Question

Calculate the standard-state entropy for the following reaction:

1 CH4(g) + 2 O2(g) ⟶ 1 CO2(g) + 2 H2O(l)

(If applicable, coefficients of one have been included for clarity.)

The standard entropy values are given in the table.

Formula	S∘ J/(K⋅mol)
CO2(g)	214
H2O(l)	189
CH4(g)	186
O2(g)	205

Answer 1

Answer –

We are given, reaction –

1 CH₄(g) + 2 O₂(g) ------> 1 CO₂(g) + 2 H₂O(l)

We know the formula for the calculating the standard-state entropy

ΔS^orxn = sum of S^o product – sum of S^o reactant

ΔS^orxn = [ S^oCO₂(g) + 2* S^o H₂O(g)] – [ ΔS^o CH₄(g) + 2*S^o O₂(g)]

            = (214 + 2*189) - (186 + 2*205)

            = - 4 J/K

Standard-state entropy is -4.0 J/K for this above reaction.