Question

$N_2+3H_2\rightleftharpoons 2{NH}_3$. Starting with one mole of nitrogen and $3$ moles of hydrogen, at equilibrium $50$% of each had reacted. If the equilibrium pressure is $P$, the partial pressure of hydrogen at equilibrium would be:

• A. $P/2$
• B. $P/3$
• C. $P/4$
• D. $P/6$

Answer 1

Answer: (A)

$P/2$

Initial mole ratio N2:H2 is 1:3

At equilibrium, 50% of N2, H2 has reacted and equilibrium pressure is P

$N_2+3H_2\to 2N{H}_{3}y3y2y$

At equilibrium, 50% of each reactant reacted, the number of moles will become half.

The mole of N2=$y-\frac{y}{2}$

The moles of H2=$3y-\frac{3y}{2}$

THE moles of NH3=$\frac{2y}{2}$

Total moles at equilibrium=$\frac{y}{2}+\frac{3y}{2}+\frac{2y}{2}=3y$

Partial pressure of H2=$\frac{MolesofH2}{Totalmoles}\times Totalpressure$=$\frac{3y}{2}\times \frac{1}{3y}\times P$=$\frac{P}{2}$