Equilibrium, temperature change, and prediction in constant volume 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 

$$\ce{3A + B -&gt; C + D}$$

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 $Q_\mathrm{p}(2T)&lt;K_\mathrm{p}(2T)$, 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)
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 $Q_{\text{p}}^{(2T)} &lt; K_\text{p}^{(2T)}$, that is:

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

Simplifying:

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

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:

$$\frac{1}{4} &lt; \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H &gt; - 2RT \ln{4}$$

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:

$$1 &lt; \exp \left(\frac{\Delta H}{2RT}\right)$$
$$\Delta H &gt; 0$$

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 $\Delta H &gt; 0$, i.e. endothermic reactions.