Question

In an experiment, $NO_2$ gas is prepared and taken into $3$ test tubes $X$, $Y$ and $Z$. $NO_2$ gas which is brown in colour dimerises into $N_2O_4$ 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? $\underset{\text{Brown}}{ {2NO2_{(g)} <=> }}$ $\underset{\text{Colourless}}{ {N2O4_{(g)}}}$ ; $\Delta H=-57.2\,kJ\,mol^{-1}$

• 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

The Correct Option is C

It is an exothermic reaction, hence the forward reaction is favoured at low temperature which means colourless $N_2O_4$ will be formed resulting in decrease in intensity of brown colour. High temperature favours backward reaction resulting in formation of $NO_2$, thus intensifying brown colour.