Calculating Average Molecular Weight and Equilibrium Constants for Gas Mixtures

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SUMMARY

This discussion focuses on calculating the average molecular weight and equilibrium constants for a gas mixture of nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4). The correct formula for average molecular weight is M(mix) = X(NO2)M(NO2) + X(N2O4)M(N2O4), where X represents mole fractions. The equilibrium constant K is derived from K = n(N2O4)/((2)n(NO2)), and the Gibbs free energy change is calculated using ΔG = -RTlnK. The discussion emphasizes the importance of accurately determining mole fractions and the initial number of moles for precise calculations.

PREREQUISITES
  • Understanding of gas laws, specifically the Ideal Gas Law (PV = nRT).
  • Familiarity with mole fractions and their calculation.
  • Knowledge of thermodynamic concepts, including Gibbs free energy and equilibrium constants.
  • Ability to perform stoichiometric calculations involving chemical reactions.
NEXT STEPS
  • Learn how to calculate mole fractions from experimental data.
  • Study the derivation and application of the Ideal Gas Law in various scenarios.
  • Explore the relationship between Gibbs free energy and equilibrium constants in chemical reactions.
  • Investigate the concept of degree of dissociation in equilibrium systems.
USEFUL FOR

Chemistry students, chemical engineers, and researchers involved in gas-phase reaction analysis and thermodynamics will benefit from this discussion.

  • #31
x + y = 1
92.011 x + 46.0055 y = 77.4395

x= -y + 1
92.011 (-y +1) + 46.0055 y = 77.4395
-92.011 y + 92.011 + 46.0055 y = 77.4395
-46.0055 y + 92.011= 77.4395
-46.0055 y = -14.5715
y = 0.316734
1- y = x
x = 0.683266

To find the degree of dissociation I did this:

n(N2O4)=77.4395/92=0.841 mol

degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24%


To find Kp I did:

n(N2O4)=77.4395/92=0.841 mol
n(NO2)= 77.4395/46=1.683 mol

Kp= n(N2O4)/(n(NO2))^2= 0.841/(1.683)^2= 0.296912

I think to find delta G standard I'd want to use this equation: delta G= -RTlnKp

and for delta S: delta S= delta G/T

Did I go about any of that right or is that not what I'm suppose to do?
 
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  • #32
tag16 said:
y = 0.316734
1- y = x
x = 0.683266

Looks OK.

To find the degree of dissociation I did this:

n(N2O4)=77.4395/92=0.841 mol

degree dissociation: x(N2O4)/n(N2O4)= 0.683266/0.841= 0.812445= 81.24%

So you have calculated composition of the mixture only to ignore it now and not use the information in following calculations?
 
  • #33
Is this right? If not I don't know how to find it.

PV = nRT --> n =PV/RT--> n = (1 − α)n + 2αn = PV/RT

α = (PV/nRT)− 1
 
  • #34
tag16 said:
n = (1 − α)n + 2αn

You have two different n in this equation - on the left, total number of moles after reaction took place, on the right - initial number of moles.

You know molar fractions of both gases present. Calculate numbers of moles of both. Use stoichiometry of the reaction to calculate initial number of moles of gas. Use definition of dissociation degree.
 

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