kp question :D some rep available :D

SO2 and 02 were mixed in the ration of 2:1. When equilibrium had reached the total pressure in the vessel was 120kpa and them ole fraction of SO2 present in the mixture was 0.9. Calculate kp

so i set it out like this
2SO2 + 02 = 2SO3
Initial 2 1 0
final 0.9

Im lost as to how to find out so2 and o2 moles at equilibrium. Can someone help

Reply 1

Eau

I think, because the partial pressure of gases as the whole will never change:

2SO2 + 02 = 2SO3

Initial: 2x : x : 0

Change: -1.1x: - 0.55x : + 1.1x

Final: 0.9x : 0.45x : 1.1x

0.9x + 0.45x + 1.1x = 120

x = 48.98 kPa

Kp = 40.2 kPa

celeritas

Reply 2

giggsy

thanks dude, but whats with -1.1x and wheres 0.55 come from?

Reply 3

giggsy

crap crap crap!!!, ive worded it wrong, theres acutally 0.9moles of S03 present its NOT S02

Reply 4

oceane

2SO2 + O2 <-> 2SO3
Start: 2x : x : 0
At eq. 2x-y : x- 0.5y : y where y=9/10
At eq. 2x-9/10 : x-9/20 : 9/10

If the mole fraction SO3 is 9/10 this means the mole fraction SO2 plus the mole fraction O2 added together equals 1/10.

So 2x - 9/10 + x - 9/20 = 1/10
3x = 29/20 so x=29/60

So at eq, u have 2*(29/60) - 9/10 = 1/15 moles of SO2.
and 29/60 - 9/20 = 1/30 moles of O2.

Kp = (partial press. SO3) squared
_________________________
(partial press. SO2) squared * (partial press. O2)

where partial pressure = mole fraction * total pressure in atmospheres

Total pressure P = 120 000Pa where 101325 Pa = 1 atm, so 1.184 atm

Kp = (9/10)^2 * P^2
__________________
(1/15)^2 * P^2 * (1/30) * P

Kp = 4617 (dimensionless ration)

Reply 5

Mustard-man

Sorry giggsy mate, looks like she beat me to it :p:

. Well if there's anything you don't understand just message me on messenger.

Reply 6

giggsy

i dont get how you work out so2 and 02 moles at final stage when S03 moles is

Reply 7

oceane

giggsy

i dont get how you work out so2 and 02 moles at final stage when S03 moles is

U don't ever know what the actual no. of moles of each is - but you're told the mole fraction is 0.9.

So nine tenths of the equilibrium mixture will be the product, SO3, and the remaining one tenth must be made up of SO2 and 02.

And u know that for every one part of O2 used up, 2 parts of SO2 are used up so u end up with the eqtn I put earlier in the thread involving the x's.

Any clearer? If not, try the M-man!