Question

For $N_2+3H_2\rightleftharpoons 2N{H}_3$, one mole of $N_2$ and three moles 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

The correct option is B $\frac{1}{\left(0.5\right)\times {\left(1.5\right)}^3}$

$N_2+3H_2\rightleftharpoons 2N{H}_3$
Initial pressure $1$ $3$ $0$

Final pressure $(1-x)$ $(3-3x)$ $2x$

$P\propto$ no. of moles

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

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

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

Thus, $K_p=\frac{1^2}{\left(0.5\right)\times {\left(1.5\right)}^3}$

Optio (B) is correct.