Respuesta :
Answer: The concentrations of [tex]Cl_2[/tex] at equilibrium is 0.023 M
Explanation:
Moles of [tex]Cl_2[/tex] = [tex]\frac{\text {given mass}}{\text {Molar mass}}=\frac{10g}{71g/mol}=0.14mol[/tex]
Volume of solution = 1 L
Initial concentration of [tex]Cl_2[/tex] = [tex]\frac{0.14mol}{1L}=0.14M[/tex]
The given balanced equilibrium reaction is,
[tex]COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)[/tex]
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
[tex]K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}[/tex]
Now put all the given values in this expression, we get :
[tex]4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}[/tex]
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of [tex]Cl_2[/tex] at equilibrium is 0.023 M
