Question

In the reaction, pcl5_pcl3+cl2, the equilibrium concentration of pcl5 and pcl3 are 0.4 and 0.2 mol/l respectively . If the value of kc is 0.5 what is the contration of cl2 in mol/l

Answer 1

Hope this helped....solution in the attachment....

Answer 2

The answer is 1 mol/l.

Given:

The equilibrium concentration of $PCl_5$ is 0.4 mol/l.

The equilibrium concentration of $PCl_3$ is 0.2 mol/l.

The value of $K_C$ is 0.5.

To Find:

The concentration of $Cl_2$

Solution:

$PCl_5$→$PCl_3+Cl_2$

The $K_C$ of this equation is given as

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

Where the terms in brackets are the equilibrium concentrations of the compounds. Hence

$0.5=\frac{[Cl_2]*0.2}{0.4} \\\ [Cl_2] = 1 mol/l$

The equilibrium concentration of $\bf Cl_2$ is 1 mol/l.

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