Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

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3 years ago