Thermodynamics and Kinetics C. Paolucci University of Notre Dame Department of Chemical & Biomolecular Engineering
What is the energy we calculated? You used GAMESS to calculate the internal (ground state) electronic energy for various molecules under the Born-Oppenheimer approximation. 2 H + + O 8+ + 10 e Eelec H 2 O Even at 0 K molecules have vibrational energy, called the zero-point vibrational energy (ZPVE). You calculated this from a frequency calculation. ZP V E = 1 3n 6 2 h i=1 ν i Paolucci Thermodynamics and Kinetics March 18, 2014 2 / 14
What is the energy we calculated? You used GAMESS to calculate the internal (ground state) electronic energy for various molecules under the Born-Oppenheimer approximation. 2 H + + O 8+ + 10 e Eelec H 2 O Even at 0 K molecules have vibrational energy, called the zero-point vibrational energy (ZPVE). You calculated this from a frequency calculation. ZP V E = 1 3n 6 2 h i=1 ν i Paolucci Thermodynamics and Kinetics March 18, 2014 2 / 14
What is the energy we calculated? You used GAMESS to calculate the internal (ground state) electronic energy for various molecules under the Born-Oppenheimer approximation. 2 H + + O 8+ + 10 e Eelec H 2 O Even at 0 K molecules have vibrational energy, called the zero-point vibrational energy (ZPVE). You calculated this from a frequency calculation. ZP V E = 1 3n 6 2 h i=1 ν i Paolucci Thermodynamics and Kinetics March 18, 2014 2 / 14
What is the energy we calculated? You used GAMESS to calculate the internal (ground state) electronic energy for various molecules under the Born-Oppenheimer approximation. 2 H + + O 8+ + 10 e Eelec H 2 O Even at 0 K molecules have vibrational energy, called the zero-point vibrational energy (ZPVE). You calculated this from a frequency calculation. ZP V E = 1 3n 6 2 h i=1 ν i E 0 =E elec + ZPVE E elec Paolucci Thermodynamics and Kinetics March 18, 2014 2 / 14
What is the energy of a molecule? E 0 is the minimum energy of a molecule. Higher energies are possible through electronic or vibrational excitations as well as translational and rotational motion. If we assume these contributions are separable E = E 0 + E elec + E vib + E rot + E trans Paolucci Thermodynamics and Kinetics March 18, 2014 3 / 14
What is the energy of a molecule? E 0 is the minimum energy of a molecule. Higher energies are possible through electronic or vibrational excitations as well as translational and rotational motion. If we assume these contributions are separable E = E 0 + E elec + E vib + E rot + E trans Paolucci Thermodynamics and Kinetics March 18, 2014 3 / 14
What is the energy of a molecule? E 0 is the minimum energy of a molecule. Higher energies are possible through electronic or vibrational excitations as well as translational and rotational motion. If we assume these contributions are separable E = E 0 + E elec + E vib + E rot + E trans Ar N N E=E 0 + E trans E=E 0 + E trans + E rot + E vib We need to specify these quantities to describe the microscopic state of the molecule. Paolucci Thermodynamics and Kinetics March 18, 2014 3 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ All thermodynamic quantities can be derived from Q Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ U = ln Q β NV Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics Macroscopically we want the average value of these quantities over all possible states of an ensemble of molecules at some specified externally imposed set of thermodynamic conditions. (ensemble average) This is the domain of statistical mechanics. The relative probability of a molecule being in a state with energy E i above E 0 is given by the Boltzmann factor P i e E iβ, β = 1 k B T Sum over all states is the (in this canonical, N, V, T free variables) partition function Q(N, V, T ) = i e E iβ G = ln Q β + V β ln Q V NT Paolucci Thermodynamics and Kinetics March 18, 2014 4 / 14
Statistical Mechanics The good news: if we know Q we can find any thermodynamic quantity. The bad news: can be difficult to calculate for some systems Paolucci Thermodynamics and Kinetics March 18, 2014 5 / 14
Statistical Mechanics The good news: if we know Q we can find any thermodynamic quantity. The bad news: can be difficult to calculate for some systems Paolucci Thermodynamics and Kinetics March 18, 2014 5 / 14
Statistical Mechanics H O H O H H The good news: if we know Q we can find any thermodynamic quantity. The bad news: can be difficult to calculate for some systems Ideal Gas Bulk Liquid Macroscopic Solid N N H O H H O H H O H No intermolecular interactions Only sum over states of individual molecule Need all inter/intramolecular states, MC/MD to get thermodynamic quantities No rotation or translation. Sum over vibrational modes Paolucci Thermodynamics and Kinetics March 18, 2014 5 / 14
Statistical Mechanics H O H O H H The good news: if we know Q we can find any thermodynamic quantity. The bad news: can be difficult to calculate for some systems Ideal Gas Bulk Liquid Macroscopic Solid N N H O H H O H H O H No intermolecular interactions Only sum over states of individual molecule Need all inter/intramolecular states, MC/MD to get thermodynamic quantities No rotation or translation. Sum over vibrational modes Here we will focus on an ideal gas. Paolucci Thermodynamics and Kinetics March 18, 2014 5 / 14
Ideal Gas For an ideal gas of N indistinguishable molecules Q = qn (V, T ) N! Once again assuming decoupling the molecular partition function is q = q trans q vib q rot q elec We can treat each of these pieces with models you have seen before. GAMESS chooses 298.15 K by default. Paolucci Thermodynamics and Kinetics March 18, 2014 6 / 14
Ideal Gas For an ideal gas of N indistinguishable molecules Q = qn (V, T ) N! Once again assuming decoupling the molecular partition function is q = q trans q vib q rot q elec We can treat each of these pieces with models you have seen before. GAMESS chooses 298.15 K by default. Paolucci Thermodynamics and Kinetics March 18, 2014 6 / 14
Ideal Gas For an ideal gas of N indistinguishable molecules Q = qn (V, T ) N! Once again assuming decoupling the molecular partition function is q = q trans q vib q rot q elec We can treat each of these pieces with models you have seen before. GAMESS chooses 298.15 K by default. Paolucci Thermodynamics and Kinetics March 18, 2014 6 / 14
Ideal Gas Translation: Particle in a 3-D box (have to choose a standard state pressure, GAMESS chooses 1 atm) Rotation: Rigid Rotor Vibration: Harmonic Oscillator Electronic: Sum over excited electronic states, usually only relevant at high T Paolucci Thermodynamics and Kinetics March 18, 2014 6 / 14
Ideal Gas Translation: Particle in a 3-D box (have to choose a standard state pressure, GAMESS chooses 1 atm) Rotation: Rigid Rotor Vibration: Harmonic Oscillator Electronic: Sum over excited electronic states, usually only relevant at high T Typical values q trans 10 30 q rot 100 q vib 1 q elec g 0 Paolucci Thermodynamics and Kinetics March 18, 2014 6 / 14
Ideal Gas Example GAMESS will give you these values making the key assumptions on the previous slide after a frequency calculation. Lets look at Argon Paolucci Thermodynamics and Kinetics March 18, 2014 7 / 14
Ideal Gas Example Ar gas Note: GAMESS scales q trans by 2.46 10 25 Paolucci Thermodynamics and Kinetics March 18, 2014 7 / 14
Ideal Gas Example Now N 2 Paolucci Thermodynamics and Kinetics March 18, 2014 7 / 14
Chemical Reactions Now we can use this information to calculate reaction energies. First lets just start with 0 K Paolucci Thermodynamics and Kinetics March 18, 2014 8 / 14
Chemical Reactions Now we can use this information to calculate reaction energies. First lets just start with 0 K Transition State Reactant 0 K Reaction Energy Product Paolucci Thermodynamics and Kinetics March 18, 2014 8 / 14
Chemical Reactions Now we can use this information to calculate reaction energies. First lets just start with 0 K HCN TS HNC HCN TS HNC Paolucci Thermodynamics and Kinetics March 18, 2014 8 / 14
Chemical Reactions HCN TS HNC E rxn = E products E reactants Paolucci Thermodynamics and Kinetics March 18, 2014 9 / 14
Chemical Reactions HCN TS HNC E rxn = E products E reactants 300 250 200 E (kj/mol) 150 100 50 ΔE rxn 0 1 2 3 Reaction Coordinate We can do thermodynamic quantities H rxn(t ) = E elec + ZP V E+ H (T ) trans + H (T ) rot + H (T ) vib Here the is product(s) minus reactant(s). H rxn = +77.65 kj/mol, endothermic reaction Paolucci Thermodynamics and Kinetics March 18, 2014 9 / 14
Gibbs Free Energy HCN TS HNC E (kj/mol) ZPVE G tot(298k) G tot(1000 K) HCN -240693.2 50.0165-0.76-160.462 HNC -240612.4 48.2648-2.593-164.758 TS -240403.5 30.9667-24.35-195.324 Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Gibbs Free Energy HCN TS HNC E (kj/mol) ZPVE G tot(298k) G tot(1000 K) HCN -240693.2 50.0165-0.76-160.462 HNC -240612.4 48.2648-2.593-164.758 TS -240403.5 30.9667-24.35-195.324 Here G tot(t ) is the Gibbs free energy contribution from translation, rotation, and vibration at the specified temperature and standard state. Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Gibbs Free Energy HCN TS HNC 250 298 K 1000 K 200 G (kj/mol) 150 100 50 ΔG rxn 0 1 2 3 Reaction Coordinate Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Gibbs Free Energy HCN TS HNC Can calculate Equilibrium constant K = q HNC q HCN e Erxn/k BT = e G rxn (T )/k BT Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Gibbs Free Energy HCN TS HNC Can calculate Equilibrium constant K = q HNC q HCN e Erxn/k BT = e G rxn (T )/k BT We did calculations at two Temperatures we could have also used the Van t Hoff Equation if the temperature range is small enough that H rxn constant ln K(T ( 2) K(T 1 ) = H rxn 1 1 ) R T 2 T 1 Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Gibbs Free Energy HCN TS HNC Can calculate Equilibrium constant K = q HNC q HCN e Erxn/k BT = e G rxn (T )/k BT We did calculations at two Temperatures we could have also used the Van t Hoff Equation if the temperature range is small enough that H rxn constant ln K(T ( 2) K(T 1 ) = H rxn 1 1 ) R T 2 T 1 For arbitrary activities (or pressures assuming Raoult s Law) G rxn = G rxn + RT ln a HNC a HCN Paolucci Thermodynamics and Kinetics March 18, 2014 10 / 14
Thermodynamic Cycle We can use this approach to calculate all sorts of thermodynamic properties under all different kinds of conditions, for example the heat of formation of methane: Paolucci Thermodynamics and Kinetics March 18, 2014 11 / 14
Transition State Theory Transition state theory assumes 1 Existence of a potential energy surface. 2 Existence of a dividing surface, point of no return. 3 Existence of a critical TS point on that surface. 4 Quasi-equilibrium between reactants and TS 5 Harmonic PES near the TS HNC H-C H-N HCN Paolucci Thermodynamics and Kinetics March 18, 2014 12 / 14
Transition State Theory Transition state theory assumes 1 Existence of a potential energy surface. 2 Existence of a dividing surface, point of no return. 3 Existence of a critical TS point on that surface. 4 Quasi-equilibrium between reactants and TS 5 Harmonic PES near the TS HNC H-C H-N HCN Paolucci Thermodynamics and Kinetics March 18, 2014 12 / 14
Transition State Theory Transition state theory assumes 1 Existence of a potential energy surface. 2 Existence of a dividing surface, point of no return. 3 Existence of a critical TS point on that surface. 4 Quasi-equilibrium between reactants and TS 5 Harmonic PES near the TS HNC H-C H-N HCN Paolucci Thermodynamics and Kinetics March 18, 2014 12 / 14
Transition State Theory Transition state theory assumes 1 Existence of a potential energy surface. 2 Existence of a dividing surface, point of no return. 3 Existence of a critical TS point on that surface. 4 Quasi-equilibrium between reactants and TS 5 Harmonic PES near the TS HNC H-C H-N HCN Paolucci Thermodynamics and Kinetics March 18, 2014 12 / 14
Transition State Theory Transition state theory assumes 1 Existence of a potential energy surface. 2 Existence of a dividing surface, point of no return. 3 Existence of a critical TS point on that surface. 4 Quasi-equilibrium between reactants and TS 5 Harmonic PES near the TS HNC H-C H-N HCN Paolucci Thermodynamics and Kinetics March 18, 2014 12 / 14
Kinetics Under these assumptions one can show ( n is change in moles) k f (T ) = k BT h q q reactants (P/RT ) n e Ea/k BT Could group all the leading terms into A(T) k f (T ) = A(T )e Ea/k BT Looks familiar right? equivalently k f (T ) = k BT h Apparent activation energy given by e G (T ) k B T E app = RT 2 ln k T Paolucci Thermodynamics and Kinetics March 18, 2014 13 / 14
Kinetics Under these assumptions one can show ( n is change in moles) k f (T ) = k BT h q q reactants (P/RT ) n e Ea/k BT Could group all the leading terms into A(T) k f (T ) = A(T )e Ea/k BT Looks familiar right? equivalently k f (T ) = k BT h Apparent activation energy given by e G (T ) k B T E app = RT 2 ln k T Paolucci Thermodynamics and Kinetics March 18, 2014 13 / 14
Kinetics Under these assumptions one can show ( n is change in moles) k f (T ) = k BT h q q reactants (P/RT ) n e Ea/k BT Could group all the leading terms into A(T) k f (T ) = A(T )e Ea/k BT Looks familiar right? equivalently k f (T ) = k BT h Apparent activation energy given by e G (T ) k B T E app = RT 2 ln k T Paolucci Thermodynamics and Kinetics March 18, 2014 13 / 14
Kinetics Under these assumptions one can show ( n is change in moles) k f (T ) = k BT h q q reactants (P/RT ) n e Ea/k BT Could group all the leading terms into A(T) k f (T ) = A(T )e Ea/k BT Looks familiar right? equivalently k f (T ) = k BT h Apparent activation energy given by e G (T ) k B T E app = RT 2 ln k T Paolucci Thermodynamics and Kinetics March 18, 2014 13 / 14
Kinetics Under these assumptions one can show ( n is change in moles) k f (T ) = k BT h q q reactants (P/RT ) n e Ea/k BT Could group all the leading terms into A(T) k f (T ) = A(T )e Ea/k BT Looks familiar right? equivalently k f (T ) = k BT h Apparent activation energy given by e G (T ) k B T E app = RT 2 ln k T Paolucci Thermodynamics and Kinetics March 18, 2014 13 / 14
Kinetics 300 250 E (kj/mol) 200 150 ΔE a,f 100 50 ΔE rxn 0 1 2 3 Reaction Coordinate k r = k f /K r = k f C HCN k f K C HNC Could stick in a mass balance, e.g unsteady CSTR dc HCN dt = C HCN,0 C HCN τ r Paolucci Thermodynamics and Kinetics March 18, 2014 14 / 14
Kinetics 300 250 E (kj/mol) 200 150 ΔE a,f 100 50 ΔE rxn 0 1 2 3 Reaction Coordinate k r = k f /K r = k f C HCN k f K C HNC Could stick in a mass balance, e.g unsteady CSTR dc HCN dt = C HCN,0 C HCN τ r Paolucci Thermodynamics and Kinetics March 18, 2014 14 / 14
Kinetics 300 250 E (kj/mol) 200 150 ΔE a,f 100 50 ΔE rxn 0 1 2 3 Reaction Coordinate k r = k f /K r = k f C HCN k f K C HNC Could stick in a mass balance, e.g unsteady CSTR dc HCN dt = C HCN,0 C HCN τ r Paolucci Thermodynamics and Kinetics March 18, 2014 14 / 14
Kinetics 300 250 E (kj/mol) 200 150 ΔE a,f 100 50 ΔE rxn 0 1 2 3 Reaction Coordinate k r = k f /K r = k f C HCN k f K C HNC Could stick in a mass balance, e.g unsteady CSTR dc HCN dt = C HCN,0 C HCN τ r Paolucci Thermodynamics and Kinetics March 18, 2014 14 / 14