equation, the rate constant decreases with increasing activation energy because an increase in activation energy increases the magnitude of \( e^{-\frac{E_a}{RT}} \) reducing the rate constant. On the other hand, the rate constant increases with increasing temperature because an increase in temperature decreases the magnitude of \( e^{-\frac{E_a}{RT}} \), thereby increasing the rate constant.

Step by step solution

01

Understanding the Arrhenius Equation

Begin by understanding the Arrhenius equation. It is an equation that shows the temperature dependence of reaction rates. The Arrhenius equation is given by \( k = A e^{-\frac{E_a}{RT}} \) where k is the rate constant, A is the Arrhenius factor or frequency factor, E_a is the activation energy, R is the universal gas constant and T is the temperature.
02

Effect of Activation Energy on Rate Constant

On increasing the activation energy (E_a), the factor exponentiating \( e \) becomes larger. Remember, the exponential function \( e^-x \) decreases as \( x \) increases. Hence, the larger the value of \( E_a \), the smaller the value of \( e^{-\frac{E_a}{RT}} \), leading to a smaller rate constant. Therefore, the rate constant decreases with an increase in activation energy.
03

Effect of Temperature on Rate Constant

When the temperature (T) increases, \( \frac{E_a}{RT} \) becomes smaller because its denominator \( RT \) is increasing. As a result, the value of \( e^{-\frac{E_a}{RT}} \) increases, which in turn increases the value of the rate constant (k). Therefore, the rate constant increases with increasing temperature.

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Most popular questions from this chapter

The rate of the reaction $$ \begin{aligned} \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q) &+\mathrm{H}_{2} \mathrm{O}(l) \\ \longrightarrow & \mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \end{aligned} $$ shows first-order characteristics-that is, rate \(=\) \(k\left[\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\right]\) - even though this is a second- order reaction (first order in \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}\) and first order in \(\mathrm{H}_{2} \mathrm{O}\) ). Explain.

What is the molecularity of a reaction?

These data were collected for the reaction between hydrogen and nitric oxide at \(700^{\circ} \mathrm{C}\) : \(2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\) $$ \begin{array}{cllc} \text { Experiment } & {\left[\mathrm{H}_{2}\right]} & {[\mathrm{NO}]} & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\ \hline 1 & 0.010 & 0.025 & 2.4 \times 10^{-6} \\ 2 & 0.0050 & 0.025 & 1.2 \times 10^{-6} \\ 3 & 0.010 & 0.0125 & 0.60 \times 10^{-6} \end{array} $$ (a) Determine the order of the reaction. (b) Calculate the rate constant. (c) Suggest a plausible mechanism that is consistent with the rate law. (Hint: Assume the oxygen atom is the intermediate.)

Consider the reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From these data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant: $$ \begin{array}{ccc} {[\mathrm{A}](\mathrm{M})} & {[\mathrm{B}](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

Explain why most metals used in catalysis are transition metals.
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