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  3. consider the equilibrium reaction shown below, whi

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

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Chemistry

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

A + B = C + D with K1 for the forward reaction and K2 for thereverse 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 backwardreaction (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 theforward rate of the catalyzed reaction at 300 K, k1cat(300K), isequal to that for the uncatalyzed reaction (k1) at 600 K.

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

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

Answer & Explanation Solved by verified expert
3.7 Ratings (276 Votes)

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
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