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Chapter 14: Problem 87

The activation energy of an uncatalyzed reaction is $95 \mathrm{~kJ} / \mathrm{mol}\(. The addition of a catalyst lowers the activation energy to \)55 \mathrm{~kJ} / \mathrm{mol}$. Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},(\mathbf{b}) 125^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The catalyst increases the rate of the reaction by a factor of approximately 243.27 at \(25^{\circ}\mathrm{C}\) and by a factor of approximately 44.21 at \(125^{\circ}\mathrm{C}\).

Step by step solution

01

Convert Temperatures to Kelvin

Convert both given temperatures to Kelvin by adding 273.15 to the temperatures in Celsius: \(T_1 = 25 + 273.15 = 298.15\,\text{K}\) \(T_2 = 125 + 273.15 = 398.15\,\text{K}\) Step 2: Calculate the rate constant of the uncatalyzed reaction and the catalyzed reaction at both temperatures using the Arrhenius equation. Note that the pre-exponential factor \(A\) remains the same, as the collision factor does not change.
02

Use Arrhenius Equation

For both reactions, the rate constant k can be calculated using the Arrhenius equation: \( k = Ae^{\frac{-E_a}{RT}} \) where \(A\) is the pre-exponential factor, \(R\) is the gas constant (8.314 J/mol K), and \(T\) is the temperature in Kelvin. For the uncatalyzed reaction: \( k_1 = Ae^{\frac{-E_a}{RT_1}} = Ae^{\frac{-95 \times 10^3}{8.314 \times 298.15}} \) For the catalyzed reaction: \( k_2 = Ae^{\frac{-E_a'}{RT_1}} = Ae^{\frac{-55 \times 10^3}{8.314 \times 298.15}} \) Repeat the calculations for \(T_2 = 398.15\,\text{K}\). Step 3: Calculate the factor by which the rate of the reaction increases due to the catalyst at each temperature.
03

Calculate Rate Increase Factor

Divide the rate constant of the catalyzed reaction (\(k_2\)) by the rate constant of the uncatalyzed reaction (\(k_1\)) at each temperature. This gives the factor by which the rate of the reaction increases due to the catalyst. At \(T_1\), the factor is: \( \frac{k_2}{k_1} = \frac{Ae^{\frac{-55 \times 10^3}{8.314 \times 298.15}}}{Ae^{\frac{-95 \times 10^3}{8.314 \times 298.15}}} \) At \(T_2\), the factor is: \( \frac{k_2}{k_1} = \frac{Ae^{\frac{-55 \times 10^3}{8.314 \times 398.15}}}{Ae^{\frac{-95 \times 10^3}{8.314 \times 398.15}}} \) Step 4: Simplify the expressions to obtain the final results.
04

Simplify Expressions

To find the rate increase factor, simplify the expressions by canceling A and applying properties of exponentials: At \(T_1\): \( \frac{k_2}{k_1} = e^{\frac{(-55 + 95) \times 10^3}{8.314 \times 298.15}} \) At \(T_2\): \( \frac{k_2}{k_1} = e^{\frac{(-55 + 95) \times 10^3}{8.314 \times 398.15}} \) Step 5: Evaluate the expressions to find the factors by which the rate of the reaction increases due to the catalyst at each temperature.
05

Evaluate Expressions

For \(T_1 = 298.15\,\text{K}\): \( \frac{k_2}{k_1}= e^{\frac{40 \times 10^3}{8.314 \times 298.15}} \approx 243.27 \) For \(T_2 = 398.15\,\text{K}\): \( \frac{k_2}{k_1}= e^{\frac{40 \times 10^3}{8.314 \times 398.15}} \approx 44.21 \) Conclusion: The catalyst increases the rate of the reaction by a factor of approximately 243.27 at 25°C and by a factor of approximately 44.21 at 125°C.

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Most popular questions from this chapter

A certain enzyme catalyzes a biochemical reaction. In water, without the enzyme, the reaction proceeds with a rate constant of $6.50 \times 10^{-4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C} .$ In the presence of the enzyme in water, the reaction proceeds with a rate constant of $1.67 \times 10^{4} \mathrm{~min}^{-1}\( at \)37^{\circ} \mathrm{C}$. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

(a) The gas-phase decomposition of sulfuryl chloride $\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right), \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\( is first order in \)\mathrm{SO}_{2} \mathrm{Cl}_{2}\(. At \)300^{\circ} \mathrm{C}$ the half-life for this process is two and a half days. What is the rate constant at this temperature? (b) At \(400^{\circ} \mathrm{C}\) the rate constant is \(0.19 \mathrm{~min}^{-1}\). What is the half-life at this temperature?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\) (b) (c) \(\mathrm{SO}_{3}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{O}(g)\)
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