$$\ce{C_{(s)} + H_2O_{(g)} <=> CO_{(g)} + H_{2(g)}}$$

$$\Delta H = +131\ \mathrm{kJ}$$

If the reaction is in a state of equilibrium under standard conditions, what can be done to increase the value of $K_\text{eq}$?

My question is would you have to increase or decrease the temperature to increase $K_\text{eq}$?

I know the basic principle that you have to decrease the temperature to increase $K_\text{eq}$, but in my notes that was for $\ce{A +B<=>C}$ and the reaction above is not in that form, Instead it has two products. Would this still apply?

  • 1
    $\begingroup$ For exothermic reactions, you decrease the temperature to increase $K$ and vice versa. This holds good in all stoichiometries. $\endgroup$ Jan 4 '14 at 5:02

You can easily see how the equilibrium constant varies with temperature using the Van’t Hoff equation: $$\frac{\mathrm{d}\ln{K_\text{eq}}}{\mathrm{d}T}=\frac{\Delta H^\circ}{R{T^2}}$$

Rearranging the equation a little bit, we get that: $$\mathrm{d}\ln K_\text{eq} = \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$

Now we can integrate each side of this equation: $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \int_{T_1}^{T_2} \frac{\Delta H^\circ}{RT^2}\mathrm{d}T$$ Because $\Delta H^\circ$ and $R$ are constants, this will be equivalent to : $$\int_{K_1}^{K_2} \mathrm{d}\ln K = \frac{\Delta H^\circ}{R} \int_{T_1}^{T_2} \frac{1}{T^2}\mathrm{d}T$$ Solving the integral we will get this : $$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)$$

For $\Delta H^\circ > 0$ we have two cases: when $T_2>T_1$ and $T_2<T_1$.
If $T_2>T_1$ then $\frac{1}{T_2}<\frac{1}{T_1}$ and hence $\frac{1}{T_2} - \frac{1}{T_1} < 0$ . Knowing that $\Delta H^\circ>0$, this means that $\frac{-\Delta H^\circ}{R}$ will also be negative, and the product of these 2 will be positive: $$\frac{-\Delta H^\circ}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)>0$$ And from the Van’t Hoff equation we get that : $$\ln\left(\frac{K_2}{K_1}\right)>0 \Leftrightarrow \frac{K_2}{K_1} > 1 \Leftrightarrow K_2 > K_1$$ i.e. The equilibrium constant increases as the temperature is increased.
Using the same principle for $T_2 < T_1$, we get that $K_2 < K_1$


As we know: $$K=Ae^{-\Delta H/RT}$$ Now as this is a monotonously increasing function in the variable T for the positive value of $\Delta H$. So you need to increase temperature to increase the equilibrium constant.


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