Question

The activation energy for the decomposition of hydrogen peroxide is 55.0 kJ/mol. When the reaction is catalyzed by the enzyme catalase, it is 11.00 kJ/mol. 2H2O2(aq) → 2H2O(l) + O2(g) Calculate the temperature that would cause the nonenzymatic catalysis to proceed as rapidly as the enzyme-catalyzed decomposition at 20.0°C. Assume the frequency factor, A, to be the same in both cases. Report your answer to 3 significant figures.

Answer 1

42.5℃

Explanation

From arrhenius equation the following relation can be derived

ln(k1/k2) = Ea/R( 1/T2 - 1/T1)

where,

k1 and k2 are rate constant 1 and rate constant 2

T1 and T2 temperature 1 and temperature 2 respectively

R = gas constant , 0.008314kJ/mol K

Ea = Activation energy

let us put

k1 = rate constant for nonenzymatic decomposition at 20℃

k2 = rate constant for nonenzymatic decomposition which is equal to rate constant of enzymatic decomposition

T1 = 20℃= 293.15K

T2 = ?

Ea = 55 kJ/mol

k2 = k1 × 5

substituting the values

ln(k1/5k1) = (55kJ/mol/ 0.008314kJ/mol K)(1/T2 - (1/293.15K))

2.303log(k1/5k1) = (55kJ/mol/0.008314kJ/molK)(1/T2 - (1/293.15K))

-1.6097 = 6615.3K(1/T2 - 0.003411K^-1)

6615.3K/T2 = 20.955

T2 = 315.69K

T2 = 42.5℃