Some properties of the Helmholtz free energy

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Doris Whitehead
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1 Some properties of the Helmholtz free energy Energy slope is T U(S, ) From the properties of U vs S, it is clear that the Helmholtz free energy is always algebraically less than the internal energy U. U(S, ) F(T, ) TS F U TS Entropy S Since F characterizes the work that can be extracted from the system, the relation F<U tells us that not all internal energy can be converted into work.

2 Helmholtz potential - link to statistical physics Link to statistical physics: F( T, V, N ) kbt ln( Z ) Where Z is the partition function: Z( T, V, N ) exp E kb ( V, N ) T F is easy to calculate in statistical physics if you know the energy spectrum E of the system of interest.

3 The Helmholtz free energy and the Euler equation From the fact that energy is a homogeneous function of the extensive parameters, we have derived the Euler equation: U TS PV + µ N +... When we apply the Legendre transformation to U in the form of the Euler equation, we obtain the Euler equation for the Helmhotz free energy: F U TS PV + µ N +...

4 The Helmholtz free energy and the Euler equation We can also write the Helmholtz free energy in its differential form: du TdS PdV + µ dn +... df du TdS SdT df SdT PdV + µ dn +...

5 The Enthalpy Legendre transformation to obtain enthalpy: ( S, P, N,... ) U PV H + Enthalpy should be used for calculation of the equilibrium state if pressure is maintained constant (the system of interest in contact with a pressure reservoir).

6 The Enthalpy Differential of enthalpy: dh TdS + VdP + µ dn +... If the pressure is constant and the number of particles in the system does not change then dp 0 and dn 0: dh TdS δq Therefore, heat added to the system at constant pressure is equal to the increase of enthalpy.

7 The Joule-Thomson process Piston 1 Piston gas gas P 1 P Fixed porous membrane The pistons move to maintain constant pressures P 1 and P in the two regions while the gas is passing through a porous membrane from one region to the other. The temperature of the gas can be changed by this process as the gas passes from one region to the other.

8 The Joule-Thomson process Let us consider a situation when one mole of gas was initially in the left region and in the final state was all transferred to the right region. Piston 1 Piston gas gas P i P f v i initial molar volume v f initial molar volume Fixed porous membrane This process is often used for refrigeration. Note that this is not an equilibrium process!

9 The Joule-Thomson process Since the process is at a constant pressure, work done on the gas by the left piston is P i v i. Work done by the gas on the right piston is P f v f. Since no heat is transferred to the gas from the environment, the internal energy of the gas changes by P i v i -P f v f : Piston 1 Piston gas gas P i P f Fixed porous membrane u f ui + Pv i i Pf v f u f + Pf v f ui + Pi vi h f hi Enthalpy of the initial and final states is the same in the Joule-Thomson process.

10 The Joule-Thomson process Fro real gases, curves of constant enthalpy on the T-P diagram look like this. T Therefore, if we go from the high initial pressure to the low final pressure, the temperature of gas will decrease. P Note that since the process is not equilibrium, the system does not go along the curve of constant enthalpy in the Joule-Thomson process. All the intermediate states of the system are non-equilibrium, only the initial and final states are equilibrium. P f P i

11 The Gibbs free energy The Gibbs potential G(T,P,N, ) is convenient to use for analysis of processes at constant pressure P and constant temperature T. Most chemical reactions are carried out at ambient pressure and temperature, so G is the potential of choice for the description of chemical reactions. G U TS + PV Substituting the Euler equation in to the expression for G: U TS PV + µ µ N 1 N G µ µ N 1 N

12 The Gibbs free energy For a single-component system: G µ 1 N1 + µ N +.. G G µ N µ g N Molar Gibbs free energy for a single-component system the chemical potential A chemical reaction: * H + O * H O or * H + 1 * O - * H O 0 Any generic 3-component reaction : ν 1 * A 1 + ν * A + ν 3 * A 3 0 Any reaction : Σ ν * A 0 Here ν are stoichiometric coefficients.

13 The Gibbs free energy: chemical reactions Differential form of the Gibbs free energy: + dg SdT + VdP µ dn Conservation of the total number of each atomic species in chemical reactions demands: dn v dn where dn is the same for all chemical components of the reaction.

14 The Gibbs free energy: chemical reactions Therefore: dg SdT + VdP + dn ν µ At constant pressure and temperature: dg dn ν µ In equilibrium, dg is at minimum at constant pressure and temperature: Since dn is arbitrary: ν µ dg dn ν µ ( T P, N, N,...) 0, 1 0 -chemical equilibrium condition

15 The Gibbs free energy: chemical reactions Let the initial mole numbers of chemicals be: 0 N Integrating dn v dn we obtain the final mole numbers: N f 0 N + 0 dn N + v dn N 0 + v N Where N is simply a numerical factor characterizing the extent of reaction

16 The Gibbs free energy: chemical reactions N f 0 N + 0 dn N + v dn N 0 + v N Where N is simply a numerical factor characterizing the extent of reaction ν µ ( T P, N, N,...) 0 ν µ ( T P, ν N, ν N,... ) 0, 1, 1 Since T, P and v are all fixed and known, the chemical equilibrium condition becomes a function of a single parameter characterizing the extent of reaction, N. Solving this equation for N, we obtain the final mole numbers for the reaction.

17 Example: Chemical Reactions in Ideal Gas In lecture 5, we considered a model of reaction in an ideal gas S N*ln(U/U 0 )+ N *ln(n /N) µ RT( ln( RT / u ) ln( N N ) 0 + µ RT / ( ln( RT / u ) ln( ) 0 + x Then we neglected a term in entropy depending on volume of the gas. If we keep this term and write chemical potential in the Gibbs energy representation: µ RT ( ln( RT / u ) ln( P) ln( ) x

18 Example: Chemical Reactions in Ideal Gas In a case of generalized ideal gas (not monoatomic), the chemical potential of the th component of a mixture of ideal gases can be written as where φ ( T ) µ RT φ ( ( T ) + ln ( P) + ln( x ) depends on the particular gas and x is molar fraction of the gas The condition of chemical equilibrium becomes: ν ln ν µ ( T P, N, N,...) 0, 1 ( x ) ( ) ν ln P ν φ ( T )

19 Example: Chemical Reactions in Ideal Gas ν ln ( x ) ( ) ν ln P ν φ ( T ) It is customary to define the equilibrium constant of a reaction K(T) ln ( K( T )) ν φ ( T ) With this definition, the equilibrium condition can be written in the form of the mass action law: x v P v K ( T )

20 Example: Chemical Reactions in Ideal Gas The mass action law in combination with the conservation of atomic species gives the equilibrium mole fractions for the reacting mixture of gases Example: Consider a reaction of oxygen and hydrogen to form water: Two moles of water are heated to T 000 K at P 1 MPa. What is the equilibrium composition of the gases in the following reaction? H O H + 1 O The equilibrium constant for this reaction is K(000) [Pa 1/ ]

21 Example: Thermal Decomposition of Water The law of mass action: x v P v K ( T ) Conservation of atomic species gives: x x H x 1/ H O 1/ O P K ( T ) N H O N N N H 1 N O N

22 Example: Thermal Decomposition of Water The total mole numbers in the reaction: N H O + N H + N O + N x H O N + N x H N + N x O 1 + N N x x H x 1/ H O 1/ O P K ( T ) 1 N 3/ 1/ P 1 1/ ( N )( + N ) K ( T )

23 Example: Thermal Decomposition of Water 1 N 1 ( N ) ( + N ) 3 K ( T ) P This is a cubic equation for N. It can either be solved numerically or analytically. The result is N xh O x 0. H 005 x 0. O 001 A simplified solution is obtained if we calculate K (T)/P first. K(T)/P (0.0877) From this it is clear that N is small and we can write: N K P ( T ) 1/

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc.

Chapter 3. Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Chapter 3 Property Relations The essence of macroscopic thermodynamics Dependence of U, H, S, G, and F on T, P, V, etc. Concepts Energy functions F and G Chemical potential, µ Partial Molar properties