Question

The following equilibrium constants were determined at $1120K:$
$2CO(g)\rightleftharpoons C(s)+C{O}_2(g);K_{P_1}=10^{-14}at{m}^{-1}$
$CO(g)+C{l}_2(g)\rightleftharpoons COC{l}_2(g);K_{P_2}=6\times 10^{-3}at{m}^{-1}$
What is the equilibrium constant $K_C$ for the following reaction at $1120K$ :
$C(s)+C{O}_2(g)+2C{l}_2(g)\rightleftharpoons 2COC{l}_2(g)$

• A. $3.31\times 10^{11}{M}^{-1}$
• B. $5.5\times 10^{10}{M}^{-1}$
• C. $5.51\times 10^6{M}^{-1}$
• D. None of these

Answer 1

Answer: (A)

$C(s)+C{O}_2(g)\rightleftharpoons 2CO(g);$...(1)

$K_{P_1}=10^{14}at{m}^{-1}$

$2CO(g)+2C{l}_2(g)\rightleftharpoons 2COC{l}_2(g);$...(2)

$K_{P_2}={\left(6\times 10^{-3}\right)}^{2}at{m}^{-2}$

Add (1) and (2), we get

$C(s)+C{O}_2(g)+2C{l}_2(g)\rightleftharpoons 2COC{l}_2(g);$

$K_P=10^{14}\times 36\times 10^{-6}=36\times 10^8$

For given reaction $n_g=-1$

$\therefore K_C=K_P(RT)=36\times 10^8\times 0.0821\times 1120$

$K_C=3.31\times 10^{11}{M}^{-1}$