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?
$2N{O}_{2\left(g\right)}\rightleftharpoons N_2{O}_{4\left(g\right)}$, $\Delta H=-57.2kJmo{l}^{-1}$

• 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)

In test tube $Z$, brown colour intensifies since high temperature favours the backward reaction.

$2N{O}_2\left(g\right)\rightleftharpoons N_2{O}_4\left(g\right);\Delta H=-57.2KJmo{l}^{-1}$

Since this is an exothermic process, heat is released, and at low temperatures forward reaction is favoured.

In test tubes, $X$ and $Y$ forward reaction is favoured and $N_2{O}_2$ is formed which is colourless. In test tube, $Z$ brown colour intensifies, as it is kept in hot water since high temperature favours the backward reaction.