Question

1. Given the values of S^o given below in J/mol K and the values of ΔH_f^o given in kJ/mol, calculate the value of ΔG^o in kJ for the combustion of 1 mole of butane to form carbon dioxide and gaseous water at 298 K.

S (C₄H₁₀(g)) = 273

S (O₂(g)) = 208

S (CO₂(g)) = 214

S (H₂O(g)) = 189

ΔH_f^o (C₄H₁₀(g)) = -123

ΔH_f^o (CO₂(g)) = -394

ΔH_f^o (H₂O(g)) = -223

2. A particular reaction has a ΔH^o value of -153 kJ and ΔS^o of -188 J/mol K at 298 K. Calculate ΔG^o at 477 K in kJ, assuming that ΔH^o and ΔS^o hardly change with temperature

Answer 1

1. Balanced chemical reaction is

2C₄H₁₀ + 13O₂   8CO₂ + 10H₂O

First calculate H for reaction

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(8v_fH⁰ CO₂) + (10_fH⁰ H₂O)] - [(2_fH⁰ C₄H₁₀) + (13_fH⁰ O₂)]

= [(8 (-394)) + (10 (-223))] - [2(-123) + (13 (0)]

= [-5382] - [-246]

H⁰ = -5136 KJ/2mol

H⁰ for reaction = -5136/2 = -2568KJ/mol

now calculate S for reaction

S⁰  =   S⁰ (Product) -   S⁰ (Reactant)

S⁰ = [8( S⁰ CO_2(g)) + (10 S⁰ H₂O_(g))] - [2( S⁰ C₄H_10(g)) + 13( S⁰ O_2(g))]

= [ 8(214) + (10 (189))] - [2(273) + 13(208)]

= [3602] - [3250]

S⁰ = 352 J/2mol.K

S⁰ = 176 J/mol.K

now calculate G⁰

G⁰ = H - TS

Where, H = -2568 KJ, S = 176 J/K = 0.176 KJ/K , T = 298 K

Substitute the value in above equation

G⁰ = [-2568] - [(0.176)(298)] = (-2568) - (52.448) = -2620.448 KJ

G⁰ for reaction = -2620.448 KJ