# Chemical equilibrium help

Chemical equilibrium help!!!!

For a reaction 2SO2(g) + O2(g) 2 SO3(g), 0.1mol of each SO2 and SO3 are mixed in a 2.0L flask at 27 degrees Celsius. After Equilibrium total pressure is 2.78atm.
Calculate a) The mole fraction of O2 at equilibrium
b) The value of Kp

I dont know how to find out the mole fraction of O2...
I did the problem finding Kp.. but I got a large number of 19.31....
And my Nt at equilibrium= n(4-alpha)

Any ideas??

Here is my work:

2SO2(g) + O2(g) 2 SO3(g)
2n 0 2n
[email protected] [email protected] [email protected]

[email protected] [email protected] 2n [email protected] Where Nt= [email protected]@[email protected]
2n([email protected]) [email protected] 2n([email protected]) [email protected] Nt= n([email protected])

2n([email protected])/n([email protected]) [email protected]/([email protected]) 2n([email protected])/ (n ([email protected]))

2([email protected])/([email protected]) Pt 1/4 Pt 2([email protected])/ ([email protected]))

Where Kp= (P SO3(g))^2 / ((P SO2)^2 * (P O2))

Kp= ((4([email protected])^2) / (([email protected])^2 ) ) 1/Pt

Where Nt at equilibrium= ( 2.78atm*2L) /( 0.08206 atm dm3 mol-1 k-1)(301.15)
Nt= .2249 moles

solving for @
.2449= .1([email protected])
@= 1.751

Plugging in @ to find Kp.... I got 19.31....... but i'ts wrong it's supposed to be 0.356

try to convert the Kp to Kc.

epenguin
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