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.

My approach:

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 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?

  • $\begingroup$ $\Delta\text{K'}_f < \Delta\text{K'}_r$ so $\text{K'}_{eq} < \text{K}_{eq}$ $\endgroup$ – MaxW Feb 19 '17 at 10:09

You know that both the forwards $k_f$ and reverse $k_r$ rate constants will increase with temperature, simply because there is more energy present and so it takes less energy to surmount the activation barrier, both from the product and reactant sides.

The question is then which is changed more by changing the temperature. The equilibrium constant is as you write $K_Q=k_f/k_r$ and now using the Arrhenius equation we get $$K_Q=\frac{A_f\exp(-\Delta E/(RT))}{A_r\exp(-(\Delta E+E_0)/(RT))}$$

where the A's are pre-exponential factors (and are constant vs temperature) and $\Delta E$ the activation barrier for the forwards reaction and $E_0$ the extra energy for the back reaction and is positive since the reaction is exothermic. (If you sketch the barriers hopefully this should be clear).

Simplifying gives $$K_Q=\frac{A_f}{A_r}\exp(E_0/(RT))$$

As T is increased $K_Q$ becomes smaller. Thus the answer is (D).


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