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Consider the following reaction equilibrium:
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$
Initially, 1 mole of $\mathrm{N}_2$ and 3 moles of $\mathrm{H}_2$ are taken in a $2 \mathrm{~L}$ flask. At equilibrium state if, the number of moles of $\mathrm{N}_2$ is 0.6 , what is the total number of moles of all gases present in the flask?
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$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$
Initially, 1 mole of $\mathrm{N}_2$ and 3 moles of $\mathrm{H}_2$ are taken in a $2 \mathrm{~L}$ flask. At equilibrium state if, the number of moles of $\mathrm{N}_2$ is 0.6 , what is the total number of moles of all gases present in the flask?
Solution:
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Verified Answer
The correct answer is:
3.2
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$
$\begin{array}{ll}1 \text { mole } 3 \text { mole } & 0 \text { moles (2 L) (initially) } \\ 0.6 \text { or } \quad(3-3 x) & 2 x \text { moles } \\ (1-x) \text { moles moles } & \text { (at equilibrium) }\end{array}$
Number of moles of $\mathrm{N}_2=0.6=1-x$
$\therefore x=1-0.6=0.4$
So, $3-3 x=3-3 \times 0.4=1.8$,
and $2 x=2 \times 0.4=0.8$
Therefore, the total number of moles at equilibrium.
$=(1+x)+(3-3 x)+2 x$
$=0.6+1.8+0.8=3.2 \mathrm{~mol}$
$\begin{array}{ll}1 \text { mole } 3 \text { mole } & 0 \text { moles (2 L) (initially) } \\ 0.6 \text { or } \quad(3-3 x) & 2 x \text { moles } \\ (1-x) \text { moles moles } & \text { (at equilibrium) }\end{array}$
Number of moles of $\mathrm{N}_2=0.6=1-x$
$\therefore x=1-0.6=0.4$
So, $3-3 x=3-3 \times 0.4=1.8$,
and $2 x=2 \times 0.4=0.8$
Therefore, the total number of moles at equilibrium.
$=(1+x)+(3-3 x)+2 x$
$=0.6+1.8+0.8=3.2 \mathrm{~mol}$
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