Question

The activation energy for the uncatalyzed decomposition of hydrogen peroxide at 20°C is 75.3 kJ/mol. In the presence of the enzyme catalase, the activation energy is reduced to 29.3 kJ/mol. Use the following form of the Arrhenius equation, RT ln1k1/k22 5 Ea2 2 Ea1 , to calculate how much larger the rate constant of the catalyzed reaction is.

Answer 1

Answer:

The rate of enzyme catalyzed reaction will increases by $1.58\times 10^{8}$  times.

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

$\log K=\log A-\frac{Ea}{2.303\times RT}$

The expression used with catalyst and without catalyst is,

$\log K_1=\log A-\frac{Ea_1}{2.303\times RT}$...(1)

$\log K_2=\log A-\frac{Ea_2}{2.303\times RT}$...(2)

On subtracting (2) from (1)

$\log \frac{K_2}{K_1}=\frac{Ea_1-Ea_2}{2.303RT}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst  

$Ea_2$ = activation energy with catalyst  = 29.3 kJ/mol = 29300 J/mol

$Ea_1$ = activation energy without catalyst  = 75.3 kJ/mol=75300 J/mol

R = gas constant =8.314 J /mol K

T = temperature = $20^oC=273+20=293K$

Now on substituting all the values in the above formula, we get

$\log \frac{K_2}{K_1}=\frac{75300 kJ/mol-29300 kJ/mol}{2.303\times 8.314 J/mol K\times 293}=1.58\times 10^{8}$

The rate of enzyme catalyzed reaction will increases by $1.58\times 10^{8}$  times.