Q14 / 160 · AP EAMCET 2003 · Chemistry

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?

  • A
    0.8
  • B
    1.6
  • Correct answer
    C
    3.2
  • D
    6.4

Solution

$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$ $\begin{array}{ll}1 \text { mole } 3 \text { mole } & 0 \text { moles (2 L) (initially) } \\ 0.6 \text { or } \quad(3-3 x) & 2 x \text { moles } \\ (1-x) \text { moles moles } & \text { (at equilibrium) }\end{array}$ 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. $=(1+x)+(3-3 x)+2 x$ $=0.6+1.8+0.8=3.2 \mathrm{~mol}$

Hence, the correct answer is option: 'C'.

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