In a flask colour less $N_{2} O_{4}$ is of equilibrium with brown coloured ${NO}_{2}$ . At equilibrium when the flask is heated at 100 $°$ C, the brown colour deepens and on cooling it becomes less coloured. The change in enthalpy, $∆$ H for this system is:

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Negative

Undefined

Zero

Positive

answer is B.

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Detailed Solution

On heating $N_{2} O_{4} \to 2 {NO}_{2}; if {NO}_{2}$ is formed more, it means $∆$ H = +ve because endothermic reactions are favored with increase in temperature.

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