I'm currently stuck on #30 for Chemistry Olympiad 2015 local exam which reads as:
For a reversible exothermic reaction, what is the effect of increasing temperature on the equilibrium constant (Keq) and on the forward rate constant (kf)?
(A) Keq and kf both increase
(B) Keq and kf both decrease
(C) Keq increases and kf decreases
(D) Keq decreases and kf increases
The answer for this problem is D.
To find the relationship between Keq and kf, I first used this theoretical reaction of
A --> B + heat
The forward rate can be defined as rate(f) = kf[A] and the backwards rate as rate(b) = kb[B]
Since we're finding the relationship between these rates with Keq, rate(f) = rate(b) because the reaction is at equilibrium. This allows the equation kf[A] = kb[B] to be set up. Dividing both sides by kb and [A] gives us
Kf/kb = [B]/[A]
And Keq = products/reactants = [B]/[A] = kf/kb
Due to the fact that the temperature is raised with an exothermic reaction, the reaction is reactant favored and more A will be formed. Since
Keq = products/reactants = [B]/[A]
Keq will decrease. However, why doesn't kf decrease? Since
Keq = Kf/Kb,
how could kf increase while Keq decrease if they're equal to each other in that equation? Shouldn't Kb increase or Kf decrease to "=" to the decreased value of Keq?