Question

In a closed container, nitrogen and hydrogen mixture initially in a mole ratio of 1:4 reached equilibrium is found that the half hydrogen is converted to ammonia. If the original pressure was 180 atm, what will be the partial pressure of ammonia at equilibrium (there is no change in temperature)?

• A. 36 atm
• B. 50 atm
• C. 54 atm
• D. 56 atm

Answer 1

Answer: (A)

36 atm

$N_2+3H_2\rightleftharpoons 2N{H}_3$
Original pressure $P=180atm$
$P=P_{N_2}+P_{N{H}_3}$
Mole fraction of nitrogen $=\frac{1}{1+4}=0.2$
Mole fraction of hydrogen $=1-0.2=0.8$
Partial pressure of nitrogen $0.2\times 180=36$ atm
Partial pressure of hydrogen $=180-36=144$ atm
Now one half of 36 atm (i.e 18 atm) of nitrogen reacts with hydrogen to produce
$2\times 18=36$ atm of ammonia