When the temperature of a system at equilibrium changes, LeChatlier's Principle predicts that the system will adjust so as to compensate for the change. If the temperature increases, the system will move in the direction of the endothermic reaction, which will cause the system to cool.

An equation very similar to the Clausius-Clapeyron equation holds for the relationship between the equilibrium constant, temperature and DH for the reaction:

ln(K₂/K₁) = -DH/R*(1/T₂ - 1/T₁)

Example: For the reaction below, K = 0.76 at 800^oC. What is the equilibrium constant for this reaction at 1000^oC?

2SO₂(g) + O₂(g) < = > 2SO₃(g)

Solution:First, we need to compute DH for the reaction. This turns out to be -197.8 kJ/mol. (If you can't remember how to do this, check the web page on computing DH for a reaction.) Once you know this, you can use the van't Hoff equation to solve for K at 1000 C. (Be careful about units: degrees K and kilojoules!)

ln(K₂) - ln(K₁) = -DH/R*(1/T₂ - 1/T₁)

ln(K₂) - ln(0.76) = - (-197.8 kJ)/(0.00831 kJ/mol*k)*(1/(1273 K) - 1/(1073))

ln(K₂) + 0.274 = -3.49

ln(K₂) = -3.75

K₂ = 0.0233

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