Question

For the reversible reaction, $N_2 \, ( g) + 3H_2 \, ( g) \leftrightharpoons 2NH_3 \, ( g )$ at $500^\circ \, C,$ the value of $K_p = 1.44 \times 10^ { - 5 }$ when partial pressure is measured in atmosphere. The corresponding value of $K_c$ with concentration in mol/L is

• A. $\frac{ 1.44 \times 10^{ - 5 }}{ ( 0.082 \times 500)^{ - 2 }}$
• B. $\frac{ 1.44 \times 10^{ - 5 }}{ ( 8.314 \times 773)^{ - 2 }}$
• C. $\frac{ 1.44 \times 10^{ - 5 }}{ ( 0.082 \times 773)^ 2}$
• D. $\frac{ 1.44 \times 10^{ - 5 }}{ ( 0.082 \times 773)^{ - 2 }}$

Answer 1

Answer: (D)

$\frac{ 1.44 \times 10^{ - 5 }}{ ( 0.082 \times 773)^{ - 2 }}$

Answer 2

$N_2 \, ( g) + 3H_2 \, ( g) \leftrightharpoons 2NH_3 \, ( g )$ \hspace20mm $\triangle = - 2$
$K_p = K_c \, (RT)^{ \triangle n}$
$K_c = \frac{ K_p }{ (RT)^{ \triangle n}} = \frac{ 1.44 \times 10^{ - 5 }}{ ( 0.082 \times 773)^{ - 2 }}$