When it's constant volume, change of $T$ (Kelvin temperature) in a reaction equilibrium system won't change the volume. Let's say we change the temperature from $T$ to $2T$ of this reaction at equilibrium

$$\ce{3A + B -> C + D}$$

Because of the change of temperature, $K_\mathrm{p}$ will change, depending on whether the reaction is exo- or endothermic. At the same time, because of constant volume and change of temperature, reaction quotient $Q_\mathrm{p}$ at $2T$ is newly formed (originally $Q_\mathrm{p}(T)=K_\mathrm{p}(T)$ in equilibrium turn to $Q_\mathrm{p}(2T)$)

Here's a question. Say, this reaction is exothermic, and $T$ is changed to $2T$. So, $K_\mathrm{p}$ should be lowered, and with $K_\mathrm{p}$ lowered, reaction also proceeds to reverse. This can also be explained by Le Chatelier's principle.

But when it is at constant volume, $Q_\mathrm{p}$ could newly form simultaneously. And if $Q_\mathrm{p}(2T)<K_\mathrm{p}(2T)$, the reaction could proceed forward even though it's exothermic with the increase in temperature.

I've done some math and it seems it only depends on how big $\Delta H$ is. So can that really happen? Or am I missing something? If what I'm saying is right, is this an expception of the Le Chatelier's principle?

(I've explained everything here, but I can explain with an example if anything is still unclear)

  • $\begingroup$ I don't understand what you mean by "$Q_\text{p}$ could newly form simultaneously". $\endgroup$
    – ralk912
    Commented Mar 15, 2018 at 17:34
  • $\begingroup$ I think the problem with this particular case is that a change in temperature at constant volume will also increase the total pressure, thus according to Le Chatelier's, the equilibrium will shift towards the side with less moles of gas (products in this case). Depending on numbers, this effect might be more predominant than the reverse shift for it being an exothermic reaction. $\endgroup$
    – ralk912
    Commented Mar 15, 2018 at 17:56

1 Answer 1


For this particular scenario, the system in equilibrium satisfies (assuming ideal gases):

$$K_{\text{p}}^{(T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}}$$

The equilibrium constant at temperature $2T$ can be calculated as:

$$K_\text{p}^{(2T)} = K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right)$$

After increasing the temperature at constant volume, a new total pressure, $p_\text{tot}^{(2T)}$ is attained. Since the molar fractions are not changed at the beginning, the reaction quotient is:

$$Q_{\text{p}}^{(2T)} = \frac{p_\text{C} p_\text{D}}{p_{\text{A}}^3 p_\text{B}} = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}}$$

So, for the reaction to proceed forward, we want $Q_{\text{p}}^{(2T)} < K_\text{p}^{(2T)}$, that is:

$$\frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(2T)\,2}}} < K_\text{p}^{(T)} \exp \left(\frac{\Delta H}{2RT}\right) = \frac{x_\text{C} x_\text{D}}{x_{\text{A}}^3 x_\text{B}}\cdot\frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$


$$\frac{1}{{p_\text{tot}^{(2T)\,2}}} < \frac{1}{{p_\text{tot}^{(T)\,2}}} \exp \left(\frac{\Delta H}{2RT}\right)$$

But $\displaystyle p_\text{tot}^{(T)} = \frac{n_\text{tot}RT}{V}$ and $\displaystyle p_\text{tot}^{(2T)} = \frac{2n_\text{tot}RT}{V}$, so:

$$\frac{1}{4} < \exp \left(\frac{\Delta H}{2RT}\right)$$ $$\Delta H > - 2RT \ln{4}$$

As long as $\Delta H$ satisfies this relationship (which will always imply a range of exothermic reactions since $T$ is always positive), this reaction will proceed when increasing the temperature at constant volume.

Notice that if this was a reaction where $\Delta n = 0$, the $p_\text{tot}$ cancels out and we get:

$$1 < \exp \left(\frac{\Delta H}{2RT}\right)$$ $$\Delta H > 0$$

Which implies that, when change in number of moles is not a factor to consider, the equilibrium will only shift to the right upon temperature increase at constant volume for reactions with $\Delta H > 0$, i.e. endothermic reactions.

  • $\begingroup$ That is exactly what I was looking forward to. Thank you very very much for your answers and solutions. I am also amazed by chemistry itself again and your skills of handling equations too. $\endgroup$
    – Zillai
    Commented Mar 16, 2018 at 7:18
  • $\begingroup$ Could I ask you one more slight thing? In the situation above, let's say we have q, reaction heat. Since it's constant volume, this q is equal to delta E( internal energy) and delta H is equal to q + delta n(mol)RT ? thanks again... $\endgroup$
    – Zillai
    Commented Mar 16, 2018 at 7:18
  • $\begingroup$ Whether it is at constant volume or not, the relationship $H = U + PV$ is the definition of $H$. So, $\Delta H = \Delta U + \Delta (PV) = \Delta U + \Delta (nRT)$ (for an ideal gas). If the process is at constant temperature, then $\Delta H = \Delta U + RT\Delta n$. If the volume is also constant, then $\Delta U = q$ as you mentioned, so $\Delta H = q + RT\Delta n$. As a side note, notice that the condition of spontaneity at constant temperature and volume is now $\Delta A < 0$, and not $\Delta G < 0$. $\endgroup$
    – ralk912
    Commented Mar 16, 2018 at 7:48

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