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At 60^(@)C, dinitrogen tetroxide is 50 p...

At `60^(@)C`, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Text Solution

Verified by Experts

`N_(2)O_(4)(g) iff 2NO_(2)(g)`
Since `N_(2)O_(4)` is 50% dissociated, then moles of `N_(2)O_(4)` left and `NO_(2)` formed are :
Moles of `N_(2)O_(4)` left = `1 - 0.5 = 0.5`
Moles of `NO_(2)` formed `= 2 xx 0.5 = 1.0`
Mole fraction of `N_(2)O_(4), x_(NO_(2)) = (0.5)/(0.5 + 1.0) = (0.5)/(1.5)`
Mole fraction of `NO_(2), x_(NO_(2)) = (1)/(0.5 + 1.0) = 1/1.5`
`P_(N_(2)O_(4)) = x_(N_(2)O_(4)) xx P = (0.5)/(1.5) xx 1 = (0.5)/(1.5)`
`P_(N_(2)) = x_(NO_(2)) xx P = (1.0)/(1.5) xx 1 = (1.0)/(1.5)`
`K_(p) = (PNO_(2))^(2)/(PN_(2)O_(4)) = ((1.0)/(1.5))^(2)/((0.5)/(1.5)) = 1.33`
`DeltaG^(@) = -2.303 RT log K_(p)`
` = -2.303 xx 8.314xx 333xx log 1.33`
` = -2.303 xx 8.314 xx 333 xx 0.1239`
` = -790.0 kJ mol^(-1)`.
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