At 60°C, dinitrogen tetroxide is 50% dissociated. Calculate the stand

1.	At 60°C, dinitrogen tetroxide is 50% dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
Answer» N₂O₄(g) ⇌ 2NO₂(g) If N₂O₄ is 50% dissociated, the mole fraction of both the substances is given by, \(X_{N_2O_4}\) = \(\frac{1-0.5}{1+0.5}\) \(X_{NO_2}\) = \(\frac{2\times0.5}{1+0.5}\) \(P_{N_2O_4}\) = \(\frac{0.5}{0.5}\)x 1 atm \(P_{NO_2}\) = \(\frac{1}{1.5}\) x 1 atm The equilibrium constant K_p is given by, K_p = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\) = \(\frac{\big(\frac{1}{1.5}\big)^2}{\frac{0.5}{1.5}}\) = 1.33 \(\therefore\) ∆_fG^Θ = −RT In K_p = −2.303 RT logK_p = −2.303 × (8.314 J K⁻¹ mol⁻¹) × (333K) log 1.33 = −2.303 × 8.14 JK⁻¹ mol⁻¹ × 333K × 0.1239 = −763.8 kJ mol⁻¹

At 60°C, dinitrogen tetroxide is 50% dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Answer»

N₂O₄(g) ⇌ 2NO₂(g)

If N₂O₄ is 50% dissociated, the mole fraction of both the substances is given by,

\(X_{N_2O_4}\) = \(\frac{1-0.5}{1+0.5}\)

\(X_{NO_2}\) = \(\frac{2\times0.5}{1+0.5}\)

\(P_{N_2O_4}\) = \(\frac{0.5}{0.5}\)x 1 atm

\(P_{NO_2}\) = \(\frac{1}{1.5}\) x 1 atm

The equilibrium constant K_p is given by,

K_p = \(\frac{(P_{NO_2})^2}{P_{N_2O_4}}\)

= \(\frac{\big(\frac{1}{1.5}\big)^2}{\frac{0.5}{1.5}}\)

= 1.33

\(\therefore\) ∆_fG^Θ = −RT In K_p

= −2.303 RT logK_p

= −2.303 × (8.314 J K⁻¹ mol⁻¹) × (333K) log 1.33

= −2.303 × 8.14 JK⁻¹ mol⁻¹ × 333K × 0.1239

= −763.8 kJ mol⁻¹

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