Question

`E_a` for the uncatalyzed reaction is 14.0 kJ. for the same reaction when catalyzed is11.9 kJ. What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at 89ºC? Assume the frequency factor is the same for each reaction. Ratio =

Answer 1

The ratio is 2:1

Step-by-step explanation:

The relation between activation energy, temperature and reaction constant is formulated as Arrhenius equation, which is:

`lnK=ln(A)-(E_(a))/(RT)`

Where

K = rate constant

A= frequency factor

T= temperature (K)

R= gas constant

Here A and R both are constant for the two given conditions

So

`lnK_(cat)=ln(A)-(E_(cat))/(RT_(cat))\\\\lnK_(uncat)=ln(A)-(E_(uncat))/(RT_(uncat))\\`

Equating the two

`lnK_(cat)+(E_(cat))/(RT_(cat))=lnK_(uncat)+(E_(uncat))/(RT_(uncat))`

`lnK_(cat)-lnK_(uncat)=(E_(uncat))/(RT_(uncat))-(E_(cat))/(RT_(cat))\\ln((K_(cat))/(K_(uncat)))=(E_(uncat))/(RT_(uncat))-(E_(cat))/(RT_(cat))`

`ln([tex](K_(cat))/(K_(uncat))=2`)=\frac{14000}{8.314X362}-\frac{11900}{8.314X362}=0.698[/tex]

Taking antilog