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* 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=([Cl_2]^2[H_2O]^2)/([HCl]^4[O_2])`

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

`K_c=((3.0)^2* (3.0)^2)/((0.80)^4* (0.20))`

`K_c=988.77=9.9* 10^2`

Therefore, the value of the equilibrium constant for this reaction is
`9.9* 10^2`

Answer 2

`\large \boxed{990}`

Step-by-step explanation:

4HCl + O₂ ⇌ 2Cl₂ + 2H₂O

E/mol·L⁻¹: 0.80 0.20 3.0 3.0

`K_{\text{eq}} = \frac{\text{[Cl$_(2)$]$^(2)$[H$_(2)$O]$^(2)$}}{\text{[HCl]$^(4)$[O$_(2)$]}} = (3.0^(2)*3.0^(2))/(0.80^(4)* 0.20) = \mathbf{990}\\\\\text{The $K_{\text{eq}}$ value is $\large \boxed{\mathbf{990}}$}`