PYQ NEET- Chemical Equilibrium-1 L-10

Question: The value of $\Delta \mathrm{H}$ for the reaction#

$$ \mathrm{X}{2(\mathrm{~g})}+4 \mathrm{Y}{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{XY}{4(\mathrm{~g})} $$ is less than zero. Formation of $X{4(\mathrm{~g})}$ will be favoured at

A) high temperature and high pressure

B) low pressure and low temperature

C) high temperature and low pressure

D) high pressure and low temperature

Answer: high pressure and low temperature#

Solution:#

$$ \begin{aligned} & \mathrm{X}{2(\mathrm{~g})}+4 \mathrm{Y}{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{XY}{4(\mathrm{~g})} \ & \Delta \mathrm{n}{\mathrm{g}}=- \text { ve and } \Delta \mathrm{H}=-\mathrm{ve} \end{aligned} $$

As $\Delta \mathrm{H}<0$ i.e., the given reaction is exothermic. According to Le-Chatelier principle, for exothermic reaction, forward reaction is favoured when temperature becomes low. Also, there are 5 gaseous moles on reactant side and 2 gaseous moles on products side. So, forward reaction is favoured when pressure of the reaction mixture becomes high. The reason is that at high pressure reaction tends to more in direction where there is lessee number of gaseous moles.