Question

Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? Δ G0f(kJ/mol) -110.9 87.6 51.3 -237.1

Answer 1

The equilibrium constant K = 1.15*10^-9

Step-by-step explanation:

Given:

ΔG°f(HNO3) = -110.9 kj/mol

ΔG°f(NO) = 87.6 kj/mol

ΔG°f(NO2) = 51.3 kj/mol

ΔG°f(HNO3) = -237.1 kj/mol

To determine:

The equilibrium constant (K) for the given reaction

Calculation:

The chemical reaction is:

`2HNO3(aq) + NO(g) \rightarrow  3 NO2(g) + H2O(l)`

The equation that relates the standard free energy change ΔG° to the equilibrium constant K is:

`\Delta G^(0)= -RTlnK`

(or)
`K = e^{-\Delta G^(0)/RT}----(1)`

where R = gas constant = 8.314 J/mol-K

T = temperature in Kelvin

`\Delta G^(0)=\sum n_(p)\Delta G^(0)f(products)-\sum n_(r)\Delta G^(0)f(reactants)`

where n(p) and n(r) are the number of moles of the products and reactants respectively

Therefore for the given reaction:

`\Delta G^(0)=[3\Delta G^(0)f(NO2)+3\Delta G^(0)f(H2O)]-[2\Delta G^(0)f(HNO3)+1\Delta G^(0)f(NO)]`

Substituting the given values for ΔG°f:

`\Delta G^(0)=[3\Delta G^(0)f(51.3)+3\Delta G^(0)f(-237.1)]-[2\Delta G^(0)f(-110.9)+1\Delta G^(0)f(87.6)]`

ΔG° = + 51 kJ

Substituting the calculated ΔG° in equation (1) at T = 298 K gives:

`K = e^(-\51000/8.314*298)=1.15*10^(-9)`