Question

One mole of $N_2\left(g\right)$ is mixed with 2 moles of $H_2\left(g\right)$ in a 4 litre vessel. If $50$% of $N_2\left(g\right)$ is converted to $H_2\left(g\right)$ by the following reaction :
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
What will be the value of $K_c$ for the following equilibrium?
${NH}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$

• A. $256$
• B. $16$
• C. $\frac{1}{16}$
• D. none of these

Answer 1

Answer: (C)

$\frac{1}{16}$

For this reaction:
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
t= 0 1 2 -
At eqb 0.5 0.5 1
So $K_c=\left(1/4{)}^2/\right(0.5/4{)}^4=256$

$\therefore$ Equilibrium constant for the following equilibrium :
${NH}_3\left(g\right)\rightleftharpoons \frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)$
$K_c^{\prime }=1/\sqrt{K_c}=1/\sqrt{256}=1/16.$