In an exothermic reaction, the value of equilibrium constant increases with the rise of temperature because,

$\Delta H < 0$ ( exothermic reactions)

$\Delta G = \Delta H - T\Delta S$

Now if $\Delta H $ and $T\Delta S$ are negative and positive respectively, the overall value of $\Delta G$ is also negative. (For positive entropy)

$\Delta G = -RT\ln K$ (where $K$ is equilibrium constant)

Since $\Delta G$ is negative, the value of $RT\ln K$ becomes positive and as $\Delta G$ increases equilibrium constant increases.

Is this logic right?


1 Answer 1


First of all, How can you assume that, $\Delta S$ is positive in any exothermic reaction? Secondly, the relation you have written $\Delta G = -RT\ln K_\text{eq}$ is not at all correct. The correct relation is $\Delta G = \Delta G^\circ + RT\ln Q_c$ (or $Q_p$ if the reactants are gaseous). Now at equilibrium $\Delta G= 0$ and $Q = K$. So, the correct relation is

$$\Delta G^\circ = -RT\ln K_\text{eq}$$

And, there are obviously differences between $\Delta G$ and $\Delta G^\circ$. So, the reasoning you have given is not totally correct.

The correct reasoning comes from Van't Hoff Equation, which states $$\frac{\mathrm d}{\mathrm dT}(\ln K_{\text{eq}}) = \frac{\Delta H}{RT^2}$$

Integrating both sides, we have

$$\int_{K_1}^{K_2}\mathrm d(\ln K_{\text{eq}} ) = \int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

So, we have $$\ln\frac{K_2}{K_1}= \frac{\Delta H}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$ Thus we see that increasing the temperature from $T_1$ to $T_2$, the equilibrium constant changes. Here $\Delta H < 0$ (exothermic reaction) and $\displaystyle\frac{1}{T_1}- \frac{1}{T_2} > 0$ so, $\displaystyle \ln\left(\frac{K_2}{K_1}\right) < 0$, which means $\displaystyle {\frac{K_2}{K_1}< 1}$, which implies $K_2 < K_1$. So, the equilibrium constant decreases.

Note: Without any mathematical equation also, the same result can be deduced. In an exothermic reaction, the products are formed along with heat production, which is the definition of exothermic reaction. That means, temperature of the system will increase due to this heat production. Now if you increase the temperature, the system will obviously want to go in the direction where this temperature minimisation is possible, which is actually the backward reaction. So, the reaction shifts backwards, and thus the equilibrium constant decreases as now the product concentration has become lesser and reactant concentration has become larger.

  • $\begingroup$ I'm sorry. I actually meant delta G^0 only.To answer your question, I am assuming Entropy to be positive.So the variation of the equilibrium constant with Delta G cannot be sought from Delta G^0 = -RTlnK? $\endgroup$
    – J_B892
    Mar 24, 2018 at 11:01
  • $\begingroup$ Actually if you look into the equation $\ce {$\Delta $G ^0 = -RTln K_{eq}}$, The LHS is constant for a reaction and independent of Temperature as well. So if you increase the temperature, the RHS says the equilibrium constant should always decrease,irrespective of reaction being exothermic or endothermic, which is not the case of always. So, better to avoid that logic. $\endgroup$
    – Soumik Das
    Mar 24, 2018 at 11:24
  • 2
    $\begingroup$ The left hand side is not independent of temperature. The Van't Hopf equation is an indication of this. $\endgroup$ Mar 24, 2018 at 12:25
  • $\begingroup$ $\Delta$$G^0$ is independent of temperature. Note that it is $G^0$ , not $G$. See this chemistry.stackexchange.com/a/91256/58497 $\endgroup$
    – Soumik Das
    Mar 24, 2018 at 12:40

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