Question

The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ ⋅ mol − 1 51.00 kJ⋅mol−1 . Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ ⋅ mol − 1 Ea=136.00 kJ⋅mol−1 Catalyzed: A ⟶ B A⟶B E a = 85.00 k J ⋅ mol − 1 Ea=85.00 kJ⋅mol−1

Answer 1

The given question is incomplete.The complete question is:

The presence of a catalyst provides a reaction pathway in which the activation energy of a reaction is reduced by 51.00 kJ /mol Uncatalyzed: A ⟶ B A⟶B E a = 136.00 kJ/mol Catalyzed: A ⟶ B A⟶B E a = 85.00 k J/mol. determine the factor by which tha catalysed reaction is faster than the uncatalysed reaction at 289.0 K if all other factors are equal.

Answer: The factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is $1.64\times 10^9$

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

$Ea_1$ = activation energy without catalyst

R = gas constant = $8.314\times 10^{-3}kJ/Kmol$

T = temperature = $289.0K$

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

$\frac{K_2}{K_1}=e^{\frac{51.00}{8.314\times 10^{-3}\times 289.0}}$

$\frac{K_2}{K_1}=e^{21.22}$

$\frac{K_2}{K_1}=1.64\times 10^9$

Therefore, the factor by which the catalysed reaction is faster than the uncatalysed reaction at 289.0 K is $1.64\times 10^9$