Question

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.

Answer 1

Answer:

$2.1\times 10^2$ 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.