Question

When 1.0 mole of $N_2$ and 3.0 moles of $H_2$ was heated in a vessel at 773K and a pressure of 3.55 atm. 30% of $N_2$ is converted into $N{H}_3$ at equilibrium. Calculate KP for the reaction -

• A. 3.1 × 10-2atm-2
• B. 4.1 × 10-2 atm-2
• C. 5.1 × 10-2atm-2
• D. 6.1 × 10-2atm-2

Answer 1

The correct option is C

5.1 × 10-2atm-2

$N_2\left(g\right)+3H_2\left(g\right)\leftrightharpoons 2N{H}_3\left(g\right)1mole3moles0atthestart1-0.33.0-0.90.6moles\text{At equilibrium}Totalmoles=0.7+2.1+0.6=3.4$

$K_P=\frac{(\frac{0.6}{3.4}\times 3.55{)}^2}{(\frac{0.7}{3.4}\times 3.55)(\frac{2.1}{3.4}\times 3.55{)}^3}$

$=5.1\times {10}^{-2}at{m}^{-2}$