Question

If the pressure of $N_2$ and $H_2$ 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) $90$

$N_2+3H_2⟶2N{H}_3$

From the above reaction,

$4$ moles of reactant produces $2$ moles of $N{H}_3$

Therefore,

$20\%$ mixture reacts to form $10\%N{H}_3$.

Thus, $80\%$ mixture and $10\%N{H}_3$, i.e., $90\%$ of mixture is left.

$\because 100\%$ mixture has $100atm$ pressure.

$\therefore$ Total pressure left $=90atm$