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)#?

Question

Carbon tetrachloride reacts at high temperatures with oxygen to produce two toxic gases, phosgene and chlorine. #C Cl_4(g) + 1/2O_2(g) rightleftharpoons COCI_2(g) + Cl_2(g)#

#K_c = 4.4 xx 10^9# for this reaction with these coefficients at #"1000 K"#

EDIT: The product should be #COCl_2#.
- Truong-Son

Naomi Aronson · Accepted Answer

#1.9 xx 10^(19)# at #"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, #"COCI"_2# does not exist; it should be #"COCl"_2#.)

Reaction #(1)# is:

#C Cl_4(g) + 1/2O_2(g) rightleftharpoons COCl_2(g) + Cl_2(g)#

#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 #(2)# is:

#2C Cl_4(g) + O_2(g) rightleftharpoons 2COCl_2(g) + 2Cl_2(g)#

therefore having:

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

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:

#color(blue)(K_c') = (4.4 xx 10^9)^2 = color(blue)(1.9 xx 10^(19))#

Kai Bowman · Answer

undefined To calculate Kc for the reaction $2CCl_4(g) + O_2(g) \rightleftharpoons 2COCl_2(g) + 2Cl_2(g)$, you need the concentrations of the reactants and products at equilibrium. Then, plug these values into the equilibrium expression:

$$ Kc = \frac{{[COCl_2]^2 \cdot [Cl_2]^2}}{{[CCl_4]^2 \cdot [O_2]}} $$

Substitute the equilibrium concentrations into this expression and calculate the value of Kc.