At $60^{0} C N_{2} O_{4}$ is 50% dissociated. At this temperature and 1 atm the standard free energy change is :

-863.8 kJ mol

^{- 1}

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-963.8 kJ.mol

^{-}

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-973.8 kJ.mol

^{- 1}

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-763.8 kJ mol

^{- 1}

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Solution

The correct option is D -763.8 kJ mol $^{- 1}$

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Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is: