For N2O4(g) 2NO2(g), Delta H deg = +57.2 kJ mol^-1 and Delta S deg = +175.7 J K^-1mol^-1 (take both as constant). Starting with pure N2O4 at 298 K and a total equilibrium pressure of 2.00 bar, find the degree of dissociation alpha (three decimals). Use R = 8.314 J mol^-1K^-1 and standard pressure 1 bar.

JEE Advanced levelNumerical answerDifficulty 3 of 3td106

For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $Δ H^{\circ} = + 57.2 kJ mo l^{- 1}$ and $Δ S^{\circ} = + 175.7 J K^{- 1} mo l^{- 1}$ (take both as constant). Starting with pure $N X_{2} O X_{4}$ at $298 K$ and a total equilibrium pressure of $2.00 bar$ , find the degree of dissociation $α$ (three decimals). Use $R = 8.314 J mo l^{- 1} K^{- 1}$ and standard pressure $1 bar$ .

Numerical answer type, tolerance 0.002.

Δ G^{\circ} = 57200 - 298 (175.7) = 4841 J

, so

K_{p} = exp (- Δ G^{\circ} / RT) = exp (- 1.954) = 0.1417

. For

α

dissociated (total moles

1 + α

K_{p} = \frac{4 α ^{2} P}{1 - α ^{2}}

, giving

α^{2} = K_{p} / (4 P + K_{p}) = 0.1417/ (8.00 + 0.1417)

and

α = 0.1319