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
1. Balanced chemical reaction is
2C4H10 +
13O2
8CO2 + 10H2O
First calculate H for
reaction
H0 =
fH0
(Product) -
fH0
(Reactant)
H0 =
[(8
vfH0
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
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 - T
S
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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