Question

Calculate the free-energy change of the following reaction at 352°C and standard pressure. Values in the table are at standard pressure and 25°C.

C2H4 + 3O2 --> 2CO2 +2H2O

	ΔHºf,(kJ/mol)	Sºf, J/mol•K	ΔGºf, kJ/mol
C2H4(g)	52.3	219.5	68.1
O2(g)	0	205.0	0
CO2(g)	-393.5	213.6	-394.4
H2O(g)	-241.8	188.7	-228.6

Answer 1

Sol :-

Given chemical reaction is :

C₂H₄ + 3O₂ ---------------> 2CO₂ + 2H₂O

As,ΔH⁰_rxn = [Sum of enthalpy of formation of products] - [Sum of enthalpy of formation of reactants]

= [2 x ΔH⁰_f of CO₂ + 2 x ΔH⁰_f of H₂O] - [ΔH⁰_f of C₂H₄ + 3 x ΔH⁰_f of O₂]

= [2 x (-393.5 KJ/mol + 2 x (-241.8 KJ/mol)] - [52.3 KJ/mol + 3 x 0.0 KJ/mol]

= [- 787 KJ/mol - 483.6 KJ/mol] - 52.3 KJ/mol

= - 1270.6 KJ/mol - 52.3 KJ/mol

= - 1322.9 KJ/mol

Similarly,

As,ΔS⁰_rxn = [Sum of entropy of formation of products] - [Sum of entropy of formation of reactants]

= [2 x S⁰_f of CO₂ + 2 x S⁰_f of H₂O] - [S⁰_f of C₂H₄ + 3 x S⁰_f of O₂]

= [2 x 213.6 J/mol.K + 2 x 188.7 J/mol.K] - [219.5 J/mol.K + 3 x 205 J/mol.K]

= [427.2 J/mol.K + 377.4 J/mol.K] - [219.5 J/mol.K + 615 J/mol.K]

= 804.6 J/mol.K - 834.5 J/mol.K

= - 29.9 J/mol.K

Relationship between ΔG⁰, ΔH⁰ and ΔS⁰ is :

ΔG⁰ = ΔH⁰ - TΔS⁰

ΔG⁰ = - 1322.9 KJ/mol - (273 + 352) K x (- 29.9 x 10^-3 KJ/mol.K)

ΔG⁰ = - 1322.9 KJ/mol + 18.6875 KJ/mol

ΔG⁰ = - 1304.2125 KJ/mol

Hence, ΔG⁰ at 352⁰C = - 1304.2125 KJ/mol