Multiple Choice
Given the concentration of H₂O changes from 0.50 M to 0.30 M between 25 s and 65 s, calculate the average rate of the reaction in M/s.
15. Chemical Kinetics
Intro to Chemical Kinetics
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If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation energy barrier have to be when sucrose is in the active site of the enzyme? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are the same.)
A
Approximately 5 kJ/mol lower
B
Approximately 35 kJ/mol lower
C
Approximately 10 kJ/mol lower
D
Approximately 100 kJ/mol lower
Verified step by step guidance1
Understand the relationship between reaction rate and activation energy using the Arrhenius equation: \( k = A e^{-\frac{E_a}{RT}} \), where \( k \) is the rate constant, \( A \) is the frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature.
Recognize that the enzyme increases the reaction rate by a factor of 1 million, meaning the catalyzed rate constant \( k_{cat} \) is 1 million times the uncatalyzed rate constant \( k_{uncat} \).
Set up the equation for the catalyzed and uncatalyzed reactions: \( k_{cat} = A e^{-\frac{E_{a,cat}}{RT}} \) and \( k_{uncat} = A e^{-\frac{E_{a,uncat}}{RT}} \).
Since \( k_{cat} = 1,000,000 \times k_{uncat} \), substitute into the Arrhenius equation: \( A e^{-\frac{E_{a,cat}}{RT}} = 1,000,000 \times A e^{-\frac{E_{a,uncat}}{RT}} \).
Solve for the difference in activation energy \( \Delta E_a = E_{a,uncat} - E_{a,cat} \) using the natural logarithm: \( \Delta E_a = -RT \ln(1,000,000) \).
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