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For the following reaction: $$\ce{W_{(s)} + 4 Br_{(g)} <=> WBr4_{(g)}}$$

At 900 K, $K_\mathrm{p}=100$ and at 2800 K, $K_\mathrm{p}=5$.

At higher temperatures, the equilibrium has shifted towards the reactants since $K_\mathrm{p}$ is significantly smaller. However, how do we determine if a reaction is endothermic or exothermic from this information? Here are my thoughts, If the reaction was endothermic, an increase in temperature would push the equilibrium to products to counteract the increase in heat. However, the equilibrium has been pushed to the reactants thus signifying the reaction is exothermic. But this assertion is on the assumption that the forward reaction is exothermic. Any thoughts?

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Think about it this way. In an exothermic reaction that is at equilibrium the forward reaction has a lower activation energy. So, if you increase T the forward reaction rate increases but the reverse rate increases even more. That's because the rates of reactions with higher activation energy are more sensitive to temperature. With the reverse rate higher more reactants will be made until the concentration reactants reaches a point where the rate of the forward reaction equals the reverse rate.

In an endothermic reaction increasing T increases the forward rate more than the reverse, producing more product until the increased product concentration causes the reverse rate to equal the forward rate.

The equilibrium constant can be thought of as a ratio of products to reactants needed to equalize the rates of two reactions with different activation energies.

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  • $\begingroup$ Just to be clear, you're referring to the link between chemical equilibrium and chemical kinetics where $K_{c}=\dfrac{k_{f}}{k_{r}}$. Since $K_{r}$ increases more than $K_{f}$ at higher temperature, $K_{c}$ decreases. So since the $K_{c}$ decreased at higher temperature, the reaction is exothermic. Correct? $\endgroup$ – Amuna Sep 1 '14 at 4:16
  • $\begingroup$ That is correct. $\endgroup$ – Brinn Belyea Sep 1 '14 at 4:46
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I think you're looking for the van't Hoff equation. Since you have the equilibrium constants for two temperatures you can integrate the equation and get $\Delta H^{0}$ from it

\begin{align} \frac{\mathrm{d} \ln K}{\mathrm{d} T} &= \frac{\Delta H^{0}}{R T^{2}} \\ \int\limits_{\ln K(T_{0})}^{\ln K(T_{1})}\mathrm{d} \ln K &= \frac{\Delta H^{0}}{R} \int\limits_{T_{0}}^{T_{1}} \frac{1}{T^{2}} \mathrm{d} T \\ \ln \frac{K(T_{1})}{K(T_{0})} &= \frac{\Delta H^{0}}{R} \left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right) \\ \Delta H^{0} &= \frac{R \ln \frac{K(T_{1})}{K(T_{0})}}{\left( \frac{1}{T_{0}} - \frac{1}{T_{1}}\right)} \end{align}

which will enable you to decide whether the reaction is exothermic or endothermic.

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