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An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respectively. A quantity of Cl 2 ( g ) is injected into the mixture, and the total pressure jumps to 263.0 Torr. The appropriate chemical equation is PCl 3 ( g ) + Cl 2 ( g ) − ⇀ ↽ − PCl 5 ( g ) Calculate the new partial pressures after equilibrium is reestablished.

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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.

Step-by-step 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=(P_(PCl_5))/(P_(PCl_3)* 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=(217.0)/(13.2* 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=((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.

User Nilesh Moradiya
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