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At 60^@C dinitrogen tetroxide is fifty p...

At `60^@C` dinitrogen tetroxide is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. 

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`N_2O_4 (9g) Leftrightarrow 2NO_2(b)`
If `N_2O_4` is 50% dissociated, the mole fraction of both the substances is given by
`x_(N_(2)O_(4))=(1-0.5)/(1+0.5), X_(NO_(2))=(2 xx 0.5)/(1+0.5)`
`P_(N2O4)=(0.5)/(1.5) xx 1 atm , P_(NO_(2))=(1)/(1.5) xx"1 atm"`
The equilibrium constant `K_P` is given by
`K_(p)=(p_(NO_(2))^2)/(P_(N_(2)O_(4))=(1.5)/((1.5)^2 (0.5))`
=1.33 atm
Since
`triangle_(r)G^@=-RT ln K_(p)`
`triangle_(r) G^@=-("8.314 JK"^(-1) mol^(-1)) xx (33K) xx (2.303) xx (0.1239)`
`=-763.8kJ mol"^(-1)`
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