Question

In an experiment, $N{O}_2$ gas is prepared and taken into $3$ test tubes $X$, $Y$ and $Z$. $N{O}_2$ gas which is brown in colour dimerises into $N_2{O}_4$ which is colourless. Test tube $X$ is kept at room temperature, $Y$ is kept in ice and $Z$ is kept in hot water. What colour changes will you observe in the test tubes and why?
$\underset{\text{Brown}}{2N{O}_{2(g)}}\rightleftharpoons \underset{\text{Colourless}}{N_2{O}_{4(g)}}$ ; $\Delta H=-57.2kJmo{l}^{-1}$

• A. In test tube $X$, brown colour intensifies since backward reaction is favoured at low temperature
• B. In test tube $Y$, brown colour intensifies since backward reaction takes place at room temperature
• C. In test tube $Z$, brown colour intensifies since high temperature favours the backward reaction
• D. Brown colour of test tubes $X$, $Y$ and $Z$ remains same since there is no effect of change in temperature on the reaction

Answer 1

Answer: (C)

It is an exothermic reaction, hence the forward reaction is favoured at low temperature which means colourless $N_2{O}_4$ will be formed resulting in decrease in intensity of brown colour. High temperature favours backward reaction resulting in formation of $N{O}_2$, thus intensifying brown colour.