Question

Consider the following reaction at 173 K: 2 N2O (g) → 2 N2 (g) + O2 (g) In one of your laboratory experiments, you determine the equilibrium constant for this process, at 173 K, is 6.678E+57. You are given a table of data that indicates the standard heat of formation (ΔHoform) of N2O is 82.0 kJ/mol. Based on this information, what is the standard entropy change (ΔSorxn) for this reaction at 173 K? ΔSorxn(J/K)=

Answer 1

2N2O(g) <-------> 2N2(g) + O2(g)

First we have to determine ∆H°rxn

∆H°rxn = ∆H°f(products) - ∆H°f(reactants)

Formation enthalpy for elements in their standard state is Zero

So , ∆H°f (products) =0

∆H°f (reactants) = 2× ∆H°f(N2O)

= 2× 82kJ/mol

= 164kJ/mol

Therefore,

∆H°rxn = 0 - 164kJ/mol

= - 164kJ/mol

We know that

Standard free energy change, ∆G° = - RTlnK

Where,

R = gas constant , 8.314(J/mol K)

T = temperature, 173K

K = equilibrium constant ,6.678×10^57

So,

∆G°rxn = - (8.314(J/mol K) × 173K × ln( 6.678×10^57))

= - ( 8.314(J/K mol) × 173K × 2.303 × log(6.678×10^57))

= -191.54kJ/mol

We know that,

∆G°rxn = ∆H°rxn - T∆S°rxn

∆S°rxn =( ∆H°rxn - ∆G°rxn)/T

= (-164kJ - (-191.54kJ))/173K

= 27.54kJ/173K

= 159.2 J/k