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Chapter 11: Problem 92

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \space\mathrm{kJ} / \mathrm{mol} .\) 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 \(29.10\space\mathrm{kJ/mol}\).

Step by step solution

01

Write the Arrhenius equation for catalyzed and uncatalyzed reactions

We have the Arrhenius equation for the rate constant (k) for both the uncatalyzed (k1) and the catalyzed (k2) reactions: \[k_1=A\space e^{\frac{-E_a}{RT}}\] \[k_2=A\space e^{\frac{-E_{a^{'}}}{RT}}\] Where: \(k_1\) - rate constant for uncatalyzed reaction \(k_2\) - rate constant for catalyzed reaction A - frequency factor (same for both reactions) \(-E_a\) - activation energy for uncatalyzed reaction \(-E_{a^{'}}\) - activation energy for catalyzed reaction R - gas constant (\(8.314\space\mathrm{J} / \mathrm{mol \cdot K}\)) T - temperature in Kelvin
02

Calculate the temperature in Kelvin and set the rate constants ratio

First, convert the given temperature from Celsius to Kelvin by adding 273.15: \[T = 37^\circ C + 273.15 = 310.15 K\] Then, set the ratio for the rate constants based on the given rate constant increase factor: \[\frac{k_2}{k_1} = 2.50 \times 10^{3}\]
03

Divide one Arrhenius equation by the other and eliminate A

Divide the Arrhenius equation for the catalyzed reaction by the one for the uncatalyzed reaction: \[\frac{k_2}{k_1} = \frac{A\space e^{\frac{-E_{a^{'}}}{RT}}}{A\space e^{\frac{-E_a}{RT}}}\] A will cancel out: \[\frac{k_2}{k_1} = \frac{e^{\frac{-E_{a^{'}}}{RT}}}{e^{\frac{-E_a}{RT}}}\] Now substitute the calculated ratio and activation energy of the uncatalyzed reaction: \[\frac{1}{2.50 \times 10^{3}} = \frac{e^{\frac{-E_{a^{'}}}{(8.314)(310.15)}}}{e^{\frac{-50\,000}{(8.314)(310.15)}}}\]
04

Solve for the activation energy of the catalyzed reaction

Rearrange the equation so that expressions with \(-E_{a^{'}}\) and \(-E_{a}\) are on the same side, and then take the natural logarithm of both sides: \[\ln\left(\frac{1}{2.50 \times 10^{3}}\right) = \frac{-E_{a^{'}}}{(8.314)(310.15)} - \frac{-50\,000}{(8.314)(310.15)}\] Next, solve for \(-E_{a^{'}}\): \[-E_{a^{'}} = [(8.314)(310.15)]\left[\ln\left(\frac{1}{2.50 \times 10^{3}}\right) + \frac{50\,000}{(8.314)(310.15)}\right]\] Now calculate \(E_{a^{'}}\): \[E_{a^{'}} = -\left[(8.314)(310.15)\left(\ln\left(\frac{1}{2.50 \times 10^{3}}\right) - \frac{50\,000}{(8.314)(310.15)}\right)\right] \approx 29.10 \space\mathrm{kJ/mol}\]
05

Express the answer

The activation energy for the catalyzed reaction, rounding to two decimal points, is approximately \(29.10\space\mathrm{kJ/mol}\).

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Most popular questions from this chapter

For enzyme-catalyzed reactions that follow the mechanism $$\mathbf{E}+\mathbf{S} \rightleftharpoons \mathbf{E} \cdot \mathbf{S}$$ $$\mathrm{E} \cdot \mathrm{S} \rightleftharpoons \mathrm{E}+\mathrm{P}$$ a graph of the rate as a function of \([\mathrm{S}]\), the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(11-7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(11-7 ?\)

The activation energy for the reaction $$\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)$$ is \(125 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Consider two reaction vessels, one containing A and the other containing \(\mathrm{B},\) with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$\begin{aligned} &k_{A}=4.50 \times 10^{-4} \mathrm{s}^{-1}\\\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1} \end{aligned}$$how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{B}] ?\)

Would the slope of a \(\ln (k)\) versus \(1 / T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus \(1 / T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.
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