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 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

-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

43.452

step-by-step explanation:

284 times .153 = 43.452