Question

For reaction, $2NOC{l}_{\left(g\right)}\leftrightharpoons 2N{O}_{\left(g\right)}+C{l}_{2\left(g\right)},K_c$ at 427${}^{\circ }$C is $3\times {10}^{-6}Lmo{l}^{-1}$. The value of $K_p$ is:

• A. $7.50\times {10}^{-5}$
• B. $2.50\times {10}^{-5}$
• C. $2.50\times {10}^{-4}$
• D. $1.75\times {10}^{-4}$

Answer 1

Answer: (D)

$1.75\times {10}^{-4}$

Answer:-

${2NOCl}_{\left(g\right)}\rightleftharpoons {2NO}_{\left(g\right)}+{{Cl}_2}_{\left(g\right)}$

$\Delta n=n_P-n_R=3-2=1$

Given:-

$K_C=3\times {10}^{-6}L{mol}^{-1}$

T = 427℃ = (427 + 273)K = 700K

As, we know that,

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

$K_P=3\times {10}^{-6}{(0.0821\times 700)}^1$

$K_P=1.72\times {10}^{-4}\approx 1.75\times {10}^{-4}$