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(D) The equilibrium involved is: $N_2O_4(g) \rightleftharpoons 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$.At $25\,^oC$,$N_2O_4$ is colorless. As it

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Pure dinitrogen tetroxide

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(D) The equilibrium involved is: $N_2O_4(g) \rightleftharpoons 2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g)$.At $25\,^oC$,$N_2O_4$ is colorless. As it is heated,it dissociates into $NO_2$,which is brown,causing the color to deepen.Above $160\,^oC$,$NO_2$ further dissociates into $NO$ and $O_2$,both of which are colorless,causing the color to fade.At $600\,^oC$,the mixture is almost colorless. According to Le Chatelier's principle,increasing the pressure shifts the equilibrium towards the side with fewer moles of gas (the formation of $NO_2$),which is brown. Thus,the color deepens again. The initial vapor was pure $N_2O_4$.

List-$I$ (Equilibrium)	List-$II$ (Conditions for the process)
$P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic)	$1. \text{High temperature}$
$Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic)	$2. \text{Low temperature}$
$R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic)	$3. \text{High temperature}$
	$4. \text{Low temperature}$
	$5. \text{Independent of pressure}$

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