Below are the values for ΔH and ΔS for two reactions.

Question

Below are the values for ΔH and ΔS for two reactions.

Reaction A: ΔH° = 10.5 kJ/mol ΔS° = 30 J/K*mol

Reaction B: ΔH° = 1.85 kJ/mol ΔS° = -113 J/K*mol

a) From the values of ΔH and ΔS, calculate ΔG^o and determine if each of the reactions would be spontaneous at 25^oC.

b) If either of the reactions is nonspontaneous at 25^oC, at what temperature might it become spontaneous?

Answer 1

∆G = ∆H - T∆S

For a reaction to be spontaneous, ∆G must be negative.

(A) ∆G = 10.5 kJ/mol - 298K * 0.030 kJ/Kmol (note to change units of ∆S to be consistent with units of ∆H)

∆G = 10.5 kJ/mol - 8.94 kJ/mol = +1.56 kJ/mol

Since this is a positive value, this reaction would not be spontaneous @ 25ºC

(B) ∆G = 1.85 kJ/mol - 298K * -0.113 kJ/Kmol

∆G = 1.85 kJ/mol - (-33.67)

∆G = 35.52 kJ/mol

Since this is a positive value, this reaction would not be spontaneous @ 25ºC

To determine temperature to become spontaneous, set ∆G = 0 and solve for T

(A) 0 = 10.5 kJ/mol - T(0.03 kJ/Kmol)

T = 10.5 kJ/mol / 0.03 kJ/Kmol)

T = > 350K = 77ºC

(B) 0 = 1.85 kJ/mol - T(-0.113 kJ/Kmol)

T = -1.85 kJ/mol / 0.113 kJ/Kmol)

T = > -16.4K = -289ºC

(Be sure to check the math on all parts)