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The value of $\Delta H$ for the reaction $X_2(g) + 4Y_2(g) \rightleftharpoons 2XY_4(g)$ is less than zero. Formation of $XY_4(g)$ will be favoured at
A
High pressure and low temperature
B
High temperature and high pressure
C
Low pressure and low temperature
D
High temperature and low pressure
Detailed Solution
$\Delta H < 0$, so the forward reaction is exothermic. By Le Chatelier's principle, an exothermic reaction is favoured by low temperature.
Moles of gaseous reactants = 1 + 4 = 5; moles of gaseous products = 2.
$\Delta n_g = 2 - 5 = -3$, so the forward reaction takes place with a decrease in the number of gaseous moles (decrease in volume).
Increasing the pressure shifts the equilibrium towards the side with fewer gaseous moles, i.e., towards $XY_4$.
Hence formation of $XY_4$ is favoured at high pressure and low temperature.
Moles of gaseous reactants = 1 + 4 = 5; moles of gaseous products = 2.
$\Delta n_g = 2 - 5 = -3$, so the forward reaction takes place with a decrease in the number of gaseous moles (decrease in volume).
Increasing the pressure shifts the equilibrium towards the side with fewer gaseous moles, i.e., towards $XY_4$.
Hence formation of $XY_4$ is favoured at high pressure and low temperature.