At what temperature, in °C, is a certain reaction at equilibrium if ∆H = +88.0 kJ/mol and ∆S = +170.2 J/mol ･ K?

Answer 1

Use the equation ∆G = ∆H - T∆S and set ∆G = 0 since that is the value at equilibrium and solve for T.

So, we have...

0 = 88.0 kJ/mol - T(0.1702 kJ/molK) note to change units of ∆S to be the same as those for ∆H

-88 kJ/mol = - 0.1702 kJ/molK (T)

T = 517K

Converting to ºC we have...

517 - 273 = 244ºC