40% of a mixture of 0.2 mol of N_2 and 0.6 | vedclass

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(A) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initial moles: $n(N_2) = 0.2 \ mol$,$n(H_2) = 0.6 \ mol$,$n(NH_3) = 0 \ mol$.

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$4:5$

Vedclass · Answer

(A) The reaction is: $N_{2(g)} + 3H_{2(g)} ightleftharpoons 2NH_{3(g)}$Initial moles: $n(N_2) = 0.2 \ mol$,$n(H_2) = 0.6 \ mol$,$n(NH_3) = 0 \ mol$.Total initial moles $(n_i) = 0.2 + 0.6 = 0.8 \ mol$.Since $40\%$ of $N_2$ reacts,the amount of $N_2$ reacted is $x = 0.40 	imes 0.2 = 0.08 \ mol$.According to the stoichiometry:$N_2$ reacted $= 0.08 \ mol$$H_2$ reacted $= 3 	imes 0.08 = 0.24 \ mol$$NH_3$ produced $= 2 	imes 0.08 = 0.16 \ mol$Moles at final state:$n(N_2) = 0.2 - 0.08 = 0.12 \ mol$$n(H_2) = 0.6 - 0.24 = 0.36 \ mol$$n(NH_3) = 0.16 \ mol$Total final moles $(n_f) = 0.12 + 0.36 + 0.16 = 0.64 \ mol$.At constant temperature and pressure,$V \propto n$,so the ratio of volumes is the ratio of total moles:Ratio $= \frac{V_f}{V_i} = \frac{n_f}{n_i} = \frac{0.64}{0.8} = \frac{64}{80} = \frac{4}{5}$.

$40\%$ of a mixture of $0.2 \ mol$ of $N_2$ and $0.6 \ mol$ of $H_2$ reacts to give $NH_3$ according to the equation:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
at constant temperature and pressure. What is the ratio of the final volume to the initial volume of gases?

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$40\%$ of a mixture of $0.2 \ mol$ of $N_2$ and $0.6 \ mol$ of $H_2$ reacts to give $NH_3$ according to the equation:$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$at constant temperature and pressure. What is the ratio of the final volume to the initial volume of gases?