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When temperature is increased, then according to Le Chatelier's principle the reaction will move in the direction where energy is absorbed. I can understand this, but there is also another fact that the $K$ value depends on temperature. So, if my temperature changes, then $K$ value should also change (van 't Hoff equation).

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  • $\begingroup$ So does k value changes ? $\endgroup$ – Yash Dec 14 '19 at 12:26
  • $\begingroup$ Find out answer in here. $\endgroup$ – Mathew Mahindaratne Dec 14 '19 at 15:02
  • $\begingroup$ Pls can u give a simple explanation, I am unable to understand $\endgroup$ – Yash Dec 14 '19 at 15:35
  • $\begingroup$ @Yash The K remains the same if and only if T remains the same. . added notes to the answer, together with the link to related Q/A. $\endgroup$ – Poutnik Dec 17 '19 at 5:26
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van 't Hoff equation

$$\frac{\mathrm d}{\mathrm dT} \ln K_\mathrm{eq} = \frac{\Delta H^⦵}{RT^2}$$

= Le Chatelier's principle in a particular context of temperature and chemical reaction equilibrium.

What the Le Chatelier's principle says qualitatively as a general principle, the van 't Hoff equation says quantitatively in context of temperature change for chemical reactions.

Van't Hoff equation is the way, how Le Chatelier's principle applies on equilibrium constants of chemical reactions at different temperature.

By other words, attempts to apply both, van 't Hoff equation and Le Chatelier's principle effectively mean applying Le Chatelier's principle twice.

Generally, all equilibrium constants are temperature dependent, it is not limited to chemical reactions.

$K$ is an equilibrium constant and does change with temperature, quantitatively according to the van't Hoff equation.

Read the Wikipedia articles about the both following the above links.


Consider the Haber-Bosch ammonia production

$$\ce{N2 + 3 H2 <=> 2 NH3}$$

The equilibrium constant is with application of the equilibrium partial pressures:

$$K=\frac{(p_{\ce{NH3}})^2 }{ p_{\ce{N2}}\cdot (p_{\ce{H2}})^3 } $$

The reaction quotient is with application of the current partial pressures:

$$Q=\frac{(p_{\ce{NH3}})^2 }{ p_{\ce{N2}}\cdot (p_{\ce{H2}})^3 } $$


Now, application of Le Chatelier's principle at higher pressure of the mixture at the same temperature:

  1. $K$ is the same, as $T$ is the same.
  2. $Q$ is at the same $T$ and composition lower (Check that $\frac{2^2}{2 \cdot 2^3} < \frac{1^2}{1 \cdot 1^3}$).
  3. Products are more preferred for $Q$ to reach $K$.
  4. The reaction decreases pressure, as products have lower pressure for the same volume, acting against increasing the pressure.
  5. As T remains the same, K remains the same. Q changes temporarily due the change of the pressure, but then it is approaching the K value back again.

Now, application of Le Chatelier's principle at higher temperature with the same pressure of the mixture:

  1. $K$ is lower as $\Delta H_\mathrm{r} < 0$, according to van 't Hoff equation.
  2. As result, reagent side of nitrogen and hydrogen is more preferred.
  3. The forward reaction is exothermic, while backward reaction is endothermic.
  4. Being more pushed toward the endothermic side, it partially consumes thermal energy provided by increasing temperature.
  5. The Q is initially the same as before T changes. But K changes, as T changes. Q is then approaching the new K value at the changed temperature.

See the related question with some nice simple charts.

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