Question

If the pressure of $N_2/H_2$ mixture in a closed apparatus is $100$ atm and $20\%$ of the mixture reacts then the pressure at the same temperature would be:

• A. $100$
• B. $90$
• C. $85$
• D. $80$

Answer 1

Answer: (B)

$90$

Solution:- (B) $90$

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

From the above reaction, $1$ volume of $N_2$ combines with $3$ volumes of $H_2$ to produce $2$ volumes of $N{H}_3$.

Therefore the pressure decreases by $10\%$, when $20\%$ of the mixture reacts.

Hence the pressure will be $90atm$.