Given the following, how do you calculate K_c for the reaction 2C Cl_4(g) + O_2(g) rightleftharpoons 2COCI_2(g) + 2Cl_2(g)?

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. $C C {l}_{4} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s C O C {I}_{2} \left(g\right) + C {l}_{2} \left(g\right)$ ${K}_{c} = 4.4 \times {10}^{9}$ for this reaction with these coefficients at $\text{1000 K}$ EDIT: The product should be $C O C {l}_{2}$. - Truong-Son

Jun 19, 2017

$1.9 \times {10}^{19}$ at $\text{1000 K}$ and whatever pressure this was for the first reaction.

You are just supposed to write each ${K}_{c}$ expression and determine the relationship between the two. Let ${K}_{c}$ be your known equilibrium constant and ${K}_{c} '$ be the one to calculate.

(By the way, ${\text{COCI}}_{2}$ does not exist; it should be ${\text{COCl}}_{2}$.)

Reaction $\left(1\right)$ is:

$C C {l}_{4} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s C O C {l}_{2} \left(g\right) + C {l}_{2} \left(g\right)$

K_c = (["COCl"_2(g)]["Cl"_2(g)])/(["CCl"_4(g)]["O"_2(g)]^"1/2")

(recall that the coefficients for each reaction participant are also their respective exponents.)

Reaction $\left(2\right)$ is:

$2 C C {l}_{4} \left(g\right) + {O}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 C O C {l}_{2} \left(g\right) + 2 C {l}_{2} \left(g\right)$

therefore having:

${K}_{c} ' = \left(\left[{\text{COCl"_2(g)]^2["Cl"_2(g)]^2)/(["CCl"_4(g)]^2["O}}_{2} \left(g\right)\right]\right)$

If we look at these equilibrium constants...

[(["COCl"_2(g)]["Cl"_2(g)])/(["CCl"_4(g)]["O"_2(g)]^"1/2")]^2 = (["COCl"_2(g)]^2["Cl"_2(g)]^2)/(["CCl"_4(g)]^2["O"_2(g)])

Therefore:

${K}_{c}^{2} = {K}_{c} '$

And so, since we have ${K}_{c}$, we can calculate this new equilibrium constant at double the mols of reactants and products to be:

$\textcolor{b l u e}{{K}_{c} '} = {\left(4.4 \times {10}^{9}\right)}^{2} = \textcolor{b l u e}{1.9 \times {10}^{19}}$