I'm currently stuck on problem #30 for the Chemistry Olympiad 2015 local exam which reads:

For a reversible exothermic reaction, what is the effect of increasing temperature on the equilibrium constant ($K_\mathrm{eq}$) and on the forward rate constant ($k_\mathrm{f}$)?

(A) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both increase

(B) $K_\mathrm{eq}$ and $k_\mathrm{f}$ both decrease

(C) $K_\mathrm{eq}$ increases and $k_\mathrm{f}$ decreases

(D) $K_\mathrm{eq}$ decreases and $k_\mathrm{f}$ increases

The answer for this problem is D.

My approach:

To find the relationship between $K_\mathrm{eq}$and $k_\mathrm{f}$, I first used this theoretical reaction of

$$\ce{A -> B + \mathrm{heat}}$$

The forward rate can be defined as

$\mathrm{rate(f)} = k_\mathrm{f}[\ce{A}]$

and the backwards rate as

$\mathrm{rate(b)} = k_\mathrm{b}[\ce{B}]$

Since we're finding the relationship between these rates with $K_\mathrm{eq}$, rate(f) = rate(b) because the reaction is at equilibrium. This allows the equation $k_\mathrm{f}\ce{[A]} = k_\mathrm{b}\ce{[B]} $ to be set up. Dividing both sides by $k_\mathrm{b}$ and [A] gives us

$k_\mathrm{f}/k_\mathrm{b} = \ce{[B]/[A]}$


$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]} = k_\mathrm{f}/k_\mathrm{b}$

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

$K_\mathrm{eq} = \mathrm{ products/reactants }= \ce{[B]/[A]}$

$K_\mathrm{eq}$will decrease. However, why doesn't $k_\mathrm{f}$ decrease? Since

$K_\mathrm{eq}= k_\mathrm{f}/k_\mathrm{b}$,

how could $k_\mathrm{f}$ increase while $K_\mathrm{eq}$ decrease if they're equal to each other in that equation? Shouldn't $k_\mathrm{b}$increase or $k_\mathrm{f}$ decrease to "=" to the decreased value of $K_\mathrm{eq}$?

  • $\begingroup$ $\Delta\text{K'}_f < \Delta\text{K'}_r$ so $\text{K'}_{eq} < \text{K}_{eq}$ $\endgroup$
    – MaxW
    Commented Feb 19, 2017 at 10:09

1 Answer 1


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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