Question

A catalyst lowers the activation energy for a certain reaction from 83.314 to 75 kj mol1 at 500 k. What will be the rate of reaction as compared to uncatalysed reaction? Assume other things being equal.

Answer 1

Answer:

The rate of reaction will increases by 0.135 times.

Explanation:

According to the Arrhenius equation,

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

The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

where,

$K_2$ = rate of reaction with catalyst

$K_1$ = rate of reaction without catalyst

$Ea_2$ = activation energy with catalyst  = 75 kJ/mol =75000 J/mol

$Ea_1$ = activation energy without catalyst  = 83.314 kJ/mol = 83,314 J/mol

R = gas constant = 8.314 J/mol K

T = temperature = $500 K$

Now put all the given values in this formula, we get

$\frac{K_2}{K_1}=e^{\frac{75000 J/mol-83314 J/mol}{8.314 J/mol K\times 500}}=0.135$

$K_2=0.135\times K_1$

Therefore, the rate of reaction will increases by 0.135 times.