Question

The molar concentrations for the reactants and products at equilibrium are found to be [HCl]=0.80 M, [O2]=0.20 M, [Cl2]=3.0 M, and [H2O]=3.0 M. What is the value of the equilibrium constant for this reaction?
4HCl(g)+O2(g)⇌2Cl2(g)+2H2O(g)

Answer 1

Answer : The value of the equilibrium constant for this reaction is $9.9\times 10^2$

Solution :  Given,

Concentration of HCl at equilibrium = 0.80 M

Concentration of $O_2$ at equilibrium = 0.20 M

Concentration of $Cl_2$ at equilibrium = 3.0 M

Concentration of $H_2O$ at equilibrium = 3.0 M

Now we have to calculate the value of equilibrium constant.

The given equilibrium reaction is,

$4HCl(g)+O_2(g)\rightleftharpoons 2Cl_2(g)+2H_2O(g)$

The expression of $K_c$ will be,

$K_c=\frac{[Cl_2]^2[H_2O]^2}{[HCl]^4[O_2]}$

Now put all the values in this expression, we get:

$K_c=\frac{(3.0)^2\times (3.0)^2}{(0.80)^4\times (0.20)}$

$K_c=988.77=9.9\times 10^2$

Therefore, the value of the equilibrium constant for this reaction is $9.9\times 10^2$

Answer 2

Answer:

$\large \boxed{990}$

Explanation:

                  4HCl +  O₂  ⇌  2Cl₂ + 2H₂O

E/mol·L⁻¹:   0.80    0.20      3.0       3.0

$K_{\text{eq}} = \dfrac{\text{[Cl _{2} ] ^{2} [H _{2} O] ^{2} }}{\text{[HCl] ^{4} [O _{2} ]}} = \dfrac{3.0^{2}\times3.0^{2}}{0.80^{4}\times 0.20} = \mathbf{990}\\\\\text{The K_{\text{eq}} value is \large \boxed{\mathbf{990}} }$