Chemical equilibrium help

  • Thread starter rock23
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  • #1
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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??
 

Answers and Replies

  • #2
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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
 
  • #3
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try to convert the Kp to Kc.
 
  • #4
epenguin
Homework Helper
Gold Member
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You probably haven't received help because your answer is too hard to read.
I don't know what @ or Nt mean.
And just above the input box there are symbols X2 and X2 which allow you to easily write things like [SO3]2 which help legibility.

The way I would find it easiest:

Total S (sulphur) is 0.1 M if I am not mistaken.
From the pressure work out the total molarity of the gas.
The difference is the molarity of O2.
 
Last edited:

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