Question

For $N_2+3H_2\rightleftharpoons 2N{H}_3$, one mole of $N_2$ and three mole of $H_2$ are at pressure of 4 atm. Equilibrium pressure is found to be 3 atm. Hence, $K_p$ is :

• A. $\frac{1}{\left(0.5\right)\times {\left(0.15\right)}^3}$
• B. $\frac{1}{\left(0.5\right)\times {\left(1.5\right)}^3}$
• C. $\frac{3\times 3}{\left(0.5\right)\times {\left(1.5\right)}^3}$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{\left(0.5\right)\times {\left(1.5\right)}^3}$

$N_2+3H_2\to 2N{H}_3$

Initial Pressure $13$

Final Pressure $(1-x)(3-3x)\left(2x\right)$

Total pressure at equilibrium $=3atm$.

$1-x+3-3x+2x=3$

$x=\frac{1}{2}=0.5$

$\therefore P_{N_2}=0.5,P_{H_2}=1.5,P_{N{H}_3}=1.0$

$K_P=\frac{1}{0.5\times (1.5{)}^3}$