An enzyme catalyzes the decomposition of hydrogen peroxide at 20°C. The activation energy for the uncatalyzed reaction is 75.3 kJ/mol. The activation energy for the catalyzed reaction is 29.3 kJ/mol. By what factor is the rate of the reaction increased by the enzyme? Use the Arrhenius equation, which is given below, to determine the ratio of rate constants, k_catalyzed/k_uncatalyzed.
(R = 8.315 J / (mol K) )
$k = A \exp \left( - \frac { E _ { a } } { R T } \right)$

Question

An enzyme catalyzes the decomposition of hydrogen peroxide at 20°C. The activation energy for the uncatalyzed reaction is 75.3 kJ/mol. The activation energy for the catalyzed reaction is 29.3 kJ/mol. By what factor is the rate of the reaction increased by the enzyme? Use the Arrhenius equation, which is given below, to determine the ratio of rate constants, k_catalyzed/k_uncatalyzed.
(R = 8.315 J / (mol K) )

k = A \exp \left( - \frac { E _ { a } } { R T } \right)

Quizplus · Accepted Answer

The Answer of An enzyme catalyzes the decomposition of hydrogen peroxide at 20°C. The activation energy for the uncatalyzed reaction is 75.3 kJ/mol. The activation energy for the catalyzed reaction is 29.3 kJ/mol. By what factor is the rate of the reaction increased by the enzyme? Use the Arrhenius equation, which is given below, to determine the ratio of rate constants, k_catalyzed/k_uncatalyzed. (R = 8.315 J / (mol K)) $k = A \exp \left( - \frac { E _ { a } } { R T } \right)$

An Enzyme Catalyzes the Decomposition of Hydrogen Peroxide at 20°C k=Aexp⁡(−EaRT)k = A \exp \left( - \frac { E _ { a } } { R T } \right)k=Aexp(−RTEa​​)

An Enzyme Catalyzes the Decomposition of Hydrogen Peroxide at 20°C $k = A \exp \left( - \frac { E _ { a } } { R T } \right)$