In the reaction,A_{2(g)} + 4B_{2(g)} rightlef | vedclass

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(A) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,a decrease in temperature shifts the equilibrium in the f

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Low temperature,high pressure

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(A) According to Le Chatelier's principle,for an exothermic reaction $(\Delta H < 0)$,a decrease in temperature shifts the equilibrium in the forward direction to release heat.
In the given reaction,the number of moles of gaseous reactants is $1 + 4 = 5$,and the number of moles of gaseous products is $2$.
Since the number of moles decreases in the forward direction $(5 	o 2)$,an increase in pressure shifts the equilibrium towards the side with fewer moles,which is the forward direction.
Therefore,the formation of $AB_{4}$ is favoured at low temperature and high pressure.

In the reaction,$A_{2(g)} + 4B_{2(g)} \rightleftharpoons 2AB_{4(g)}$,$\Delta H < 0$,the formation of $AB_{4}$ is favoured at

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