Chemistry: 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?

Carbon and three oxygen atomsExplanation:Ap3x, thanks for brainliest...

For the equilibrium pcl5(g) pcl3(g) + cl2(g),

shyann78

05.01.2022

Answer : The equilibrium concentration of PCl_3 is, 1.0 M

Explanation : Given,

Equilibrium concentration of PCl_5 = 0.26 M

Volume of solution = 1.00 L

Equilibrium constant (K_c) = 4.0

The balanced equilibrium reaction will be,

$PCl_5\rightleftharpoons PCl_3+Cl_2$

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

From the reaction we conclude that the concentration of PCl_3 and Cl_2 are equal.

Let the concentration of PCl_3 be 'X'.

So, concentration of Cl_2 = X

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

$4.0=\frac{(X)\times (X)}{0.26}$

$4.0=\frac{(X)^2}{0.26}$

X=1.0M

Thus,

The concentration of PCl_3 at equilibrium = X = 1.0 M

The concentration of Cl_2 at equilibrium = X = 1.0 M

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