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Grade: 12th pass

                        

At some temperature and under a pressure of 4 atm PCL5 is 10% dissociated calculate the pressure at with PCL5 will be 20% dissociated temperature remaining same

3 years ago

Answers : (1)

Arun
25211 Points
							

We have,                             PCl5        ↔          PCl3        +             Cl2

Mole before dissociation      1                              0                          0

Mole after dissociation        1 – x                       x                             x

Given, x = 0.1 at 4 atm pressure

Therefore,    KP            =             {[nPCl3 × nCl2]/nPCl5} × [P/∑n]Δn

=             [x.x/(1 – x)]×[P/(1+x)]1

=             Px2/(1 – x2)

=             4×(0.1)2/{1 – (0.1)2}

=             0.040 atm

Again when x is desired at 0.2, KP remaining constant and thus

KP            =             Px2/(1 – x2)

Or,             0.040     =             {P×(0.2)2}/{1 – (0.2)2}

Or,       P             =             0.96 atm  

3 years ago
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