Multiple Choice
For two measured rate constants for the same reaction at different temperatures: Temperature / °C | k / s⁻¹ - 199 | 4.50 × 10⁻⁵ - 258 | 3.2 × 10⁻³ What is the activation energy (in kJ mol⁻¹) for this reaction?
15. Chemical Kinetics
Arrhenius Equation
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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. If an enzyme increases the rate of the hydrolysis reaction by a factor of 1 million, how much lower does the activation barrier have to be when sucrose is in the presence of the enzyme?
A
Approximately 90 kJ/mol lower
B
Approximately 70 kJ/mol lower
C
Approximately 30 kJ/mol lower
D
Approximately 50 kJ/mol lower
Verified step by step guidance1
Understand that the problem involves the effect of an enzyme on the activation energy of a reaction. The enzyme lowers the activation energy, which increases the reaction rate.
Use the Arrhenius equation to relate the rate constant (k) to the activation energy (Ea): k = A * exp(-Ea / (R * T)), where A is the pre-exponential factor, R is the gas constant, and T is the temperature in Kelvin.
Recognize that the rate of the reaction increases by a factor of 1 million due to the enzyme. This means k_with_enzyme = 1,000,000 * k_without_enzyme.
Set up the equation using the Arrhenius equation for both scenarios (with and without enzyme) and equate the ratio of the rate constants to 1,000,000: exp(-Ea_with_enzyme / (R * T)) / exp(-Ea_without_enzyme / (R * T)) = 1,000,000.
Solve for the difference in activation energy (ΔEa = Ea_without_enzyme - Ea_with_enzyme) using the natural logarithm: ΔEa = R * T * ln(1,000,000). Calculate this value to find how much lower the activation barrier is with the enzyme.
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