Dissociation of phosphorous penta chloride

Dissociation of  phosphorous penta chloride

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Dissociation of phosphorous penta chloride

Dissociation of PCl₅ takes place accprding to the eqⁿ

PCl₅    PCl₃ +Cl₂

Let ‘a’ moles of phosphorous penta chloride be taken in a closed vessel of Volume ‘V’  litre. It is heated till equilibrium is established, x moles of PCl₅ are dissociated into  PCl₃  &  Cl₂ .  Therefore x moles of PCl₅ give x moles of PCl₃ & x moles of Cl₂.

PCl₅                      PCl₃                +                      Cl₂

Initial Conc. (mole/l)             a/V                                    O                                             O

Concⁿ at eq. (Mol/l)            ( a-x)  /V                             x/V                                         x/V

Applying law of mass action,

Kc =[  PCl₃ ] [ Cl₂ ] /[  PCl₅ ]

Kc =[ ( x/V) ( x/V)]/[ ( a-x)/V]

Kc =x2 /  ( a-x) V

Factors affecting Equilibrium constant-

(i)        Effect of Pressure

            When pressure is increased, the volume is decreased Hence equilibrium will shift in backward direction.  For Kc to remain constt. x will decrease. Hence dissociation of phosphorous penta chloride decreases & equilibrium will shift in the backward direction.

If pressure is decreased, the equilibrium will shift in the forward direction & dissociation of PCl5 increases.

(ii)       Effect of Temp.

PCl₅   PCl₃ +Cl₂  –    Qk. Cal.

Dissociation of PCl₅ is an endothermic reacⁿ. For endothermic reaction, increase in temp. increases the value of equi. constt. & the equilibrium is shifted in the forward direction.

(iii)     Effect of Concⁿ–

If concⁿ of any of the reactants or products is changed , the equilibrium is disturbed but Kc remains constt.

If the concⁿ of PCl₅ is increased then concentration of PCl3 & Cl₂ also increase. Then equilibrium will shift in forward direction & dissociation of PCl5 increases..

(iv)  Effect of  inert gas–  At constt volume no effect of adding inert gas. At constt pressure volume increases on adding inert gas therfore  eqilibrium will shift in forward direction

Q-    3 moles of PCl₅ is heated in a flask of 4 litre volume. At equilibrium it dissociates to give 40%  PCl₃ &  Cl_{2 .} Calculate Kc.

Solⁿ     Amount of dissociation ‘ x ‘= 3 x 40 /100 =1.2 mole

V= 4 litre

PCl₅                    PCl₃                +                      Cl₂

Initial Conc. (mole/l)             3 /V                                   0                                              0

Concⁿ at eq. (Mol/l)             3-1.2 /V                               1.2 /V                                          1.2/V                                                                                                           =1.8/V

Kc =[  PCl₃ ] [ Cl₂ ] /[  PCl₅ ]

Kc =[ ( 1.2/V) ( 1.2/V)]/[1.8/V]

=( 1.2 x 1.2)/( 1.8 x 4)

Kc     =0.2

1.   0.34 moles of PCl₅ is heated in a closed vessel of 1 litre  capacity at 500 K. At equilibrium 0.1 moles of PCl₅ is dissociated. calculate Kc.

Ans.           v=1 litre

PCl₅                                 PCl₃                +                      Cl₂

Initial Conc. (mole/l)             0.34/V                                   O                                             O

Concⁿ at eq. (Mol/l)             0.34-0.1 /V                             0.1/V                                 0.1 /V                                                                                                                 =0.24/V

Kc =[  PCl₃ ] [ Cl₂ ] /[  PCl₅ ]

Kc =[ ( 0.1/V) ( 0.1/V)]/[ 0.24/V]

=(0.1 x 0.1)/(0.24 x 1)

=0.042 Ans

Q –625.5 gm. PCl₅ is heated in a vessel of 5 litre capacity at equilibrium 45% of PCl₅ is dissociated into PCl₃ & Cl2. calculate Kc.=?

Ans.    m of PCl₅  = 31+5+35.5

=208.5

Initial Concⁿ of PCl₅ = 625.5/208.5 =3

Amt. of dissociation of PCl_{5 =(3 x 45)/100}

=1.35  mole

PCl₅                                 PCl₃                +                      Cl₂

Initial Conc. (mole/l)           3 /V                                  O                                             O

Concⁿ at eq. (Mol/l)             3-1.35/V                         1.35  /V                                  1.35 /V                                                                                                                =1.65/V

Kc =[  PCl₃ ] [ Cl₂ ] /[  PCl₅ ]

Kc =[ ( 1.35/V) ( 1.35/V)]/[ ( 1.65)/V]

=(1.35 x 1.35)/(1.65 x 5)

Kc  =0.22 Ans

Relation between degree of dissociation & Vapour density-

Suppose degree of dissociation at equilibrium is x

PCl₅                                 PCl₃                +                      Cl₂

Initial no. of moles               1                                         O                                             O

mole at equi                        1-x                                          x                                              x

Initial no. of moles = 1+0+0=1

Total moles at Equilibrium = 1-x+x+x= 1+x

‘D’ is the vapour density when there is no dissociation

‘d’ is the observed vapour density at a  temp. when degree of dissociation is x

No.of moles ∝  1/vapour density

total moles at equilibrium /initial moles =D/d

(1+x) /1 =D/d

x = (D-d) /d

1. The V. density of PCl₅ at 600K is 61 . Calculate the degree of dissociation of PCl₅ at this temp. Mol wt. of PCl₅= 208.5

Ans.    d=61

D =mol,wt /2 =208.5/2 =104.25

(1+x) /1 =D/d

( 1+x) /1 = 104.25 /61

x =0.709

% of x =70.9 % Ans.