Physical Chemistry Homework: Equilibrium Constant and Gas Properties

  • #1
krootox217
51
2

Homework Statement


I have the following task:

A chemical reaction 2 A(g) <-> B(g) + C(g)
has an equilibrium constant defined by: ln=0.8+ 1050/T−1,3.105/T²
a) What is K at 400K? Calculate ΔrG0
In a volume of 5L you get 2 moles of A and 0.5 moles of B. Reaction still proceeds at 400K. Consider the gas as ideal.
b) What are the amount of compounds when equilibrium is reached? What is the partial pressure of each compounds? What is the average molar mass of the gas mixture?
c) Calculate the kinetic energy of each gaseous compound.

I already have the solutions, but I can't find the right one for b)
K = 13.633
at equilibrium, Molar fraction of A = 0.12 ; B = 0.54 ; C = 0.34
Therefore Partial pressure of A = 2.10^5 Pa ; B = 8,98.10^5 Pa ; C = 5,65.10^5 Pa
average molar mass of mixture : 39.2 g/mol
Ek A = 1,497 kJ ; B = 6,734 kJ ; C = 4,24 kJ


I get the right value for K, and then I tried it this way:

A=2mol-2x
B=0.5mol+x
C=x

13.6=K=(product)/(starting material)=((x)*(0.5+x))/(2-2x) --> x=0.8542

and then I calculated 2mol-2*X= 0.2916 mol
0.2916mol/2.5 mol = 0.12 = molar fraction of A
and I got the right fractions for B and C, and also the right partial pressures

but how do i calculate the average molar mass of the mixture?

Homework Equations


See above

The Attempt at a Solution


See above
 
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  • #2
Weighted average. For given T, P conditions density of an ideal gas depends on its molar mass. For every density there exist an equivalent molar mass - for example for air (approx 80% N2 and 20%O2) the average molar mass is 28.8 g/mol.
 
  • #3
To get the average molar mass, you need to know that molar masses of each individual species. Do they give you that info in the problem statement?
 
  • #4
No, they don't. Therefore I assume they made a mistake, thanks for the help
 
  • #5
Maybe they just expect you to express it algebraically.

By the way, in this equation: 13.6=K=(product)/(starting material)=((x)*(0.5+x))/(2-2x)

that (2-2x) should be squared.
 
  • #6
I don't think so since they gave an exact value in the solutions.

And yes, I did it correctly but made the miskake when I posted it here, thank you
 
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