Answer:
[tex]\large \boxed{990}[/tex]
Explanation:
4HCl + O₂ ⇌ 2Cl₂ + 2H₂O
E/mol·L⁻¹: 0.80 0.20 3.0 3.0
[tex]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 $\large \boxed{\mathbf{990}}$}[/tex]
Answer : The value of the equilibrium constant for this reaction is [tex]9.9\times 10^2[/tex]
Solution : Given,
Concentration of HCl at equilibrium = 0.80 M
Concentration of [tex]O_2[/tex] at equilibrium = 0.20 M
Concentration of [tex]Cl_2[/tex] at equilibrium = 3.0 M
Concentration of [tex]H_2O[/tex] at equilibrium = 3.0 M
Now we have to calculate the value of equilibrium constant.
The given equilibrium reaction is,
[tex]4HCl(g)+O_2(g)\rightleftharpoons 2Cl_2(g)+2H_2O(g)[/tex]
The expression of [tex]K_c[/tex] will be,
[tex]K_c=\frac{[Cl_2]^2[H_2O]^2}{[HCl]^4[O_2]}[/tex]
Now put all the values in this expression, we get:
[tex]K_c=\frac{(3.0)^2\times (3.0)^2}{(0.80)^4\times (0.20)}[/tex]
[tex]K_c=988.77=9.9\times 10^2[/tex]
Therefore, the value of the equilibrium constant for this reaction is [tex]9.9\times 10^2[/tex]
