Question

The following equilibrium constants were determined at 1120 K: $2CO\left(g\right)\rightleftharpoons C\left(s\right)+C{O}_2\left(g\right)K{p}_1={10}^{-14}at{m}^{-1}$
$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_{2}K{p}_2=6\times {10}^{-3}at{m}^{-1}$
What is the equilibrium constant $K_p$ for the following reaction at 1120 K. $C\left(s\right)+C{O}_2\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2COC{l}_2\left(g\right)$

• A. $36\times {10}^{4}at{m}^{-1}$
• B. $3.6\times {10}^{9}at{m}^{-1}$
• C. $3.6\times {10}^{5}at{m}^{-1}$
• D. $3.6\times {10}^{10}at{m}^{-1}$

Answer 1

The correct option is B $3.6\times {10}^{9}at{m}^{-1}$

$2CO\left(g\right)\rightleftharpoons C\left(s\right)+C{O}_2\left(g\right);K{p}_1={10}^{-14}at{m}^{-1}......\left(1\right)$
$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons COC{l}_2\left(g\right);K{p}_2=6\times {10}^{-3}at{m}^{-1}.....\left(2\right)$
Multiply eqn (2) by (2), we get
$2CO\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2COC{l}_2\left(g\right)$ $K^{\prime }{p}_2=(6\times {10}^{-3}{)}^{2}at{m}^{-1}$ ...(3)
reverse eqn (1) we get
$C\left(O\right)+C{O}_2\left(g\right)\rightleftharpoons 2CO$ $K{p}_1^1=\frac{1}{{10}^{-14}}at{m}^{-1}$ ...(4)
add eqn (3) and (4)
$C\left(s\right)+C{O}_2\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2COC{l}_2\left(g\right)$
$K=K{p}_2^1\times K{p}_1^1$
$K=(6\times {10}^{-3}{)}^2\times \frac{1}{{10}^{-14}}$
$,K=\frac{36\times {10}^{-6}}{{10}^{-14}}$
$K=36\times {10}^8$
$K=3.6\times {10}^{9}at{m}^{-1}$