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

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%