Question

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 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

Answer 1

`1.9 xx 10^(19)` at `"1000 K"` and whatever pressure this was for the first reaction.

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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))`