## Gibbs energy minimization and the NIST webbook

| categories: optimization | View Comments

Gibbs energy minimization and the NIST webbook

# Gibbs energy minimization and the NIST webbook

John Kitchin

## Contents

In Post 1536 we used the NIST webbook to compute a temperature dependent Gibbs energy of reaction, and then used a reaction extent variable to compute the equilibrium concentrations of each species for the water gas shift reaction.

Today, we look at the direct minimization of the Gibbs free energy of the species, with no assumptions about stoichiometry of reactions. We only apply the constraint of conservation of atoms. We use the NIST Webbook to provide the data for the Gibbs energy of each species.

As a reminder we consider equilibrium between the species , , and , at 1000K, and 10 atm total pressure with an initial equimolar molar flow rate of and .

H2

H2O

CO

CO2

function nist_wb_constrained_min

T = 1000; %K
R = 8.314e-3; % kJ/mol/K

P = 10; % atm, this is the total pressure in the reactor
Po = 1; % atm, this is the standard state pressure


## We define species like this.

We are going to store all the data and calculations in vectors, so we need to assign each position in the vector to a species. Here are the definitions we use in this work.

1  CO
2  H2O
3  CO2
4  H2
species = {'CO' 'H2O' 'CO2' 'H2'};


## Heats of formation at 298.15 K

Hf298 = [
-110.53  % CO
-241.826 % H2O
-393.51  % CO2
0.0]; % H2


## Shomate parameters for each species

   A             B           C          D          E       F          G       H
WB = [25.56759  6.096130     4.054656  -2.671301  0.131021 -118.0089 227.3665   -110.5271  % CO
30.09200  6.832514     6.793435  -2.534480  0.082139 -250.8810 223.3967   -241.8264    % H2O
24.99735  55.18696   -33.69137    7.948387 -0.136638 -403.6075 228.2431   -393.5224   % CO2
33.066178 -11.363417  11.432816  -2.772874 -0.158558 -9.980797 172.707974    0.0]; % H2


## Shomate equations

t = T/1000;
T_H = [t; t^2/2; t^3/3; t^4/4; -1/t; 1; 0; -1];
T_S = [log(t); t; t^2/2; t^3/3; -1/(2*t^2); 0; 1; 0];

H = WB*T_H; % (H - H_298.15) kJ/mol
S = WB*T_S/1000; % absolute entropy kJ/mol/K

Gjo = Hf298 + H - T*S; % Gibbs energy of each component at 1000 K


## Gibbs energy of mixture

this function assumes the ideal gas law for the activities

    function G = func(nj)
Enj = sum(nj);
G = sum(nj.*(Gjo'/R/T + log(nj/Enj*P/Po)));
end


## Elemental conservation constraints

We impose the constraint that all atoms are conserved from the initial conditions to the equilibrium distribution of species. These constraints are in the form of A_eq x = beq, where x is the vector of mole numbers. CO H2O CO2 H2

Aeq = [ 1    0    1    0   % C balance
1    1    2    0   % O balance
0    2    0    2]; % H balance

% equimolar feed of 1 mol H2O and 1 mol CO
beq = [1   % mol C fed
2   % mol O fed
2]; % mol H fed


## Constraints on bounds

This simply ensures there are no negative mole numbers.

LB = [0 0 0];


## Solve the minimization problem

x0 = [0.5 0.5 0.5 0.5]; % initial guesses

options = optimset('Algorithm','sqp');
x = fmincon(@func,x0,[],[],Aeq,beq,LB,[],[],options)

Local minimum found that satisfies the constraints.

Optimization completed because the objective function is non-decreasing in
feasible directions, to within the default value of the function tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.

x =

0.4549    0.4549    0.5451    0.5451



## Compute mole fractions and partial pressures

The pressures here are in good agreement with the pressures found in Post 1536 . The minor disagreement (in the third or fourth decimal place) is likely due to convergence tolerances in the different algorithms used.

yj = x/sum(x);
Pj = yj*P;

for i=1:numel(x)
fprintf('%5d%10s%15.4f atm\n',i, species{i}, Pj(i))
end

    1        CO         2.2745 atm
2       H2O         2.2745 atm
3       CO2         2.7255 atm
4        H2         2.7255 atm


## Computing equilibrium constants

We can compute the equilibrium constant for the reaction . Compared to the value of K = 1.44 we found at the end of Post 1536 , the agreement is excellent. Note, that to define an equilibrium constant it is necessary to specify a reaction, even though it is not necessary to even consider a reaction to obtain the equilibrium distribution of species!

nuj = [-1 -1 1 1]; % stoichiometric coefficients of the reaction
K = prod(yj.^nuj)

K =

1.4359


end

% categories: optimization
% tags: Thermodynamics, reaction engineering


## Finding equilibrium composition by direct minimization of Gibbs free energy on mole numbers

| categories: optimization | View Comments

Finding equilibrium composition by direct minimization of Gibbs free energy on mole numbers

# Finding equilibrium composition by direct minimization of Gibbs free energy on mole numbers

John Kitchin

Adapted from problem 4.5 in Cutlip and Shacham

## Problem setup

Ethane and steam are fed to a steam cracker at a total pressure of 1 atm and at 1000K at a ratio of 4 mol H2O to 1 mol ethane. Estimate the equilibrium distribution of products (CH4, C2H4, C2H2, CO2, CO, O2, H2, H2O, and C2H6).

Solution method: We will construct a Gibbs energy function for the mixture, and obtain the equilibrium composition by minimization of the function subject to elemental mass balance constraints.

function main

R = 0.00198588; % kcal/mol/K
T = 1000; % K

% we store the species names in a cell array since they have different
% lengths (i.e. different number of characters in names).
species = {'CH4' 'C2H4' 'C2H2' 'CO2' 'CO' 'O2' 'H2' 'H2O' 'C2H6'};

%
% $G_^\circ for each species. These are the heats of formation for each % species. Gjo = [4.61 28.249 40.604 -94.61 -47.942 0 0 -46.03 26.13]; % kcal/mol  ## The Gibbs energy of a mixture We start with . Recalling that we define , and in the ideal gas limit, , and that . Since in this problem, P = 1 atm, this leads to the function .  function G = func(nj) Enj = sum(nj); G = sum(nj.*(Gjo/R/T + log(nj/Enj))); end  ## Linear equality constraints for atomic mass conservation The total number of each type of atom must be the same as what entered the reactor. These form equality constraints on the equilibrium composition. We express these constraints as: where is a vector of the moles of each species present in the mixture. CH4 C2H4 C2H2 CO2 CO O2 H2 H2O C2H6 Aeq = [0 0 0 2 1 2 0 1 0 % oxygen balance 4 4 2 0 0 0 2 2 6 % hydrogen balance 1 2 2 1 1 0 0 0 2]; % carbon balance % the incoming feed was 4 mol H2O and 1 mol ethane beq = [4 % moles of oxygen atoms coming in 14 % moles of hydrogen atoms coming in 2]; % moles of carbon atoms coming in  ## Bounds on mole numbers No mole number can be negative, so we define a lower bound of zero for each mole number. In principle there are upper bounds too, but it is unecessary to define them here. LB = [0 0 0 0 0 0 0 0 0]; % no mole numbers less than zero  ## initial guess for the solver these are the guesses suggested in the book. x0 = [1e-3 1e-3 1e-3 0.993 1 1e-4 5.992 1 1e-3]; % initial guess  ## Setup minimization fmincon does not solve this problem with the default solver, and warns you to pick another one, either 'interior-point' or 'sqp'. Both work for this problem. options = optimset('Algorithm','sqp'); [x fval] = fmincon(@func,x0,[],[],Aeq,beq,LB,[],[],options); for i=1:numel(x) fprintf('%5d%10s%10.3g\n',i,species{i},x(i)) end  Local minimum possible. Constraints satisfied. fmincon stopped because the size of the current step is less than the default value of the step size tolerance and constraints are satisfied to within the default value of the constraint tolerance. 1 CH4 0.0664 2 C2H4 8.88e-008 3 C2H2 2.76e-021 4 CO2 0.545 5 CO 1.39 6 O2 4.56e-021 7 H2 5.35 8 H2O 1.52 9 C2H6 1.6e-007  ## Verify the equality constraints were met you can see here the constraints were met. Aeq*x' beq  ans = 4 14 2 beq = 4 14 2  ## Results Interestingly there is a distribution of products! That is interesting because only steam and ethane enter the reactor, but a small fraction of methane is formed! The main product is hydrogen. The stoichiometry of steam reforming is ideally . Even though nearly all the ethane is consumed, we don't get the full yield of hydrogen. It appears that another equilibrium, one between CO, CO2, H2O and H2, may be limiting that, since the rest of the hydrogen is largely in the water. It is also of great importance that we have not said anything about reactions, i.e. how these products were formed. This is in stark contrast to Post 1589 where stoichiometric coefficients were necessary to perform the minimization. ## Water gas shift equilibrium The water gas shift reaction is: . We can compute the Gibbs free energy of the reaction from the heats of formation of each species. Assuming these are the formation energies at 1000K, this is the reaction free energy at 1000K. G_wgs = Gjo(4) + Gjo(7) - Gjo(5) - Gjo(8)  G_wgs = -0.6380  ## WGS equilibrium constant at 1000K exp(-G_wgs/R/T)  ans = 1.3789  ## Equilibrium constant based on mole numbers One normally uses activities to define the equilibrium constant. Since there are the same number of moles on each side of the reaction all factors that convert mole numbers to activity, concentration or pressure cancel, so we simply consider the ratio of mole numbers here. x(4)*x(7)/(x(5)*x(8))  ans = 1.3789  Clearly, there is an equilibrium between these species that prevents the complete reaction of steam reforming. ## Summary This is an appealing way to minimize the Gibbs energy of a mixture. No assumptions about reactions are necessary, and the constraints are easy to identify. The Gibbs energy function is especially easy to code. end % categories: optimization % tags: Thermodynamics, reaction engineering % post_id = 1630; %delete this line to force new post; % permaLink = http://matlab.cheme.cmu.edu/2011/12/25/finding-equilibrium-composition-by-direct-minimization-of-gibbs-free-energy-on-mole-numbers/;  Read and Post Comments ## Constrained optimization | categories: optimization | View Comments constrained_minimization ## Contents ## Constrained optimization John Kitchin adapted from http://en.wikipedia.org/wiki/Lagrange_multipliers. function constrained_minimization  clear all; close all  ## Introduction Suppose we seek to minimize the function subject to the constraint that . The function we seek to maximize is an unbounded plane, while the constraint is a unit circle. In Post 1602 we setup a Lagrange multiplier approach to solving this problem. Today, we use the builtin function fmincon in Matlab to solve the same problem. We plot these two functions here. x = linspace(-1.5,1.5); [X,Y] = meshgrid(x,x); figure; hold all surf(X,Y,X+Y); shading interp;  ## now plot the circle on the plane. First we define a circle in polar coordinates, then convert those coordinates to x,y cartesian coordinates. then we plot f(x1,y1). theta = linspace(0,2*pi); [x1,y1] = pol2cart(theta,1); plot3(x1,y1,x1+y1) view(38,8) % adjust view so it is easy to see  ## Construct the function to optimize and the nonlinear constraint function the original function, provided that the constraint is met!  function f = func(X) x = X(1); y = X(2); f = x + y; end function [c,ceq] = nlconstraints(X) % function for the nonlinear constraints. % We have to define two outputs for fmincon: the nonlinear % inequality contraints, and the equality constraints. % % C(X) <= 0 % % Ceq(X) = 0 x = X(1); y = X(2); c = []; ceq = x^2 + y^2 - 1; end  ## Now we solve for the problem fmincon takes a lot of options. We define all of them here, even though in this example we do not have any linear inequality or equality constraints, and we don't define bounds on the solution. It is necessary to put these empty placeholders in the fmincon call, however, so that it knows what the nonlinear constraints are. A = []; B = []; % the linear inequality constraints: A*X <= B Aeq = []; Beq = []; % the linear equality constraints: Aeq*X = B LB = []; UB = []; % LB <= X <= UB X0 = [1, 1]; % initial guess [X,FVAL,EXITFLAG,OUTPUT,LAMBDA] = fmincon(@func,X0,A,B,Aeq,Beq,LB,UB,@nlconstraints) OUTPUT.message plot3(X(1),X(2),FVAL,'bo','markerfacecolor','b')  Warning: The default trust-region-reflective algorithm does not solve problems with the constraints you have specified. FMINCON will use the active-set algorithm instead. For information on applicable algorithms, see Choosing the Algorithm in the documentation. Local minimum possible. Constraints satisfied. fmincon stopped because the predicted change in the objective function is less than the default value of the function tolerance and constraints are satisfied to within the default value of the constraint tolerance. X = 0.7071 0.7071 FVAL = 1.4142 EXITFLAG = 5 OUTPUT = iterations: 5 funcCount: 15 lssteplength: 1 stepsize: 5.9440e-005 algorithm: [1x44 char] firstorderopt: 7.9553e-006 constrviolation: 2.3986e-011 message: [1x776 char] LAMBDA = lower: [2x1 double] upper: [2x1 double] eqlin: [1x0 double] eqnonlin: 0.7071 ineqlin: [1x0 double] ineqnonlin: [1x0 double] ans = Local minimum possible. Constraints satisfied. fmincon stopped because the predicted change in the objective function is less than the default value of the function tolerance and constraints are satisfied to within the default value of the constraint tolerance. Stopping criteria details: Optimization stopped because the predicted change in the objective function, 6.856786e-010, is less than options.TolFun = 1.000000e-006, and the maximum constraint violation, 2.398637e-011, is less than options.TolCon = 1.000000e-006. Optimization Metric Options abs(steplength*directional derivative) = 6.86e-010 TolFun = 1e-006 (default) max(constraint violation) = 2.40e-011 TolCon = 1e-006 (default)  Note the warning about the algorithm change. Matlab automatically detected that it could not use the default algorithm because of the nonlinear constraints. Matlab also automatically selected a better algorithm for you. There is a lot of output, that mostly tells us the function worked as expected. Note that the solution above is not a global minimum. There is another solution at the bottom of the circle. Let's try another initial guess. X0 = [-1, -1]; [X,FVAL,EXITFLAG,OUTPUT,LAMBDA] = fmincon(@func,X0,A,B,Aeq,Beq,LB,UB,@nlconstraints) plot3(X(1),X(2),FVAL,'ro','markerfacecolor','r')  Warning: The default trust-region-reflective algorithm does not solve problems with the constraints you have specified. FMINCON will use the active-set algorithm instead. For information on applicable algorithms, see Choosing the Algorithm in the documentation. Local minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance. X = -0.7071 -0.7071 FVAL = -1.4142 EXITFLAG = 1 OUTPUT = iterations: 4 funcCount: 15 lssteplength: 1 stepsize: 1.5019e-006 algorithm: [1x44 char] firstorderopt: 1.0880e-008 constrviolation: 2.2982e-012 message: [1x787 char] LAMBDA = lower: [2x1 double] upper: [2x1 double] eqlin: [1x0 double] eqnonlin: 0.7071 ineqlin: [1x0 double] ineqnonlin: [1x0 double]  This is the global minimum (by inspection). ## Summary the built in function fmincon is more flexible than what we did in Post 1602 since it includes inequality constraints. Of course, that flexibility comes at some cost, you have to know the expected syntax for each kind of constraint. But that is easy to find out with the Matlab documentation. end % categories: optimization  Read and Post Comments ## Using Lagrange multipliers in optimization | categories: optimization | View Comments lagrange_multiplier ## Contents ## Using Lagrange multipliers in optimization John Kitchin adapted from http://en.wikipedia.org/wiki/Lagrange_multipliers. function lagrange_multiplier  clear all; close all  ## Introduction Suppose we seek to maximize the function subject to the constraint that . The function we seek to maximize is an unbounded plane, while the constraint is a unit circle. We plot these two functions here. x = linspace(-1.5,1.5); [X,Y] = meshgrid(x,x); figure; hold all surf(X,Y,X+Y); shading interp;  ## now plot the circle on the plane. First we define a circle in polar coordinates, then convert those coordinates to x,y cartesian coordinates. then we plot f(x1,y1). theta = linspace(0,2*pi); [x1,y1] = pol2cart(theta,1); plot3(x1,y1,x1+y1) view(38,8) % adjust view so it is easy to see  ## Construct the Lagrange multiplier augmented function To find the maximum, we construct the following function: where , which is the constraint function. Since , we aren't really changing the original function, provided that the constraint is met!  function gamma = func(X) x = X(1); y = X(2); lambda = X(3); gamma = x + y + lambda*(x^2 + y^2 - 1); end  ## Finding the partial derivatives The minima/maxima of the augmented function are located where all of the partial derivatives of the augmented function are equal to zero, i.e. , , and . the process for solving this is usually to analytically evaluate the partial derivatives, and then solve the unconstrained resulting equations, which may be nonlinear. Rather than perform the analytical differentiation, here we develop a way to numerically approximate the partial derivates.  function dLambda = dfunc(X)  function to compute the partial derivatives of the function defined above at the point X. We use a central finite difference approach to compute .  % initialize the partial derivative vector so it has the same shape % as the input X dLambda = nan(size(X)); h = 1e-3; % this is the step size used in the finite difference. for i=1:numel(X) dX=zeros(size(X)); dX(i) = h; dLambda(i) = (func(X+dX)-func(X-dX))/(2*h); end   end  ## Now we solve for the zeros in the partial derivatives The function we defined above (dfunc) will equal zero at a maximum or minimum. It turns out there are two solutions to this problem, but only one of them is the maximum value. Which solution you get depends on the initial guess provided to the solver. Here we have to use some judgement to identify the maximum. X1 = fsolve(@dfunc,[1 1 0],optimset('display','off')) % let's check the value of the function at this solution fval1 = func(X1) sprintf('%f',fval1)  I don't know why the output in publish doesn't show up in the right place. sometimes that happens. The output from the cell above shows up at the end of the output for some reason. ## Add solutions to the graph plot3(X1(1),X1(2),fval1,'ro','markerfacecolor','b')  ## Summary Lagrange multipliers are a useful way to solve optimization problems with equality constraints. The finite difference approach used to approximate the partial derivatives is handy in the sense that we don't have to do the calculus to get the analytical derivatives. It is only an approximation to the partial derivatives though, and could be problematic for some problems. An adaptive method to more accurately estimate the derivatives would be helpful, but could be computationally expensive. The significant advantage of the numerical derivatives is that it works no matter how many variables there are without even changing the code! There are other approaches to solving this kind of equation in Matlab, notably the use of fmincon. 'done'  ans = done  end % categories: optimization  X1 = 0.7071 0.7071 -0.7071 fval1 = 1.4142 ans = 1.414214  Read and Post Comments ## The Gibbs free energy of a reacting mixture and the equilibrium composition | categories: optimization | View Comments The Gibbs free energy of a reacting mixture and the equilibrium composition # The Gibbs free energy of a reacting mixture and the equilibrium composition John Kitchin Adapted from Chapter 3 in Rawlings and Ekerdt. ## Contents ## Introduction In this post we derive the equations needed to find the equilibrium composition of a reacting mixture. We use the method of direct minimization of the Gibbs free energy of the reacting mixture. The Gibbs free energy of a mixture is defined as where is the chemical potential of species , and it is temperature and pressure dependent, and is the number of moles of species . We define the chemical potential as , where is the Gibbs energy in a standard state, and is the activity of species if the pressure and temperature are not at standard state conditions. If a reaction is occurring, then the number of moles of each species are related to each other through the reaction extent and stoichiometric coefficients: . Note that the reaction extent has units of moles. Combining these three equations and expanding the terms leads to: The first term is simply the initial Gibbs free energy that is present before any reaction begins, and it is a constant. It is difficult to evaluate, so we will move it to the left side of the equation in the next step, because it does not matter what its value is since it is a constant. The second term is related to the Gibbs free energy of reaction: . With these observations we rewrite the equation as: Now, we have an equation that allows us to compute the change in Gibbs free energy as a function of the reaction extent, initial number of moles of each species, and the activities of each species. This difference in Gibbs free energy has no natural scale, and depends on the size of the system, i.e. on . It is desirable to avoid this, so we now rescale the equation by the total initial moles present, and define a new variable , which is dimensionless. This leads to: where is the initial mole fraction of species present. The mole fractions are intensive properties that do not depend on the system size. Finally, we need to address . For an ideal gas, we know that , where the numerator is the partial pressure of species computed from the mole fraction of species times the total pressure. To get the mole fraction we note: This finally leads us to an equation that we can evaluate as a function of reaction extent: we use a double tilde notation to distinguish this quantity from the quantity derived by Rawlings and Ekerdt which is further normalized by a factor of . This additional scaling makes the quantities dimensionless, and makes the quantity have a magnitude of order unity, but otherwise has no effect on the shape of the graph. Finally, if we know the initial mole fractions, the initial total pressure, the Gibbs energy of reaction, and the stoichiometric coefficients, we can plot the scaled reacting mixture energy as a function of reaction extent. At equilibrium, this energy will be a minimum. We consider the example in Rawlings and Ekerdt where isobutane (I) reacts with 1-butene (B) to form 2,2,3-trimethylpentane (P). The reaction occurs at a total pressure of 2.5 atm at 400K, with equal molar amounts of I and B. The standard Gibbs free energy of reaction at 400K is -3.72 kcal/mol. Compute the equilibrium composition. function main  clear all; close all u = cmu.units; R = 8.314*u.J/u.mol/u.K; P = 2.5*u.atm; Po = 1*u.atm; % standard state pressure T = 400*u.K; Grxn = -3.72*u.kcal/u.mol; %reaction free energy at 400K yi0 = 0.5; yb0 = 0.5; yp0 = 0.0; % initial mole fractions yj0 = [yi0 yb0 yp0]; nu_j = [-1 -1 1]; % stoichiometric coefficients  ## Define a function for We will call this difference Gwigglewiggle.  function diffG = Gfunc(extentp) diffG = Grxn*extentp; sum_nu_j = sum(nu_j); for i = 1:length(yj0) x1 = yj0(i) + nu_j(i)*extentp; x2 = x1./(1+extentp*sum_nu_j); diffG = diffG + R*T*x1.*log(x2*P/Po); end end  ## Plot function There are bounds on how large can be. Recall that , and that . Thus, , and the maximum value that can have is therefore where . When there are multiple species, you need the smallest to avoid getting negative mole numbers. epsilonp_max = min(-yj0(yj0>0)./nu_j(yj0>0)) epsilonp = linspace(0,epsilonp_max,1000); plot(epsilonp,Gfunc(epsilonp)) xlabel('\epsilon''') ylabel('Gwigglewiggle')  epsilonp_max = 0.5000  ## Find extent that minimizes Gwigglewiggle fminbnd is not defined for the units class, so we wrap the function in a function handle that makes it output a double. Alternatively, we could make the function fully dimensionless. We save that for a later post. f = @(x) double(Gfunc(x)) epsilonp_eq = fminbnd(f,0,epsilonp_max) hold on plot(epsilonp_eq,Gfunc(epsilonp_eq),'ro')  f = @(x)double(Gfunc(x)) epsilonp_eq = 0.4696  ## Compute equilibrium mole fractions yi = (yi0 + nu_j(1)*epsilonp_eq)/(1 + epsilonp_eq*sum(nu_j)) yb = (yb0 + nu_j(2)*epsilonp_eq)/(1 + epsilonp_eq*sum(nu_j)) yp = (yp0 + nu_j(3)*epsilonp_eq)/(1 + epsilonp_eq*sum(nu_j))  yi = 0.0572 yb = 0.0572 yp = 0.8855  ## alternative way to compute mole fractions y_j = (yj0 + nu_j*epsilonp_eq)/(1 + epsilonp_eq*sum(nu_j))  y_j = 0.0572 0.0572 0.8855  ## Check for consistency with the equilibrium constant K = exp(-Grxn/R/T)  K = 108.1300  . yp/(yi*yb)*Po/P  ans = 108.0702  ## an alternative form of the equilibrium constant We can express the equilibrium constant like this :$K = \prod\limits_j a_j^{\nu_j}\$, and compute it with a single command.

prod((y_j*P/Po).^nu_j)

ans =

108.0702



These results are very close, and only disagree because of the default tolerance used in identifying the minimum of our function. you could tighten the tolerances by setting options to the fminbnd function.

## Summary

In this post we derived an equation for the Gibbs free energy of a reacting mixture and used it to find the equilibrium composition. In future posts we will examine some alternate forms of the equations that may be more useful in some circumstances.

end

% categories: optimization
% tags: thermodynamics, reaction engineering