Given $Δ_{f} G^{\circ} (NO X_{2}) = 51.3 kJ mo l^{- 1}$ and $Δ_{f} G^{\circ} (N X_{2} O X_{4}) = 97.8 kJ mo l^{- 1}$ at $298 K$ , consider $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ (take $R = 8.314 J mo l^{- 1} K^{- 1}$ , standard pressure $1 bar$ ). Find the degree of dissociation of $N X_{2} O X_{4}$ at equilibrium when the total pressure is $2.0 bar$ (two decimals).

Numerical answer type, tolerance 0.01.

Answer: 0.13

Δ G^{\circ} = 2 (51.3) - 97.8 = 4.8 kJ mo l^{- 1}

, so

K_{p} = exp (- 4800/ (8.314 \times 298)) = 0.1441

. For degree of dissociation

α

K_{p} = 4 α^{2} P / (1 - α^{2})

, so

α^{2} = K_{p} / (K_{p} + 4 P) = 0.1441/ (0.1441 + 8) = 0.01769

and

α = 0.1330

, about 0.13.