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Chapter 14: Problem 93

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\) \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) .\) In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Short Answer

Expert verified
The difference in activation energies for the uncatalyzed and enzyme-catalyzed reactions is approximately \( -48,300\, J/mol\). The negative sign indicates that the activation energy of the enzyme-catalyzed reaction is lower than that of the uncatalyzed reaction.

Step by step solution

01

Recall the Arrhenius Equation

We will use the Arrhenius equation, which describes the temperature dependence of reaction rates: \[k = Ae^{\frac{-E_a}{RT}}\] where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol K), and T is the temperature in Kelvin.
02

Convert Temperature to Kelvin

We need to convert the given temperature from Celsius to Kelvin. Temperature in Kelvin can be calculated using the formula: \[T(K) = T(°C) + 273.15\] For the given temperature of 25°C, we have: \[T(K) = 25 + 273.15 = 298.15\;K\]
03

Solve for Activation Energies

We first solve for activation energy in both situations (uncatalyzed and enzyme-catalyzed reactions) using the Arrhenius equation. Let \(E_{a1}\) be the activation energy for the uncatalyzed reaction and \(E_{a2}\) be the activation energy for the enzyme-catalyzed reaction: For the uncatalyzed reaction: \[k_1 = Ae^{\frac{-E_{a1}}{RT}}\] For the enzyme-catalyzed reaction: \[k_2 = Ae^{\frac{-E_{a2}}{RT}}\] We are given that the collision factor (A) is the same in both situations. So, we can divide the second equation by the first one: \[\frac{k_2}{k_1} = e^{\frac{E_{a1} - E_{a2}}{RT}}\] Our next objective is to isolate the term \((E_{a1} - E_{a2})\).
04

Isolate the Difference in Activation Energies

To find the difference in activation energies, we need to isolate the term \((E_{a1} - E_{a2})\). We can do this by taking the natural logarithm of both sides of the equation: \[\ln\left(\frac{k_2}{k_1}\right) = \frac{E_{a1} - E_{a2}}{RT}\] Now, we can solve for \((E_{a1} - E_{a2})\): \[(E_{a1} - E_{a2}) = RT\ln\left(\frac{k_2}{k_1}\right)\]
05

Calculate the Difference in Activation Energies

Now, we substitute the given values for \(k_1\), \(k_2\) and the temperature (in Kelvin) into the equation to calculate the difference in activation energies: \[(E_{a1} - E_{a2}) = (8.314\, J/mol\, K)(298.15\, K)\ln\left(\frac{1.0 \times 10^{6}\, s^{-1}}{0.039\, s^{-1}}\right)\] \[(E_{a1} - E_{a2}) \approx -48,300\, J/mol\] The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately -48,300 J/mol. The negative sign indicates that the activation energy of the enzyme-catalyzed reaction is lower than that of the uncatalyzed reaction.

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Most popular questions from this chapter

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

The reaction $2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\( \)\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$ was studied with the following results:

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)+\mathrm{I}^{-}(a q) \end{aligned} $$ (a) Write the chemical equation for the overall process. (b) Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.

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