Question

Consider the equilibrium reaction shown below, which is first order with respect to each [A], [B], [C], and [D].

A + B = C + D with K1 for the forward reaction and K2 for the reverse reaction.

Use the following data:

For T=300K we have Keq= 9*10^-3 and K1=21.3 M^-1 sec^-1

For T=350K we have Keq= 1.6*10^-2 and K1= 3,696 M^-1 sec^-1

A) Find the activation energies for the forward and backward reaction (Eaf and Eab) and the pre-expontential factors (Af and Ab) in the Arrhenius rate expressions for k1 and k2.

B) A catalyst is added to the solution. It is found that the forward rate of the catalyzed reaction at 300 K, k1cat(300K), is equal to that for the uncatalyzed reaction (k1) at 600 K.

i) What is the equilibrium constant Keq(cat) of the catalyzed reaction at 300 K?

ii) What are the forward and backward activation energies (Ea(fc) and Ea(bc)) in the presence of the catalyst?

Answer 1

from Arhenius equation

ln(K2/K1)= (Ea/R)*(1/T1-1/T2)

T1= 300K   T2= 350K K1= 21.3/M.Sec K2=3696

ln(3696/21.3)= (Ea/R)*(1/300-1/350)= (Ea/R)*0.000476

5.2= (Ea/R)*0.000476

Ea/R= 5.2/0.000476=10924.37   Ea= 10924.37*8.314= 90825.21 Joules/mole

if the rate constant for the back ward reactions are K-1 at 300K K-2 at 350 K

Keq=K1/K-1 at 300K

K-1= K1/Keq= 21.3/9*10-3=2367 /M.sec at 350 K, K-2= 3696/1.6*10^-3 =2310*10³

for back ward reaction ln(2310*1000/2367)= (Ea/R)*0.000476, Ea= 120228 Joules/mole

B) at 600K, forward rate constant K1 for Uncatalyzed reaction can be calculated from Arhenius equation by Letting T1= 300K T2= 600K and Ea= 90825.21 Joules/mole

ln(K1/21.3)= (90825.21/8.314)*(1/300-1/600)=18.2 K1/21.3= 80783469 , K1= 80783469*21.3=1.721*10⁹

similarly K=1 at 600K   ln(K-1/2367)= (120228/8.314)*(1/300-1/600)=24

K-1= 6.94*10¹³

Keq= 1.721*10⁹/6.94*10¹³ =2.48*10-5