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Chapter 14: Problem 85

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 versus enzyme-catalyzed reaction is approximately -43.8 kJ/mol. This result indicates that the enzyme lowers the activation energy required for the reaction to proceed, which is consistent with the role of enzymes as catalysts.

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation relates the rate constant (k) of a reaction to its activation energy (Ea) and temperature (T), as follows: \[ k = Ae^{-\frac{Ea}{RT}} \] where A is the pre-exponential factor, R is the universal gas constant (8.314 J/mol K), and T is the temperature in Kelvin.
02

Convert the given temperature to Kelvin

We are given the temperature in Celsius (25°C). To use the Arrhenius equation, we need the temperature in Kelvin. To convert Celsius to Kelvin, add 273.15: \[ T = 25 + 273.15 = 298.15 \ \mathrm{K} \]
03

Set up two separate Arrhenius equations for uncatalyzed and enzyme-catalyzed reactions

We have the rate constants (k) for both the uncatalyzed and enzyme-catalyzed reactions. We will set up two separate Arrhenius equations, one for each reaction: \[ k_{1} = Ae^{-\frac{Ea_{1}}{RT}} \text{ (for the uncatalyzed reaction)} \] \[ k_{2} = Ae^{-\frac{Ea_{2}}{RT}} \text{ (for the enzyme-catalyzed reaction)} \] where \(k_{1}\) and \(k_{2}\) are the rate constants for the uncatalyzed and enzyme-catalyzed reactions, respectively, while \(Ea_{1}\) and \(Ea_{2}\) are their activation energies.
04

Divide one equation by the other and solve the resulting equation for the difference in activation energies

We want to find the difference in activation energies (\(Ea_{2} - Ea_{1}\)). To do this, divide the equation for the enzyme-catalyzed reaction (k2) by the equation for the uncatalyzed reaction (k1): \[\frac{k_{2}}{k_{1}} = \frac{Ae^{-\frac{Ea_{2}}{RT}}}{Ae^{-\frac{Ea_{1}}{RT}}} \] The pre-exponential factors (A) cancel out: \[\frac{k_{2}}{k_{1}} = e^{\frac{Ea_{1}-Ea_{2}}{RT}} \] We will now solve this equation for \(Ea_{2} - Ea_{1}\). First, take the natural logarithm of both sides: \[\ln{\left(\frac{k_{2}}{k_{1}}\right)} = \frac{Ea_{1} - Ea_{2}}{RT} \] Now, solve for \(Ea_{2} - Ea_{1}\): \[ Ea_{2} - Ea_{1} = -RT\ln{\left(\frac{k_{2}}{k_{1}}\right)} \]
05

Substitute the given values and calculate the difference in activation energies

Using the given rate constants (\(k_{1} = 0.039 \ \mathrm{s}^{-1}\) and \(k_{2} = 1.0 \times 10^6 \ \mathrm{s}^{-1}\)), the temperature in Kelvin (298.15 K), and the gas constant R (8.314 J/mol K): \[ Ea_{2} - Ea_{1} = -8.314 \ \mathrm{J/mol \ K} \times 298.15 \ \mathrm{K} \times \ln{\left(\frac{1.0 \times 10^6 \ \mathrm{s}^{-1}}{0.039 \ \mathrm{s}^{-1}}\right)} \] \[ Ea_{2} - Ea_{1} \approx -43.8 \ \mathrm{kJ/mol} \] The difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction is approximately -43.8 kJ/mol. This result indicates that the enzyme lowers the activation energy required for the reaction to proceed, which is consistent with the role of enzymes as catalysts.

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Most popular questions from this chapter

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{A}]^{x}\) . Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of \(8 .\)

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

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In a hydrocarbon solution, the gold compound \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3}\) decomposes into ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and a different gold compound, (CH \(_{3} ) \mathrm{AuPH}_{3} .\) The following mechanism has been proposed for the decomposition of \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} :\) $$ \quad Step \quad1.\left(\mathrm{CH}_{3}\right)_{3} \mathrm{AuPH}_{3} \frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au}+\mathrm{PH}_{3} $$ $$ Step\quad \quad2.\left(\mathrm{CH}_{3}\right)_{3} \mathrm{Au} \stackrel{k_{2}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}+\left(\mathrm{CH}_{3}\right) \mathrm{Au} \quad$$ $$ Step\quad 3 :\left(\mathrm{CH}_{3}\right) \mathrm{Au}+\mathrm{PH}_{3} \stackrel{k_{3}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) \mathrm{AuPH}_{3}$$ (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary steps? (d) What is the ratedetermining step? (e) What is the rate law predicted by this mechanism? (f) What would be the effect on the reaction rate of adding \(\mathrm{PH}_{3}\) to the solution of \(\left(\mathrm{CH}_{3}\right)_{3}\) AuPH \(_{3} ?\)

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 M^{-1} \mathrm{s}^{-1}\) . (a) Based on the units for \(k,\) is the reaction first or second order in \(\mathrm{NO}_{2} ?\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M},\) how would you determine how long it would take for the concentration to decrease to 0.025 \(\mathrm{M}\) ?
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