Chemical Equilibrium Analysis Software- Tutorial with Examples

Tutorial (TOC) > Daemons > Equilibrium > Examples

Ex-1 Evaluate the equilibrium constant, expressed as ln K for the reaction CO+0.5O2<----->CO2 at 100 kPa and (a) 298 K, and (b) 1000 K. What-if scenario: (c) How would the answer in part (b) change if the pressure is increased to 1 MPa?

Fig. 1.1 Image of the composition panel. As you select a reactant species, all the probable products species are added to the products panel.

Solution Launch the system state chemical equilibrium daemon located at TEST.Daemons.System.State page.

On the composition panel, set up the stoichiometric reaction. Select CO and O2 as the reactants and sepcify their molar amounts as 1 kmol and 0.5 kmol respectively. After enter the amount in the appropriate cell, press the Enter key to register the change. Select CO2 as the only products.

In the states panel, evaluate the products states at 100 kPa and 298 K, by entering these properties, selecting the Products button, and pressing the Calculate button. In the composition panel, ln(K) is displayed as 103.82. Such a high value of ln(K) indicates that at 298 K, the reaction goes to completion, that is, only CO2 can be expected to be present.

Now calculate state-2 with p2=p1 and T2=1000 K. As shown in the figure above, ln(K) is calculated as 23.523. In Table G.3 (visit TEST.Daemons.Basics.Tables.Equilibrium page), ln(K) for the inverse reaction is listed as -103.762 and -23.529 respectively.

Evaluate state-3 with p3=10 MPa, and T3=T2. A Calculate produces ln(K) as 23.523, verifying the fact that equilibrium constant is not a function of pressure.

Ex-2 Pure oxygen (O2) is heated at a constant pressure of 100 kPa to a temperautre of 2500 deg-C. (a) Determine the (a) mass and (b) mole fraction of oxygen atom in the dissociated mixture. (c) What-if scenario: How would the answers change if the pressure was reduced to 50 kPa?

Solution Oxygen undergoes a constant-pressure dissociation process. We can use either the steady-flow or the closed-process equilibrium daemon.

Launch the IG chemical equilibrium daemon located at State.System page.

On the composition panel, set up the reactants (including amount, default is 1 kg) and select the products.

Switch to the state panel, select Products radio button, enter p1 and T1, and Calculate. The state including its composition is calculated. While the state properties are displayed on the state panel, the composition can be found on the composition panel as shown below. The mass and mole fractions of O are calculated as 0.0244 and 0.0477 respectively.

For the what-if scenario, change p1 to 50 kPa, press the Enter key, and Super-Calculate to produce x_O as 0.0345 and y_O as 0.0667.

Ex-3 A mixture of 1 kmol of H2O and 3 kmol of O2 is heated to 3500 K at a pressure of 1 atm. Determine the equilibrium composition of this mixture, assuming the products to contain H2O, OH, O2 and H2 only.

Solution Launch the system state chemical equilibrium daemon located at TEST.Daemons.System.State page.

On the composition panel, set up the reactants and products. For the reactants, also enter the amount in kmols (by double clicking the default amount, editing the value, and pressing the Enter key).

Now select the Products radio button in the state panel, enter p1=1 atm, T1=2500 K, and Calculate. The state is displayed as shown below.

The molar composition can be found in the composition panel (see image below) as 0.4783H2O + 0.1123H2 + 0.8188OH + 2.8515O2.

Ex-4 Determine the stoichiometric equlibrium flame temperature of methane burning with air at a constant pressure of 100 kPa.

Solution Using the open, steady, separated combustion daemon obtain the stoichiometric composition of the reactants as CH4 + 2O2 + 7.5238N2. (Start the combustion daemon at open.steady.specific.combustion.separated.IGmixture page, select methane as the fuel and Theoretical Air from the action menu. Now select Air->O2,N2 from the action menu to obtain the balanced reaction).

Launch the open steady equilibrium daemon located at Open, Steady, Specific, Combustion, Equilibrium branch. On the composition panel, set up the reactants by adding CH4, O2 and N2 in known molar amounts. Also select relevant species for the products (you can select "standard combustion products" as a shortcut). Calculate state-1 as the reactants state (with the Reactants button selected) with p1=100 kPa and T1=25 deg-C, and an assumed mass flow rate mdot1= 1 kg/s. Note the enthalpy calculated as h1=-985.02 kJ/kg of reactants mixture. In the state panel calculate State-2 with p2=p1. In the device panel, select state-1 as the i-state and state-2 as the e-state. Enter Wdot_ext=0. Go back to the state-panel and use a guessed value for T2, say, 1900 deg-C. Press the Enter key and then Super-Calculate. This will update the device panel. Qdot is negative, indicating that the adiabatic temperature must be higher. Increase T2 and Super-Calculate again. Continue until yo uget Qdot is close to zero. At that point T2 is 1965 deg-C. Therefore, the equilibrium temperature (for the chosen products) must be quite close to 1965 deg-C.

In the near future, this daemon will be able to handle input such as h2=h1. By entering Qdot=0, or h2=h1, T2 can be directly calculated without any need for iterations.

Ex-5 A 1 kg block of aluminum at 600 K is brought in thermal contact with another identical block at 300 K in an isolated chamber. Show that entropy of the combined system is maximized at equilibrium. What-if Scenario: How would the answers change if the second block was made of copper?

# TEST-codes for Ex-1;
# TEST>Daemons>Systems>Closed>Process>Generic>
# NonUniform>Nonmixing>SL/SL Model;

States    {
    State-1: Aluminum(Al), Aluminum(Al);
    Given:       { T1= 600.0 K;   Vel1= 0.0 m/s;   z1= 0.0 m;   m1= 1.0 kg;   Model1= 1.0 UnitLess;   }

    State-2: Aluminum(Al), Aluminum(Al);
    Given:       { T2= 300.0 K;   Vel2= 0.0 m/s;   z2= 0.0 m;   m2= 1.0 kg;   Model2= 2.0 UnitLess;   }

    State-3: Aluminum(Al), Aluminum(Al);
    Given:       { T3= 400.0 K;   Vel3= 0.0 m/s;   z3= 0.0 m;   m3= "m1" kg;   Model3= 1.0 UnitLess;   }

    State-4: Aluminum(Al), Aluminum(Al);
    Given:       { Vel4= 0.0 m/s;   z4= 0.0 m;   m4= "m2" kg;   Model4= 2.0 UnitLess;   }
    }

Analysis    {
    Process-A: b-State = State-1, State-2; f-State = State-3, State-4;
    Given: { Q= 0.0 kJ;   W= 0.0 kJ;   T_B= 25.0 deg-C;   }
    }

Solution The problem describes a closed process involving a composite system made of two blocks of solids. Therefore, launch the Non-Mixing, Generic, Closed Process Daemon with the SL/SL model. Evaluate the four states as described in the TEST-codes. Notice that State-1 and State-2 describe the bA and bB states. Although the working substances for both sub-systems are identical, for sub-system A select the material from the left menu and for sub-system B use the right menu. Also assume an arbitrary equilibrium temperature for system A as T3=400 K.

In the process panel, load the bA, bB, fA and fB states. Enter Q=W=0. Super-Calculate to find T4=500 K and Delta_S=0.095 kJ/K. Now change the guess for T3 to 425 K, Super-Calculate and find T4=475 K and Delta_S to be 0.103 kJ/K. Similarly, a T3= 460 K produces T4=440 K and Delta_S to be 0.105 kJ/K. Finally, a choice of T3=450 K produces T4=450 K and Delta_S=0.106 kJ/K . Entropy of the system is, thus, maximized when temperatures of the two sub-stystems become equal.

For the what-if study, change the second working substance to copper in State-2 and Super-Calculate. Change T3 and Super-Calculate T4 and Delta_S. Entropy is maximized when T3=T4= 510 K. Note that the second working fluid can be changed only for State-2 or State-4 since a change of working fluid automatically recalculates the current state using the selected working substance .

Ex-6 Determine the more likely of the two compositions - 1 kmol of CO2 versus 0.95CO2+.05CO+.05O - at (a) 100 kPa and 300 K, and (b) 100 kPa and 3000 K.

Solution The question can be answered by evaluating the Gibb's free energy for the two mixtures. We can use several different daemons - system or flow state daemons with IG mixture, closed or open combustion daemons with IG or PG model - to evaluate g of a gas mixture.

# TEST-codes for Ex-6;
# TEST>Daemons>Systems>Closed>Process>
# Specific>Combustion>Premixed>IdealGas;

States    {
    State-1: Reactants > mixture;
    Given:       { p1= 100.0 kPa;   T1= 26.85 deg-C;   Vel1= 0.0 m/s;   z1= 0.0 m;   Model1= 1.0 UnitLess;   }

    State-2: Products > mixture;
    Given:       { p2= "p1" kPa;   T2= "T1" deg-C;   Vel2= 0.0 m/s;   z2= 0.0 m;   Model2= 3.0 UnitLess;   }
    }

Fig. 6.1 Image of the products state in Ex. 6

Launch the premixed closed process combustion daemon located at Closed, Process, Specific, Combustion, Premixed, IdealGas branch. Set up the reaction by entering CO2 as the reactant and CO2, CO and O as the products. Enter the amounts, 1 kmol for CO2 in the reactants block, 0.95 kmol for CO2 in the products, and select Balance Reaction to set up the reaction.

In the state panel, evaluate the states (reactants and products) as described in the TEST-codes above.

Compare g1=-10400 kJ/kg with g2=-9867 kJ/kg. Clearly, State-1, the reactants, is a more stable state since its Gibb's free energy is lower. Now change T1 to 3000 K and Super-Calculate . The new values for g1 and g2 are -28255 kJ/kg and -28388 kJ/kg respectively. The products mixture is, therefore, a more likely state at 3000 K.