(C) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initially: $1 \ mol \ N_2, 3 \ mol \ H_2, 0 \ mol \ NH_3$.At equilibrium: $(1

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(C) The reaction is: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$Initially: $1 \ mol \ N_2, 3 \ mol \ H_2, 0 \ mol \ NH_3$.At equilibrium: $(1-x) \ mol \ N_2, (3-3x) \ mol \ H_2, 2x \ mol \ NH_3$.Given that at equilibrium,moles of $N_2 = 0.6$.So,$1-x = 0.6 \implies x = 0.4$.Now,calculate moles of each component at equilibrium:Moles of $N_2 = 0.6 \ mol$.Moles of $H_2 = 3 - 3(0.4) = 3 - 1.2 = 1.8 \ mol$.Moles of $NH_3 = 2(0.4) = 0.8 \ mol$.Total moles at equilibrium = $0.6 + 1.8 + 0.8 = 3.2 \ mol$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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Consider the following reaction equilibrium:
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$
Initially,$1 \ mol$ of $N_2$ and $3 \ mol$ of $H_2$ are taken in a $2 \ L$ flask. At equilibrium,if the number of moles of $N_2$ is $0.6$,what is the total number of moles of all gases present in the flask?

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A
$0.8$
B
$1.6$
C
$3.2$
D
$6.4$

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Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)

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The equilibrium constant for the following reaction is $1.6 \times 10^{5}$ at $1024 \, K$:
$H_{2(g)} + Br_{2(g)} \longleftrightarrow 2HBr_{(g)}$
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