At $60^{o} C$ , dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one temperature.

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Solution

Dissociation of $N_{2} O_{4}$ , proceeds as,
$N_{2} O_{4} \to 2 N O_{2}$
Mole fractions, dissociated is $50$ is $0.5$ ,
$= \frac{(1 - 0.5)}{(1 + 0.5)} = 0.5 / 1.5 = 0.33$ Therefore, $p = 0.33 \times 1 a t m . = 0.33.$
Similarly for $N O_{2} = 2 \times \frac{0.5}{(1 + 0.5)} = 1 / 1.5 = p = 0.66$
Equilibrium const $K p = p (N O_{2})_{2} / p (N_{2} O_{4}) = (0.66) 2 / (0.33) = 1.33$
We know that, $Δ G = - R T$ ln $K_{p}$
Substituting the values, $= - 8.314 \times 333 \times 2.303 \times (0.1239) = - 768.3$ $k J / m o l$

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