For a particular reaction at 233.0 °C, Δ𝐺=−449.05 kJ , and Δ𝑆=957.70 J/K .

Question

Calculate Δ𝐺 for this reaction at −58.6 °C.

Δ𝐺=_______KJ

Answer 1

∆G = ∆H - T∆S

∆G = -449.05 kJ

∆S = 957.70 J/K = 0.95770 kJ/K

T = 233C + 273 = 506K

∆H = ?

Solving for ∆H we have..

-449.05 kJ = ∆H - (506K)(0.95770 kJ/K) = ∆H - 484.60

∆H = 35.55 kJ

Assuming that ∆H and ∆S are not significantly affected by temperature, we can now solve for ∆G@ -58.6ºC

∆G = ∆H - T∆S

∆G = ?

∆H = 35.55 kJ

∆S = 0.9577 kJ/K

T = -58.6C + 273 = 214.4K

Solving for ∆G we have ...

∆G = 35.55 kJ - (214.4K)(0.9577 kJ/K)

∆G = 35.55 kJ - 205.33 kJ

∆G = -169.78 kJ