Question

Starting with one mole of nitrogen and 3 moles of hydrogen, at equilibrium $50\%$ of each had reacted according to the reaction: $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
If the equilibrium pressure is $P$, the partial pressure of hydrogen at equilibrium would be:

• A. $\frac{P}{2}$
• B. $\frac{P}{3}$
• C. $\frac{P}{4}$
• D. $\frac{P}{6}$

Answer 1

The correct option is A $\frac{P}{2}$

$N_2+{3H}_2⟶2{NH}_3$

$x$ $3x$ $0$ $⟶$ Initial

$(x-\frac{x}{2})(3x-\frac{3x}{2})\left(\frac{2x}{2}\right)⟶$ Eqn ($50$% of $N_2,H_2$ has reacted)

Total moles of eqn $=x-\frac{x}{2}+3x-\frac{3x}{2}+x=3x$

Now, $P_{H_2}=\frac{Molesof{H}_2}{Totalmoles}\times$ Total pressure

$=\left(\frac{3x-3x/2}{3x}\right)\times P=\frac{P}{2}$