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My professor gave us 2 different ways to calculate this problem. I did it using the Clausis-Clapeyron equation.

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However, I don't quite get the second method because I thought that if the change in enthalpy and entropy of formation are in standard states, the temperature (T) in the Gibbs Free Energy equation meant to be 298K? And then making the temperature in the second equation 1100K, to find the equilibrium constant?

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In fact, Gibbs free energy is not defined at exact temperature. It is defined for standard states of all compounds in equation. So you really can use this solution of problem. However, it is a very simplified solution, because you assume that standard entropy and enthalpy do not change with temperature. This is not entirely correct assumption but if you took change of enthapy and entropy with temperature into account, it would be much more complicated. You can find information about Gibbs free energy here http://www.chem1.com/acad/webtext/thermeq/TE4.html#3A

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    $\begingroup$ To add to what you said, if the OP used the Clausius Clapeyron equation as his first method, he must also have assumed that the enthalpy change did not vary with temperature. The two methods are mathematically equivalent, both if the temperature dependence (of the entropy change and enthaply change) is neglected and if the temperature dependence is included. $\endgroup$ – Chet Miller May 22 '16 at 11:50

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