I try to follow the equilibrium calculations within COFE and Gibbs reactor for steam reformation.
CH4+H2O->CO+3H2
p=1.9 bar = 1.9*10^5 Pa
T=923K
x(H2O)=0.75
x(CH4)=0.25
I took simulated equilibrium mole fractions (x) from COFE simulation results, multiplied them with the fugacity coefficients (f) from COFE simulation results and calculated equilibirum constant Kx from those values.
Kx=(x(H2)*f(H2))^3*x(CO)*f(CO)/(x(H2O)*f(H2O)*x(CH4)*f(CH4))
Now I tried to optain the same Kx value via Gibbs energy.
For each component I took enthalpyF and entropyF from COFE simualtion results.
(enthalpyF seems to be enthalpy of formation at standard conditions + enthalpy delta at reaction conditions, same for entropy)
From those values I calculated reaction enthalpy and reaction entropy.
From those I obtained Gibbs energy Gr = Hr-T*Sr
From Gr I got K+=e^(-Gr/)(R*T)
From K+ I obtained Kx by: Kx=K+*(101325Pa/p Simulation)^2
However, Kx values calculated both ways do not match.
Kx calculated from component concentrations is 1,15
Kx calculated from Gibbs energy is 0,32.
Did I miss anything here?
Gibbs Reactor equilibrium calculations
Moderator: jasper
Gibbs Reactor equilibrium calculations
Last edited by Stephan on 10 February 2021, 08:46, edited 1 time in total.
Re: Gibbs Reactor equilibrium calculations
I am unable to rework your numbers. You took pure compounds heat of formation and abs entropy? These are at reference conditions - in this case you miss the Cp contributions.
Re: Gibbs Reactor equilibrium calculations
I reworked the post.jasper wrote:I am unable to rework your numbers. You took pure compounds heat of formation and abs entropy? These are at reference conditions - in this case you miss the Cp contributions.
Re: Gibbs Reactor equilibrium calculations
Taking the pure compound entropy and enthalpy contributions (at proper T,P, for composition of 1?) you still miss out on the excess terms.
Re: Gibbs Reactor equilibrium calculations
Perhaps use the partial molar enthalpies (at proper T,P,X) and entropies instread? These are equal to the mole number derivatives, or enthalpyF.Dmoles and entropyF.Dmoles (available from the calculator).
Re: Gibbs Reactor equilibrium calculations
Does the Gibbs reactor use enthalpyF.Dmoles and entropyF.Dmoles?
Do those include the excess terms?
Do those include the excess terms?
Re: Gibbs Reactor equilibrium calculations
No - it uses mixture enthalpyF and entropyF and mimimizes Gibbs energy based on that (at constant T&P; at constant P&H it maximizes EntropyF, both multiplied by total flow to give J/s or J/K/s units) using a directional search (which yes, requires enthalpyF.Dmoles and entropyF.Dmoles). Yes - those include excess terms.