Entropy&FreeEnergy - Maple Help

Entropy and Free Energy

 Overview In thermodynamics free energy is the energy that is available to the system at constant temperature to perform useful work.  At constant pressure the free energy of the system is known as the Gibbs free energy, and it depends on the enthalpy and entropy of the system.  Enthalpy is the system's internal energy plus the energy from the product of its pressure and volume, and entropy is the system's energy, unavailable for performing useful work, per unit temperature.   In this lesson we will explore the change in free energy for one of the simplest combustion reactions, the combustion of carbon monoxide to form carbon dioxide.   Figure 1: Combustion reaction of carbon monoxide plus oxygen to form carbon dioxide.

Free Energy

Consider the chemical reaction

CO (g)   +  1/2  O2     CO2 (g)

The changes in enthalpy ΔH, entropy ΔS, and free energy ΔG at standard temperature and pressure (298 K and 1 atm) can be calculated from the enthalpies, entropies, and Gibbs free energies of the reactants and the products, which can be computed from quantum theory.  We will compute these enthalpies using the Thermodynamics command in the Quantum Chemistry package.

First, we set the Digits to 15 and load the Quantum Chemistry package

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 $\left[{\mathrm{AOLabels}}{,}{\mathrm{ActiveSpaceCI}}{,}{\mathrm{ActiveSpaceSCF}}{,}{\mathrm{AtomicData}}{,}{\mathrm{BondAngles}}{,}{\mathrm{BondDistances}}{,}{\mathrm{Charges}}{,}{\mathrm{ChargesPlot}}{,}{\mathrm{CorrelationEnergy}}{,}{\mathrm{CoupledCluster}}{,}{\mathrm{DensityFunctional}}{,}{\mathrm{DensityPlot3D}}{,}{\mathrm{Dipole}}{,}{\mathrm{DipolePlot}}{,}{\mathrm{Energy}}{,}{\mathrm{ExcitationEnergies}}{,}{\mathrm{ExcitationSpectra}}{,}{\mathrm{ExcitationSpectraPlot}}{,}{\mathrm{ExcitedStateEnergies}}{,}{\mathrm{ExcitedStateSpins}}{,}{\mathrm{FullCI}}{,}{\mathrm{GeometryOptimization}}{,}{\mathrm{HartreeFock}}{,}{\mathrm{Interactive}}{,}{\mathrm{Isotopes}}{,}{\mathrm{MOCoefficients}}{,}{\mathrm{MODiagram}}{,}{\mathrm{MOEnergies}}{,}{\mathrm{MOIntegrals}}{,}{\mathrm{MOOccupations}}{,}{\mathrm{MOOccupationsPlot}}{,}{\mathrm{MOSymmetries}}{,}{\mathrm{MP2}}{,}{\mathrm{MolecularData}}{,}{\mathrm{MolecularGeometry}}{,}{\mathrm{NuclearEnergy}}{,}{\mathrm{NuclearGradient}}{,}{\mathrm{OscillatorStrengths}}{,}{\mathrm{Parametric2RDM}}{,}{\mathrm{PlotMolecule}}{,}{\mathrm{Populations}}{,}{\mathrm{RDM1}}{,}{\mathrm{RDM2}}{,}{\mathrm{RTM1}}{,}{\mathrm{ReadXYZ}}{,}{\mathrm{Restore}}{,}{\mathrm{Save}}{,}{\mathrm{SaveXYZ}}{,}{\mathrm{SearchBasisSets}}{,}{\mathrm{SearchFunctionals}}{,}{\mathrm{SkeletalStructure}}{,}{\mathrm{Thermodynamics}}{,}{\mathrm{TransitionDipolePlot}}{,}{\mathrm{TransitionDipoles}}{,}{\mathrm{TransitionOrbitalPlot}}{,}{\mathrm{TransitionOrbitals}}{,}{\mathrm{Variational2RDM}}{,}{\mathrm{VibrationalModeAnimation}}{,}{\mathrm{VibrationalModes}}{,}{\mathrm{Video}}\right]$ (2.1)

The enthalpies, entropies, and free energies of O2 and CO2 have been precomputed in Table 1.

Table 1: Enthalpies and free energies in kJ/mol and entropies in kJ/mol/K for the combustion of CO

 CO O2 CO2 H ? $-3.94506117$ x 105 kJ/mol $-4.94876985$ x 105 kJ/mol S ? $0.39305376802$ kJ/mol/K $0.4111337849$ kJ/mol/K G ? $-3.94623306$ x 105 kJ/mol $-4.94999564$ x 105 kJ/mol

The enthalpy, entropy, and free energy of CO can be computed as follows.  First, we define the geometry at the equilibrium geometry, computed in the Enthalpy lesson

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 ${\mathrm{mol_CO}}{≔}\left[\left[{"C"}{,}{0}{,}{0}{,}{0}\right]{,}\left[{"O"}{,}{0}{,}{0}{,}{1.13485718}\right]\right]$ (2.2)

Finally, we compute a table of thermodynamic properties for CO including the enthalpy, entropy, and free energy.  By default, the command uses a temperature of 298 K.

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 ${\mathrm{thermo_CO}}{≔}{table}{}\left(\left[{{\mathrm{\theta }}}_{{C}}{=}{2.74504846}{}⟦{K}⟧{,}{{\mathrm{\theta }}}_{{B}}{=}{2.74504846}{}⟦{K}⟧{,}{\mathrm{enthalpy}}{=}{-}{2.97353168}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{zpe}}{=}{13100.56194694}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{entropy}}{=}{389.04949897}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧{,}{\mathrm{heat_capacity}}{=}{20.81001215}{}⟦\frac{{J}}{{\mathrm{mol}}{}{K}}⟧{,}{\mathrm{free_energy}}{=}{-}{2.97469163}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{energy}}{=}{-}{2.97355647}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧{,}{\mathrm{electronic_energy}}{=}{-}{2.97388047}{}{{10}}^{{8}}{}⟦\frac{{J}}{{\mathrm{mol}}}⟧\right]\right)$ (2.3)

(a) What is the computed enthalpy of CO in kJ/mol?

(b) What is the computed entropy of CO in kJ/mol/K?

(c) What is the computed free energy  of CO in kJ/mol?

We can convert the enthalpy, entropy, and free energy from J to kJ using Maple's convert command

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 ${{H}}_{{\mathrm{CO}}}{≔}{-}{297353.16764525}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (2.4)
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 ${{S}}_{{\mathrm{CO}}}{≔}{0.38904950}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}{}{K}}⟧$ (2.5)
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 ${{G}}_{{\mathrm{CO}}}{≔}{-}{297469.16275336}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (2.6)

(d) Using the enthalpy of CO in (a) and the precomputed enthalpy values in Table 1, calculate the change in enthalpy in the combustion of 1 mol of CO.

The precomputed values in Table 1 are as follows

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 ${{H}}_{{\mathrm{O2}}}{≔}{-}{394506.11705451}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$
 ${{H}}_{{\mathrm{CO2}}}{≔}{-}{494876.98482204}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (2.7)

Therefore, we can compute the change in enthalpy from

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 ${\mathrm{DeltaH}}{≔}{-}{270.75864954}{}⟦\frac{{\mathrm{kJ}}}{{\mathrm{mol}}}⟧$ (2.8)

(e) Using the entropy of CO in (b) and the precomputed entropy values in Table 1, calculate the change in entropy in the combustion of 1 mol of CO.

(f) Using the free energy of CO in (c) and the precomputed free energy values in Table 1, calculate the change in free energy in the combustion of 1 mol of CO.

(g) Using the change in enthalpies and entropies and the Gibbs free energy equation, compute the change in free energy in the combustion of 1 mol of CO.

(h) Do you results for the free energy change in (f) and (g) agree?

(i) Using the change in enthalpies and entropies and the Gibbs free energy equation, estimate the temperature below which the reaction is spontaneous.

 >