Question

Calculate the temperature (K) at which the equation as written is at equilibrium. Express your answer to three significant figures.

CaO(s) + CO2(g) → CaCO3(s)

∆G°f (kJ/mol)
CaO(s)-604.0
CO2(g)-394.4
CaCO3(s)-1128.8

∆H°f (kJ/mol)
CaO(s)-635.1
CO2(g)-393.5
CaCO3(s)-1206.9

∆H°f (kJ/mol)
CaO(s)39.7
CO2(g)213.7
CaCO3(s)92.9

Here are the values I got:

∆G°rxn = -130.4 kJ
∆H°rxn = -178.3 kJ
∆S°rxn = -160.5 kJ

Solving for T,

T= (∆H-∆G)/∆S

plugging in the values I get: 298K

however the answer returned states that it's 1110K.

Answer 1

You have made two errors.
a. dS is given in J/mol and not kJ/mol so change that to 0.1605 kJ/mol

b. Equilibrium is when dG = 0
dG = dH - TdS
0 = dH - TdS
TdS = dH and
T = dH/dS = -178.3/-0.1605 = 1110.9 K so to three s.f. that would be 1.11E1 K.