Question

$K_p$ for the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3$ at ${400}^{\circ }C$ is $1.64\times {10}^{-4}$.

Calculate $K_C$:

• A. $0.3(mole{)}^{2}litr{e}^{-2}$
• B. $0.4(mole{)}^{2}litr{e}^{-2}$
• C. $0.5(mole{)}^{2}litr{e}^{-2}$
• D. $0.6(mole{)}^{2}litr{e}^{-2}$

Answer 1

The correct option is C $0.5(mole{)}^{2}litr{e}^{-2}$

Given:
$K_p=1.64\times {10}^{-4}$
$T={400}^{o}C=673K$

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

$\Delta n=-2$

and

$K_P=K_C(RT{)}^{\Delta n}$

$1.64\times {10}^{-4}=K_C(0.0821\times 673{)}^{-2}$

$K_C=0.5(mole{)}^{2}litr{e}^{-2}$