Question

Two moles of $PCl_{5}$ were heated to 327°C in a closed 2L vessel and when equilibrium was achieved, $PCl_{5}$ was found to be 40% dissociated. Calculate the equilibrium constants Kp and Kc for this reaction.

Answer 1

The ionization reaction is as follows:  

$P{ Cl }_{ 5 }(g) \rightleftharpoons  P{ Cl }_{ 3 }(g) + { Cl }_{ 2 }(g)$

$Initial\quad \quad \quad \quad \quad \quad \quad 2\quad mol\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0\quad mol\quad \quad \quad \quad \quad \quad \quad \quad 0\quad mol$$Change\quad \quad \quad \quad \quad 2\alpha \quad mol\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 2\alpha \quad mol\quad \quad \quad \quad \quad \quad \quad 2\alpha \quad mol$

$Equilibrium\quad \quad 2(1-\alpha )\quad mol\quad \quad \quad \quad \quad \quad \quad 2\alpha \quad mol\quad \quad \quad \quad \quad \quad \quad 2\alpha \quad mol$

Where, degree of dissociation, $\alpha$ = 40% = 0.4

Volume of equilibrium mixture, V = 2L  

At equilibrium the molar concentrations of the components of the mixture are  

$\left[ P{ Cl }_{ 5 }(g) \right] =\frac { 2(1-\alpha ) }{ V } =\frac { 2(1-0.4) }{ 2 }=0.6mol.{ L }^{ -1 }$

$\left[ P{ Cl }_{ 3 }(g) \right] =\frac { 2\alpha  }{ V } =\frac { 2\times0.4 }{ 2 }=0.4mol.{ L }^{ -1 }$

$\left[ { Cl }_{ 2 }(g) \right]=\frac { 2\alpha  }{ V }=\frac { 2\times0.4 }{ 2 }=0.4mol.{ L }^{ -1 }$

Equilibrium constant,

${ { K }_{ c }=\frac { \left[ P{ Cl }_{ 3 }(g) \right] \left[ { Cl }_{ 2 }(g) \right]  }{ \left[ P{ Cl }_{ 5 }(g) \right]  }$

$=\frac { 0.4\times0.4 }{ 0.6 }mol.{ L }^{ -1 }$

${ K }_{ c }=0.26mol.{ L }^{ -1 }$

Therefore, equilibrium constant of the reaction is 0.26 $mol.{ L }^{ -1 }$

Answer 2

Answer:

PCl

5

⇌PCl

3

+Cl

2

2moles − −

2(1−α) 2α 2α α=40% =0.4

V=2lit

[PCl

5

]=

V

2(1−α)

=

2

2(1−0.4)

=0.6mol/L

[PCl

3

]=[Cl

2

]=

2

2(0.4)

=0.4mol/L

K

C

=

0.6

0.4×0.4

=0.27molL

−1