Question

2 moles of PCl5 were heated in a sealed 5L container to constant temperature. If the degree of dissociation of PCl5 at this temperature is 0.4, the Kc of the reaction PCl5 = PCl3 + Cl2 is?

Answer 1

PCl5

PCl3

Cl2

Initial

2

0

0

Change

-0.8

0.8

0.8

Equilibrium

1.2

0.8

0.8

Molar concentration

0.6

0.4

0.4

The expression for the equilibrium constant is Kc=[PCl5][PCl3][Cl2]=0.60.4×0.4=0.266.

Answer 2

Answer:

The $K_{c}$ of the reaction at constant temperature is 0.11

Explanation:

Given,

• The moles of PCl₅ = 2
• The volume of the container = 5 L
• The degree of dissociation α = 0.4

Solution:

The given reaction is :

                               PCl₅           ⇄        PCl₃       +      Cl₂

Initial                    2 moles              0 moles          0 moles

At equilibrium        2-2α                     2α                    2α

α = 0.4                 2 - (2×0.4)              2×0.4                2×0.4

                              1.2                         0.8                    0.8

Formula:

The equilibrium constant $K_{c} = \frac{[PCl_{3} ] [Cl_{2} ]}{PCl_{5} }$

Substitute the values in the above equation

$K_{c} = \frac{(\frac{0.8}{5} )( \frac{0.8}{5}) }{\frac{1.2}{5} }$

$K_{c} = \frac{0.16 * 0.16}{0.24}$

$K_{c} = \frac{0.0256}{0.24}$

$K_{c} = 0.1066\\ \\K_{c} = 0.11$

Answer: The $K_{c}$ of the reaction at constant temperature = 0.11

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