Question

for the uncatalyzed reaction is for the same reaction when catalyzed is kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at ? Assume that the frequency factor A is the same for each reaction. Answer

Answer 1

The question is incomplete, the complete question is :

$E_a$ for the following uncatalyzed reaction is 14.0 kJ.

$O_3(g) + O(g)\rightarrow 2 O_2$

What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C? Assume that the frequency factor A is the same for each reaction.  

Answer:

2.3340 is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C.

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 constant of reaction with catalyst

$K_1$ = rate constant of reaction without catalyst

$Ea_2$ = activation energy with catalyst  =11.9 kJ = 11,900 J

$Ea_1$ = activation energy without catalyst  = 14.0 kJ = 14,000 J

R = gas constant = 8.314 J/ mol K

T = temperature = $25^oC=273+25=298 K$

A =  Arrhenius constant

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

$\frac{K_2}{K_1}=e^{\frac{14,000 - 11,900 Jl}{8.314 J/mol K\times 298 K}}=2.3340$

2.3340 is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C.