[NISHANT]

Q:1 mole of N2 and 3 moles of PCl5 are placed in a 100 litre vessel and was heated at 500K.The eq. pressure is 2.05atm.Assuming ideal behaviour,calculate degree of dissocaition of PCl5 and kp of the reaction that occurs.

here it is very clear that N2 will not take part in the reaction.so if we calculate the initial preesure and the partial pressure of N2 and PCl5,partial preesure of N2 or I could say pressure due to the N2 molecules would not change in the course of the reaction.Partial pressure of PCl5 would change and we can easily find the kp of the reaction.
But in the sloution of the book this is not given and even if we calculate all the things by the method which I have supposed Partial pressure of PCl5 AT EQ. COMES OUT TO BE NEGATIVE??

CAN YOU PLEASE SPOT THE MISTAKE

[Mitch]

please read forum rules. It's the first post of all forums.

[Donaldson Tan]

PCl5 <-> PCl3 + Cl2
let x be amount of PCl5 dissociated
amount of PCl5 at equilibrium = 3-x
amount of PCl3 at equilibrium = x
amount of Cl2 at equilibirum = x

V = 100L = 0.1m3
T = 500K

P = nRT/V

partial P of N2 = (1)R(500)/(0.1) = 5000R
partial P of PCl5 = (3-x)R(500)/(0.1) = 5000R(3-x)
partial P of PCl3 = (x)R(500)/(0.1) = 5000R(x)
partial P of Cl2 = (x)R(500)/(0.1) = 5000R(x)

given eqbm P is 2.04atm = 207665Pa
(make sure all your quantities are in SI units)

5000R + 5000R(3-x) + 5000R(x) + 5000R(x) = 207665
5000R(1+3-x+x+x) = 207665
5000R(4+x) = 207665
4+x = 207665/5000R = 4.9955
x = 0.9955 ~ 1 mole

[NISHANT]

but if we calculate the pressure due to N2 in the beginning,it comes out to be different,but N2 has not taken part in the reaction then why so?

[AWK]

From this data, assuming ideal gas behaviour and reaction:
PCl5 = PCl3 + Cl2
we can calculate the total moles of N2 + PCl5 + PCl3 + CL2 in the mixture (4.9963)
Hence, at the equilibrium we have  0.9963 moles of Cl2,  0.9963 moles of PCl3 and 2.0037 moles of PCl5
 N2 do not take part in reaction but modifies partial pressures of PCl5, PCl3 and Cl2