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#### NISHANT

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« on: January 03, 2005, 11:21:42 AM »
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

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« Reply #1 on: January 03, 2005, 03:11:23 PM »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
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#### Donaldson Tan

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« Reply #2 on: January 03, 2005, 08:22:34 PM »
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
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#### NISHANT

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« Reply #3 on: January 04, 2005, 10:03:24 AM »
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

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