Chemistry: The molar concentrations for the reactants and products at equilibrium are found to be [ccl4]=1.0 m, [o2]=0.3 m, [cocl2]=4.0 m, and [cl2]=2.0 m. what is the value of the equilibrium constant for this reaction? 2ccl4(g)+o2(g)⇌2cocl2(g)+2cl2(g) express your answer numerically using two significant figures.

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The molar concentrations for the reactants and

alexianunez15

30.12.2021

$2.1\times 10^2$ is the value of the equilibrium constant for this reaction.

Explanation:

$2CCl_4(g)+O_2(g)\rightleftharpoons 2COCl_2(g)+2Cl_2(g)$

Molar concentrations of CCl_4 at equilibrium = [CCl_4]=1.0M

Molar concentrations of O_2 at equilibrium = [O_2]=0.3M

Molar concentrations of COCl_2 at equilibrium = [COCl_2]=4.0M

Molar concentrations of Cl_2 at equilibrium = [Cl_2]=2.0M

The equilibrium constant is given as:

$K_c=\frac{[COCl_2]^2[Cl_2]^2}{[CCl_4]^2[O_2]}=\frac{(4.0 M)^2\times (2.0M)^2}{(1.0M)^2\times (0.3 M)}$

$K_c=213.33\approx 2.1\times 10^2$

$2.1\times 10^2$ is the value of the equilibrium constant for this reaction.

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