Question

The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.498 moles of CO and 0.498 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.

Answer 1

Answer:

$[COCl_2]_{eq}=0.424M$

$[Cl_2]_{eq}=0.074M$

$[CO]_{eq}=0.074M$

Explanation:

Hello,

In this case, one could solve the problem by stating the law of mass action for the undergoing chemical reaction which is:

$CO(g)+ Cl_2(g)\rightarrow COCl_2(g)$

As shown below:

$Keq=\frac{[COCl_2]}{[CO][Cl_2]}=77.5$

Now, the initial concentration of carbon monoxide equals that of chlorine which results:

$[CO]_0=[Cl_2]_0=\frac{0.498mol}{1L}=0.498M$

Next, as the change $x$ is introduced due to the chemical reaction, considering the stoichiometry, the law of mass action turns out:

$Keq=77.5=\frac{x}{(0.498M-x)(0.498M-x)}$

Solving for $x$ by using the quadratic equation one obtains:

$x_1=0.424M\\x_2=0.585M$

Therefore, the feasible result is 0.424M as the other one will lead to negative concentrations at the equilibrium. Thus, the equilibrium concentrations are:

$[COCl_2]_{eq}=x=0.424M$

$[Cl_2]_{eq}=0.498M-x=0.498M-0.424M=0.074M$

$[CO]_{eq}=0.498M-x=0.498M-0.424M=0.074M$

Best regards.