Question

In Haber process $30L$ of Dihydrogen and $30L$ of dinitrogen were taken for reaction which yielded only $50\%$of the expected product . what will be the composition of gaseous mixture under the aforesaid condition in the end?

• A. $20L$ ammonia $10L$ nitrogen , $30L$ hydrogen
• B. $20L$ ammonia , $125L$ nitrogen , $25L$ ydrogen
• C. $20L$ ammonia , $20L$ nitrogen , $20L$hydrogen
• D. $10L$ ammonia , $25L$nitrogen , $15L$ hydrogen

Answer 1

The correct option is D $10L$ ammonia , $25L$nitrogen , $15L$ hydrogen

solution:

$N_2+3H_2\rightleftharpoons 2N{H}_3$

30 30 0

30-x 30-x 2x

$2x=10$

x = 5;

$N_2=30-5=25L$

$H_2=30-3X5=15L$

30 L of H_2 will react with 10 L of N_2 to produce 20 L of $N{H}_3$ (Here $H_2$ is the limiting reactant). As the yield is only 50%, 15 L of $H_2$ will react with 5 L of $N_2$ to produce 10 L of $N{H}_3$

Reaction mixture will contain,

15 L of $H_2$ (30L-15L), and 10 L of $N{H}_3$

hence the correct opt: D