Question

Consider the following reaction: S↔P where the rate constant for the forward reaction is k₁, and the rate constant for the reverse reaction is k₂, and Keq= [P]/[S]

Which of the following would be affected by an enzyme? Please answer yes or no and give a short explanation (5-20 words maximally)

a) decreased K_eq

b) increased k₁

c) increased Keq

d) increased Δ G^#

e) decreased Δ G^#

f) increased k₂

g) more negative Δ G⁰

Answer 1

Given reaction is S↔P

Rate constant for the forward reaction is k₁,

and the rate constant for the reverse reaction is k₂,

and Keq = [P]/[S]

According to the Le-Charterlier's principle,a catalyst speeds up both forward & backward reactions to the same extent but does not have any effect on Equilibrium point.

Here the catalyst is an enzyme

a) not decreased K_eq ----- since catalyst will not change the equilibrium constant but it speeds the reaction.

b) increased k₁           ------ as catalyst added it speeds up both forward & backward reactions to the same extent

                                            there by k₁ increases also k₂ increases

c)not increased Keq       ------ since catalyst will not change the equilibrium constant but it speeds the reaction.

d)not increased ΔG^#          ----- since by adding catalyst the K_eq will not change.

                                                      We have ΔG^# = -RTlnK_eq

                                           As K_eq will not change ΔG^# will not change

e) not decreased Δ G^#           --------- since by adding catalyst the K_eq will not change.

                                                      We have ΔG^# = -RTlnK_eq

                                           As K_eq will not change ΔG^# will not change

f) increased k₂                ------ as catalyst added it speeds up both forward & backward reactions to the same extent

                                            there by k₁ increases also k₂ increases

g) more negative ΔG⁰    ----- will not happen