contestada

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 4.0 at 228°C. If pure PCl5 is placed in a 1.00-L container and allowed to come to equilibrium, and the equilibrium concentration of PCl5(g) is 0.26 M, what is the equilibrium concentration of PCl3?

Respuesta :

Answer : The equilibrium concentration of [tex]PCl_3[/tex] is, 1.0 M

Explanation : Given,

Equilibrium concentration of [tex]PCl_5[/tex] = 0.26 M

Volume of solution = 1.00 L

Equilibrium constant [tex](K_c)[/tex] = 4.0

The balanced equilibrium reaction will be,

[tex]PCl_5\rightleftharpoons PCl_3+Cl_2[/tex]

The expression of equilibrium constant for the reaction will be:

[tex]K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}[/tex]

From the reaction we conclude that the concentration of [tex]PCl_3[/tex] and [tex]Cl_2[/tex] are equal.

Let the concentration of [tex]PCl_3[/tex] be 'X'.

So, concentration of [tex]Cl_2[/tex] = X

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

[tex]4.0=\frac{(X)\times (X)}{0.26}[/tex]

[tex]4.0=\frac{(X)^2}{0.26}[/tex]

[tex]X=1.0M[/tex]

Thus,

The concentration of [tex]PCl_3[/tex] at equilibrium = X = 1.0 M

The concentration of [tex]Cl_2[/tex] at equilibrium = X = 1.0 M

Otras preguntas