If 1 mole of aqueous nitric acid is formed,c | vedclass

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(C) To produce $2 \ moles$ of $HNO_{3(aq)}$ from reaction $(iii)$,we need $3 \ moles$ of $NO_{2(g)}$.Heat from reaction $(iii) = -140 \ kJ$.To get $3

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$493 \ kJ$

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(C) To produce $2 \ moles$ of $HNO_{3(aq)}$ from reaction $(iii)$,we need $3 \ moles$ of $NO_{2(g)}$.Heat from reaction $(iii) = -140 \ kJ$.To get $3 \ moles$ of $NO_{2(g)}$ from reaction $(ii)$,we need $\frac{3}{2}$ of the reaction $(ii)$:$\Delta H_2' = \frac{3}{2} 	imes (-112) = -168 \ kJ$.This requires $3 \ moles$ of $NO_{(g)}$. To get $3 \ moles$ of $NO_{(g)}$ from reaction $(i)$,we need $\frac{3}{4}$ of the reaction $(i)$:$\Delta H_1' = \frac{3}{4} 	imes (-904) = -678 \ kJ$.Total heat for $2 \ moles$ of $HNO_{3(aq)} = -678 + (-168) + (-140) = -986 \ kJ$.Total heat for $1 \ mole$ of $HNO_{3(aq)} = \frac{-986}{2} = -493 \ kJ$.Therefore,the total heat released is $493 \ kJ$.

Reaction	$\Delta H \ (kJ)$
$(i) \ 4NH_{3(g)} + 5O_{2(g)} \to 4NO_{(g)} + 6H_2O_{(l)}$	$-904$
$(ii) \ 2NO_{(g)} + O_{2(g)} \to 2NO_{2(g)}$	$-112$
$(iii) \ 3NO_{2(g)} + H_2O_{(l)} \to 2HNO_{3(aq)} + NO_{(g)}$	$-140$

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