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Chapter 12: Problem 92

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Short Answer

Expert verified
The activation energy for the catalyzed reaction is approximately \(19.85 \mathrm{~kJ/mol}\).

Step by step solution

01

Write the Arrhenius equation

The Arrhenius equation relates the rate constant (k) of a reaction to the activation energy (Ea), temperature (T), and the frequency factor (A): \[k = Ae^{\frac{-Ea}{RT}}\] Here, R is the gas constant (8.314 J/mol·K).
02

Set up a ratio for the catalyzed and uncatalyzed reactions

We are given the rate constant for the catalyzed reaction is 2.50 x 10^3 times greater than the uncatalyzed reaction. Since the frequency factors for both reactions are the same, we can remove them from the equation and set up a ratio: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-Ea_{cat}}{RT}}}{e^{\frac{-Ea_{uncat}}{RT}}}\]
03

Solve for the activation energy of the catalyzed reaction

First, substitute the given values into the ratio: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-Ea_{cat}}{RT}}}{e^{\frac{-Ea_{uncat}}{RT}}} = 2.50 \times 10^3\] Now, simplify the equation by dividing one exponent by the other (subtracting the exponents): \[e^{\frac{-(Ea_{cat} - Ea_{uncat})}{RT}} = 2.50 \times 10^3\] Take the natural logarithm of both sides to isolate the exponent: \[\frac{-(Ea_{cat} - Ea_{uncat})}{RT} = \ln(2.50 \times 10^3)\] Now, solve for Ea_cat: \[Ea_{cat} = Ea_{uncat} - RT \times \ln(2.50 \times 10^3)\]
04

Plug in the given values and calculate Ea_cat

Given, Ea_uncat = 50.0 kJ/mol and T = 37°C (convert to Kelvin: 37+273 = 310 K). Also, convert activation energy from kJ/mol to J/mol: 50 kJ/mol × 1000 = 50000 J/mol. Plug in these values and solve for Ea_cat: \[Ea_{cat} = 50000 J/mol - (8.314 \frac{J}{mol \cdot K})(310 K) \times \ln(2.50 \times 10^3)\] \[Ea_{cat} ≈ 19848 J/mol\]
05

Report the answer

Finally, convert Ea_cat back to kJ/mol: 19848 J/mol × 0.001 = 19.85 kJ/mol (approximately). The activation energy for the catalyzed reaction is about 19.85 kJ/mol.

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Most popular questions from this chapter

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M\), concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{~min}^{-1}\). 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 \(\mathrm{A}\) to decrease to \(2.50 \times 10^{-3} M ?\)

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 \(1.25\) -L container, how many molecules of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) remain after \(12.5 \mathrm{~h}\) ?

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Define stability from both a kinetic and thermodynamic perspective. Give examples to show the differences in these concepts.
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