Question

CO2

H2O(l)

C6H12O6(s)

O2(g)

∆H (kJ/mol)

-393.5

-285.8

-1273.3

S° (J/mol ∙ K)

213.6

69.9

209.2

205.0

Instruction: Calculate the ∆H°, ∆S°, and ∆G°. Is the reaction spontaneous at standard condition?

Answer 1

$6CO_{2(g)} +6H_2O_{(l)} \to C_6H_{12}O_{6(s)} +6O_{2(g)}$

$∆H° = \textsf{Enthalpy of formation of }C_6H_{12}O_6 ​ = 6(−393.5)+6(−285.9)−(−2816)=−1260\ KJmol ^{−1}\\ \quad\\ ∆S°= 6(213.6)+6(69.9)−(209.2)-6(205.0)=261.8J/mol.K$