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

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

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

$1.9 xx 10^(19)$ 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 ...

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

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:

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

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

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