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avanturin [10]
4 years ago
11

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 c

ombustion of 1 mole of propane to form carbon dioxide and gaseous water at 298 K. S (C3H8(g)) = 271 S (O2(g)) = 204 S (CO2(g)) = 214 S (H2O(g)) = 184 ΔHfo (C3H8(g)) = -100 ΔHfo (CO2(g)) = -398 ΔHfo (H2O(g)) = -226
Chemistry
1 answer:
Fantom [35]4 years ago
7 0

Answer:

-2024 kJ

Explanation:

The combustion of propane is

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

And, ΔG° = ΔH° - TΔS°

Where T is the temperature in K.

ΔH° = ∑n*H°f, reactants - ∑n*H°f, products

Where n is the number of moles in the stoichiometry reaction. H°f, O₂(g) = 0 because it's a substance formed by only one element.

ΔH° = [4*(-226) +3*(-398)] - [-100] = -1998 kJ

ΔS° = ∑n*S°, reactants - ∑n*S°, products

ΔS° = [4*(184) + 3*214] - [5*204 + 271] = 87 J/K = 0.087 kJ/K

So

ΔG° = -1998 - 298*0.087

ΔG° = -2024 kJ

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