Question

For the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3,$ Initially $N_2:H_2$ were taken in the ratio of 1 : 3. Up to the point of equilibrium 50% each reactant has been reacted. If total pressure at equilibrium is P. The partial pressure of ammonia would be -

• A. P/3
• B. P/6
• C. P/4
• D. P/8

Answer 1

The correct option is A P/3

$\begin{array}{ccccc}& N_2+& 3H_2& \rightleftharpoons & 2N{H}_3\\ Initially& 1& 3& & 0\\ Atequilibrium& 1-0.5& 3-1.5& & 2\times 0.5\\ Totalmoles& =(1-0.5)+(3-1.5)+1=3& & & \end{array}$
$P_{N{H}_3}=\frac{molesofN{H}_3}{totalno.ofmoles}\left(P\right)$
$P_{N{H}_3}=\frac{1}{3}P$