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Chapter 11: Problem 73

A catalyst lowers the activation energy of a reaction from \(215 \mathrm{~kJ}\) to \(206 \mathrm{~kJ} .\) By what factor would you expect the reaction-rate constant to increase at \(25^{\circ} \mathrm{C} ?\) Assume that the frequency factors \((\mathrm{A})\) are the same for both reactions. (Hint: Use the formula \(\ln k=\ln \mathrm{A}-E_{2} / R T .\) )

Short Answer

Expert verified
Answer: The reaction-rate constant increases by a factor of approximately 6.41 after using the catalyst at 25°C.

Step by step solution

01

Write down the given information

We are given: - Initial activation energy, E1 = 215 kJ/mol - New activation energy after using a catalyst, E2 = 206 kJ/mol - Temperature, T = 25°C
02

Convert the temperatures to Kelvin

T = 25°C + 273.15 = 298.15 K
03

Convert the activation energies to J/mol

Since the activation energy is given in kJ/mol, we multiply by 1,000 to convert to J/mol. E1 = 215 kJ/mol * 1,000 J/kJ = 215,000 J/mol E2 = 206 kJ/mol * 1,000 J/kJ = 206,000 J/mol
04

Write the Arrhenius equation for both activation energies

ln(k1) = ln(A) - E1 / (R*T) ln(k2) = ln(A) - E2 / (R*T)
05

Calculate the factor by which the reaction rate increases

To find the factor by which the reaction rate constant increases, we can divide the second equation by the first equation. (ln(k2) - ln(k1)) = (E1 - E2) / (R*T) So, ln(k2/k1) = (E1 - E2) / (R*T) Now, solve for k2/k1: k2/k1 = exp[((E1 - E2) / (R*T))] Here, we use the gas constant R = 8.314 J/(mol*K). k2/k1 = exp[((215,000 - 206,000) / (8.314 * 298.15))]
06

Calculate the final value for the increase in reaction rate constant

k2/k1 = exp[(9,000) / (2,482)] ≈ 6.41 Therefore, the reaction-rate constant would increase by a factor of approximately 6.41 after using the catalyst at 25°C.

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Most popular questions from this chapter

For the reaction $$2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ the experimental rate expression is rate \(=k[\mathrm{NO}]^{2} \times\left[\mathrm{H}_{2}\right] .\) The following mechanism is proposed: $$\begin{array}{cc}2 \mathrm{NO} \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{2} & \text { (fast) } \\ \mathrm{N}_{2} \mathrm{O}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{N}_{2} \mathrm{O} & \text { (slow) } \\ \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \longrightarrow \mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} & \text { (fast) } \end{array}$$ Is this mechanism consistent with the rate expression?

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