Question

1. Given the values of So given below in J/mol K and the values of ΔHfo...

1. Given the values of So given below in J/mol K and the values of ΔHfo given in kJ/mol, calculate the value of ΔGo 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

ΔHfo (C4H10(g)) = -123

ΔHfo (CO2(g)) = -394

ΔHfo (H2O(g)) = -223

2. A particular reaction has a ΔHo value of -153 kJ and ΔSo of -188 J/mol K at 298 K. Calculate ΔGo at 477 K in kJ, assuming that ΔHo and ΔSo hardly change with temperature

Homework Answers

Answer #1

1. Balanced chemical reaction is

2C4H10 + 13O2   8CO2 + 10H2O

First calculate H for reaction

H0 = fH0 (Product) - fH0 (Reactant)

H0 = [(8vfH0 CO2) + (10fH0 H2O)] - [(2fH0 C4H10) + (13fH0 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 CO2(g)) + (10 S0 H2O(g))] - [2( S0 C4H10(g)) + 13( S0 O2(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

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