Respuesta :
The question is incomplete, the complete question is :
[tex]E_a[/tex] for the following uncatalyzed reaction is 14.0 kJ.
[tex]O_3(g) + O(g)\rightarrow 2 O_2[/tex]
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,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
The expression used with catalyst and without catalyst is,
[tex]\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}[/tex]
[tex]\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}[/tex]
where,
[tex]K_2[/tex] = rate constant of reaction with catalyst
[tex]K_1[/tex] = rate constant of reaction without catalyst
[tex]Ea_2[/tex] = activation energy with catalyst =11.9 kJ = 11,900 J
[tex]Ea_1[/tex] = activation energy without catalyst = 14.0 kJ = 14,000 J
R = gas constant = 8.314 J/ mol K
T = temperature = [tex]25^oC=273+25=298 K[/tex]
A = Arrhenius constant
Now put all the given values in this formula, we get
[tex]\frac{K_2}{K_1}=e^{\frac{14,000 - 11,900 Jl}{8.314 J/mol K\times 298 K}}=2.3340[/tex]
2.3340 is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 25°C.
