Question

In at experiment, $2N_{2} + 6H_{2}4NH_{3}$gas is prepared and taken into 3 test tubes X, Y and Z. $K_{c}$gas which is brown in colour dimerises into $PCl_{5(g)}$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 whay?

$PCl_{5}$ $PCl_{5(g)}$

• 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.

Answer 2

: 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 $NO_{2,}$ thus intensifying brown colour.