교과서 질문
Consider this energy diagram:
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b. Label the reactants, products, and intermediates.
15. Chemical Kinetics
Energy Diagrams
객관식
If an enzyme increases the rate of the hydrolysis reaction of sucrose into glucose and fructose by a factor of 7 million, how much lower must the activation barrier be?
A
Approximately 50 kJ mol⁻¹ lower
B
Approximately 70 kJ mol⁻¹ lower
C
Approximately 10 kJ mol⁻¹ lower
D
Approximately 30 kJ mol⁻¹ lower
검증된 단계별 안내1
Understand that the rate of a reaction is related to the activation energy by the Arrhenius equation: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
Recognize that the enzyme increases the reaction rate by a factor of 7 million, which means the new rate constant \( k' = 7,000,000 \times k \).
Set up the equation for the rate constants before and after the enzyme is added: \( k' = A e^{-\frac{E_a'}{RT}} \) and \( k = A e^{-\frac{E_a}{RT}} \).
Divide the equation for \( k' \) by the equation for \( k \) to eliminate \( A \): \( \frac{k'}{k} = e^{-\frac{E_a'}{RT}} / e^{-\frac{E_a}{RT}} = e^{-\frac{E_a' - E_a}{RT}} \).
Solve for the change in activation energy \( \Delta E_a = E_a - E_a' \) using the natural logarithm: \( \ln(7,000,000) = \frac{\Delta E_a}{RT} \). Rearrange to find \( \Delta E_a = RT \ln(7,000,000) \).
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