Talking about an exothermic reaction in equilibrium it is said that heat is on the product side, so according to Le Chatelier principle, the reaction should move backward to counteract the increase in the number of products. But it is also stated that in an exothermic reaction when the temperature increases, the value of equilibrium constant decreases, which means the product is less than reactant, then shouldn't the reaction move in the forward direction? Please help me with wherever I'm going wrong.

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    $\begingroup$ "Equilibrium constant decreases" means "reaction moves backward". $\endgroup$ – Ivan Neretin Feb 28 '18 at 11:21

As per Le Chatelier's principle, when the temperature is increased, for an exothermic reaction , the reaction shifts in the backward direction. This is clear from the van't Hoff's equation:

$$ \log\left( \frac{K'_\mathrm{p}}{K_\mathrm{p}}\right) = \frac{\Delta H^0}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

where $T_1$ corresponds to $K_p$ and $T_2$ corresponds to $K'_p$ . Also $T_2 > T_1$.

This is because $K'_\mathrm{p}$ becomes less than $K_\mathrm{p}$. Hence the reaction goes in the reverse direction.


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