Using the appropriate thermodynamic values in the table below, calculate the lowest temperature at which the reaction is spontaneous under standard conditions.

Question

A mixture of CO(g) and H₂(g) is produced by passing steam over hot charcoal:

Using the appropriate thermodynamic values in the table below, calculate the lowest temperature at which the reaction is spontaneous under standard conditions.

H₂O(g) + C(s) --> H₂ (g) + CO(g)

Compound	ΔGf (kJ/mol)	ΔHf (kJ/mol)	S (J/mol • K)
H2O(g)	–228.6	–241.8	188.8
H2O(ℓ)	–237.2	–285.8	69.9
C(s)	0.0	0.0	5.7
H2(g)	0.0	0.0	130.6
CO(g)	–137.2	–110.5	197.7
O2(g)	0.0	0.0	205.0

Answer 1

This is just like your previous question. Wish I had seen this first, so now I have to do it all over again.

H₂O(g) + C(s) --> H₂ (g) + CO(g)

∆Gº = ∆Hº - T∆S

∆Hºrxn = ∑∆Hºf products - ∑∆Hºf reactants = (0 + -110.5) - (-241.8 + 0) = -110.5 + 241.8 = 131.3 kJ/mol

∆Sºrxn = ∑∆Sºf products - ∑∆Sºf reactants = (130.6 + 197.7) - (188.8 + 5.7) = 328.3 - 194.5 = 133.8 J/Kmol

Setting ∆Gº = 0 for a reaction to be spontaneous, we now solve for T (temperature):

0 = ∆Hº - T∆Sº

T = ∆Hº / ∆Sº (remember to change units of ∆Sº to kJ/Kmol)

T = 131.3 kJ/mol / 0.1338 kJ/Kmol

T = 981K

T = 981K - 273 = 708ºC

(be sure to check all of the math)