Question

The following equilibrium constants were determined at $1120K$:
$2CO\left(g\right)\rightleftharpoons C\left(s\right)+{CO}_2\left(g\right);K_{p_1}={10}^{-14}{atm}^{-1}$
$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right);K_{p_2}={6\times 10}^{-3}{atm}^{-1}$
What is the equilibrium constant $K_c$ for the following reaction at $1120K$?
$C\left(s\right)+{CO}_2\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2CO{Cl}_2\left(g\right)$

• 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

The correct option is A ${3.31\times 10}^{11}{M}^{-1}$

$2CO\left(g\right)\rightleftharpoons C\left(s\right)+{CO}_2\left(g\right);K_{p_1}={10}^{-14}{atm}^{-1}$ so for reverse reaction:

$C\left(s\right)+{CO}_2\left(g\right)\rightleftharpoons 2CO\left(g\right);K_{p^{\prime }}=1/{10}^{-14}{atm}^{-1}={10}^{14}$

$CO\left(g\right)+C{l}_2\left(g\right)\rightleftharpoons CO{Cl}_2\left(g\right);K_{p_2}={6\times 10}^{-3}{atm}^{-1}$ so for

$2CO\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2CO{Cl}_2\left(g\right);K_{p^{\prime \prime }}={{6\times 10}^{-3}}^2=36\times {10}^{-6}$ so for

$C\left(s\right)+{CO}_2\left(g\right)+2C{l}_2\left(g\right)\rightleftharpoons 2CO{Cl}_2\left(g\right)$

$K_p=K_p^{\prime }\times K_p^{\prime \prime }=36\times {10}^8$ and $K_c=K_{p}RT=3.31\times {10}^{11}.$