# In an endothermic reaction at equilibrium, what is the effect of raising the temperature?

Dec 17, 2016

Well, the principle of $\text{old Le Chatelier}$ makes a clear prediction.........

#### Explanation:

$\text{For a reaction at equilibrium that is subject to an external}$
$\text{perturbation, the equilibrium will move in a direction so as to }$
$\text{oppose the external perturbation.}$

The important qualifier in the above spray is that $\text{oppose "!=" counteract}$, and we speak of an initial change, an initial movement of the equilibrium in the face of change. Especially with temperature changes, the equilibrium MAY evolve to give other conditions of operation.

And (finally!!) to answer your question. We have the endothermic reaction at equilibrium:

$A + B + \Delta r i g h t \le f t h a r p \infty n s C + D$

Clearly, the $\Delta$ symbol represents energy in. If the temperature is raised at equilibrium, the initial response is to try to oppose the perturbation, and the equilibrium does this by going towards the right as written, i.e. by producing greater concentrations of $C$ and $D$ utilizing the added energy.

I am sorry to be so long-winded, but there are a lot of assumptions and qualifications to be made in this treatment.