For the formation of NH_{3(g)} from its const | vedclass

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(A) The formation of $NH_3$ follows Le Chatelier's principle for the reaction: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)} + 	ext{Heat}$.Numb

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high pressure and low temperature

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(A) The formation of $NH_3$ follows Le Chatelier's principle for the reaction: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)} + 	ext{Heat}$.Number of moles of reactants $= 1 + 3 = 4$.Number of moles of products $= 2$.Since the number of moles of products is less than the number of moles of reactants,an increase in pressure will shift the equilibrium in the forward direction.As the reaction is exothermic (heat is released),a decrease in temperature will shift the equilibrium in the forward direction.Therefore,high pressure and low temperature are the favourable conditions for the formation of $NH_3$.

List-$I$ (Equilibrium)	List-$II$ (Conditions for the process)
$P. A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ (Endothermic)	$1. \text{High temperature}$
$Q. 2AB_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ (Exothermic)	$2. \text{Low temperature}$
$R. 2AB_{3(g)} \rightleftharpoons A_{2(g)} + 3B_{2(g)}$ (Endothermic)	$3. \text{High temperature}$
	$4. \text{Low temperature}$
	$5. \text{Independent of pressure}$

For the formation of $NH_{3(g)}$ from its constituent elements,the favourable conditions for its formation are

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