For a particular reaction at 132.0 °C, Δ𝐺=−64.76 kJ , and Δ𝑆=676.54 J/K . Calculate Δ𝐺 for this reaction at −77.1 °C.

Answer 1

∆G = ∆H - T∆S

We can use this equation to find ∆H:

∆H = ∆G + T∆S

∆G = -64.76 kJ/mol

T = 132 + 273 = 405K

∆S = 676.54 J/Kmol = 0.677 kJ/Kmol (change units to match those of ∆G)

∆H = -64.76 + (405)(0.677) = -64.76 + 274

∆H = + 209.4 kJ/mol

Now we can use this to find ∆G at -77.1ºC (196K)

∆G = ∆H - T∆S

∆G = 209.4 kJ/mol - (196K)(0.677 kJ/Kmol)

∆G = 209.4 - 132.6

∆G = 76.8 kJ/mol