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

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Solution

$N_{2} O_{4} (g) ⇋ 2 N O_{2} (g)$
1 mol 0 mol --------------- initially
(1-0.5) mol 2x0.5 mol ----------- at eqbm

Total number of moles $= 0.5 + 1 = 1.5 m o l$

$P_{N_{2} O_{4}} = \frac{0.5}{1.5} \times 1 a t m = \frac{1}{3} a t m$

$P_{N O_{2}} = \frac{1}{1.5} \times 1 a t m = \frac{2}{3} a t m$

According to low of chamical equilibrium

$K_{p} = \frac{P_{N O_{2}}^{2}}{P_{N_{2} O_{4}}} = \frac{(\frac{2}{3})^{2}}{(\frac{1}{3})} = 1.33 a t m$

We know

$△ G^{\circ} = - 2.303 R T l o g K_{p} - - - (1)$

Substituting the volume equation (1) we get

$△ G^{\circ} = - 2.303 \times (0.082 J K^{- 1} m o l^{- 1}) \times (333 K) \times l o g (1.33)$

$= - 7.79 a t m . l i t$

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