Question

When a vapour, at atmospheric pressure was gradually heated from $25{}^{\circ }\text{C}$ it's colour was found to deepen at first and then fade as the temperature was raised above $180{}^{\circ }\text{C}$. At $600{}^{\circ }\text{C}$, the vapour was almost colourless, but its colour deepened when the pressure was raised at this temperature. The vapour was:

• A. Bromine
• B. A mixture of nitrogen dioxide and dinitrogen tetraoxide
• C. Pure nitrogen dioxide
• D. Pure dinitrogen tetroxide

Answer 1

The correct option is D Pure dinitrogen tetroxide

$\underset{\text{(colourless)}}{N_2{O}_4}\stackrel{160{}^{\circ }\text{C}}{\rightleftharpoons }\underset{\text{(Brown)}}{2N{O}_2}\stackrel{600{}^{\circ }\text{C}}{\rightleftharpoons }\underset{\text{(colourless)}}{2NO+O_2}$
The colour change is due to the conversion of colourless $N_2{O}_4$ to brown coloured $N{O}_2$. But beyond $160{}^{\circ }\text{C}$, $N{O}_2$ gets converted to a colourless mixture of $NO$ and $O_2$ and the colour fades away. But if the pressure is increased at $600{}^{\circ }\text{C}$ then according to Le-Chatlier's principle, equilibrium will again shift towards brown coloured $N{O}_2$ and colour again darkens. Thus, originally the vapours contained pure $N_2{O}_4$.