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 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,

$4$ mole of reactant produces $2$ mole 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 $=90atm$