THERMODYNAMICS

Spontaneous Processes. An important part of experimental chemistry deals with spontaneous reactions, that is, reactions that take place without having to continually supply energy from outside the system. It is also important to remember that the term spontaneous doesn't necessarily mean a fast reaction rate. One goal of thermodynamics is to be able to predict whether a reaction will take place when a given set of reactants is brought together. Thermodynamics only tells us whether or not the reaction will occur, but it tells us nothing about how fast.

The release of heat by a reaction was once thought to be an indication that the reaction was spontaneous. The sign of the enthalpy change, H, by itself is not an adequate guide to spontaneity because while some spontaneous reactions are known to be exothermic ( H is –), many endothermic reactions ( H is +) are known to be spontaneous as well.

In addition to the heat absorbed or released in a spontaneous process, another factor called entropy must be considered. Entropy is a measure of the disorder or randomness of a system. The entropy (S) is a state function that increases in value as the disorder or randomness of the system increases. Entropy has the units J/K·mol. Intuitively we consider a system to be "ordered" if it is arranged according to some plan or pattern. The system is "disordered" when its parts are helter-skelter and their arrangement is random.

Examples of systems where entropy increases, that is Ssys = (Sfinal - Sinitial ) = + quantity

Also remember in gases, as volume increases (by pressure decrease at a constant temp or by temp. increase at a constant pressure) entropy increases.

If a reaction produces

a) an increase in the number of particles, S is positive. b) an increase in moles of gaseous compounds, S is positive.

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EXERCISE: Changes in Entropy

Predict the sign of Ssys for each of the following reactions using just the qualitative ideas discussed above.

a. H2O2(l) H2O(l) + O2(g) + – b. H (aq) + OH (aq) H2O(l)

c. CaO(s) + CO2(g) CaCO3(s) + – – d. Ag (aq) + 2CN (aq) Ag(CN)2 (aq)

Entropy Changes in Chemical Reactions. Not only can the sign S for a chemical reaction be estimated, but the actual value of S can also be calculated. The absolute value of the entropy S of one mole of an element or compound can be determined by very careful experimentation. Appendix 3 lists experimental values of the absolute entropy of a number of substances in their standard states at 1 atm and 25°C. Recall that the degree superscript " ° " refers to the standard state of the substance. These values are called absolute or sometimes standard entropies.

For a reaction: aA + bB cC + dD the standard entropy change of the reaction is given by:

S°rxn = nS° (products) – mS° (reactants) where m and n are stoichiometric coefficients. When applied to the above reaction, we get

S°rxn = [c S°(C) + d S°(D)] – [a S°(A) + b S°(B)]

Calculation of the Entropy Change for a Reaction

Use absolute entropies to calculate the standard entropy of reaction :

a) H2(g) + O2(g) H2O(l) S° rxn =

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b) Using tabulated values, calculate S° for the following reaction: N2(g) + 3H2(g) 2NH3(g) S° rxn =

The Second Law. The second law of thermodynamics is concerned with predicting the direction of spontaneous change. It states that the entropy of the universe increases in a spontaneous change. The term universe used here refers to a system and all of it's surroundings. For an isolated system:

Suniv = Ssys + Ssurr 0 where Ssys stands for the entropy change of the system, and Ssurr is the entropy change of the surroundings. For any spontaneous change:

Ssys + Ssurr > 0 and for a reaction at equilibrium (no net change):

Ssys + Ssurr = 0

Note that the entropy of the system can decrease in a spontaneous process, but the entropy of the surroundings must increase enough for the total entropy change of the universe to increase.

The solubility of silver chloride is so low that it precipitates spontaneously from many solutions. The entropy change of the system is negative for this process.

+ – Ag (aq) + Cl (aq) AgCl(s); Ssys = negative; H° = –65 kJ

Since S decreases in this spontaneous reaction, shouldn't the reaction be nonspontaneous?

In order to predict whether a reaction is spontaneous, both Ssys and Ssurr must be considered, not just Ssys.

Since the entropy change of the system is negative, the only way for the inequality to be true is if Ssurr is positive and greater in amount than Ssys. In this case, this is a reasonable assumption because the reaction is exothermic. This means that heat is released to the surroundings, which causes increased thermal motion and disorder of molecules in the surroundings. Therefore, the sum of Ssys and Ssurr is greater than zero, even though Ssys is negative.

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Gibbs Free Energy. Quite often it is not possible to calculate Ssurr. This makes the second law difficult to apply when it is in the form given in the previous section. The Gibbs free energy, expressed in terms of enthalpy and entropy, refers only to the system, yet can be used to predict spontaneity. The Gibbs free energy change ( G) for a reaction carried out at constant temperature and pressure is given by:

G = H – T S where both H and S refer to the system.

The Gibbs free energy change is equal to the maximum possible work (w) that can be obtained from a process. Any process that occurs spontaneously can be utilized to perform useful work. The Gibbs free energy of the system will decrease ( G < 0) in a spontaneous process, and will increase ( G > 0) in a nonspontaneous process. The free energy criteria (at constant temperature and pressure) are summarized as follows:

If G < 0, the forward reaction is spontaneous. If G = 0, the reaction is at equilibrium. If G > 0, the forward reaction is nonspontaneous. The reverse reaction will have a negative G and will be spontaneous.

The standard free-energy change of reaction, , is the free-energy change for a reaction when it occurs under standard state conditions, when reactants in their standard states are converted to products in their standard states.

Calculation of G°rxn. The free energy change for a reaction can be calculated in two ways. When both H and S are known, then G = H – T S will give the free energy change at the temperature T. G°rxn can also be calculated from standard free energies of formation in a manner analogous to the calculation of H° by using enthalpies of formation of the reactants and products. The standard free energies of formation ( G°f) of selected compounds are tabulated in Appendix of the text. Just as for the standard enthalpies of formation, the free energies of formation of elements in their standard states are equal to zero.

For a general reaction aA + bB cC + dD

The standard free energy change is given by

G°rxn = [c G°f(C) + d G°f(D)] – [a G°f(A) + b G°f(B)]

In general: G°rxn = n G°f(products) – m G°f(reactants) where n and m are stoichiometric coefficients.

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EXERCISE:

a) Calculate G°rxn at 25°C for 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(s) using Appendix and given: G°f(Fe2O3) = –741.0 kJ/mol

b) Calculate G°rxn for the following reaction 3NO2(g) + H2O(l) 2HNO3(l) + NO(g)

Given the following free energies of formation:

Enthalpy and Entropy Factors that affect the sign of G°rxn

The temperature at which H° = T S°, (means when G° = 0), can be calculated from the equation T = H°/ S°. At this temperature, the system is at equilibrium. For a phase transition, G = 0 when the two phases coexist in equilibrium. For example, at the boiling point (Tbp) the liquid and vapor phases are in equilibrium, and

Hvap – Tbp Svap = 0

When both H° and S° are positive, above this temperature, the reaction favors the products at equilibrium. When both H° and S° are negative, below this temperature, the reaction favors the products at equilibrium.

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EXERCISE:

a) Hydrated lime Ca(OH)2 can be reformed into quicklime CaO by heating.

Ca(OH)2(s) CaO(s) + H2O(g)

At what temperatures is this reaction spontaneous under standard conditions (that is, where H2O is formed at 1 atm pressure)? Given the following data on Ca(OH)2 that is not in the Appendix: H°f(Ca(OH)2) = –986.2 kJ/mol ; S°(Ca(OH)2) = 83.4 J/K·mol

Temperatures above ______.

b) The heat of fusion of water ( Hfus) at 0°C is 6.02 kJ/mol. What is Sfus for 1 mole of H2O at the melting point?

Sfus = J/K·mol

c) Calculate G°rxn for the following reaction at 298 K: O3(g) O2(g) + O(g) Given H° = 106.5 kJ and S° = 127.3 J/K.

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FREE ENERGY AND EQUILIBRIUM

The relationship between G and G° is:

G = G° + RT ln Q where R is the gas constant (8.314 J/K·mole), T is the absolute temperature, and Q is the reaction quotient. For a certain reaction at a given temperature the value of G° is constant, but the value of Q depends on the composition of the reacting mixture; therefore G will depend on Q. To calculate G, first find G°, then calculate Q from the given concentrations of reactants and products, and substitute into the preceding equation.

Please remember the following about reaction quotient:

For a gas-phase reaction, the thermo-dynamic reaction quotient Q is identical to Q . p

N (g) + 3 H (g)  2 NH (g) 2 2 3

Q = Q = (P of NH )2 / { P of N x ( P of H )3 } p 3 2 2

For a reaction that occurs in aqueous solution, Q is identical to Q . c

2+ 2+ Zn(s) + Cu (aq)  Zn (aq) + Cu(s)

Q = Q = [ zn 2+] / [ Cu 2+ ] c

Q is different from Q or Q when the reaction involves gases and aqueous solutions: p c

2+ + H (g) + Cu (aq)  2 H (aq) + Cu(s) 2

+ 2 2+ Q = [H ] / { ( P of H2 ) x [Cu ] }

At equilibrium, Q = K, and therefore G = 0. The equation then becomes:

0 = G° + RT ln K or G° = – RT ln K

Three possible relationships exist between G° and K, because G° can be negative, positive, or zero.

1. When G° is negative, ln K is positive and K > 1. The products are favored over reactants at equilibrium. The extent of reaction is large. 2. When G° is positive, ln K is negative and K < 1. The reactants are favored over products at equilibrium. The extent of reaction is small.

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3. When G° = 0, ln K is zero and K = 1. The reactants and products are equally favored at equilibrium.

a) The standard free energy change for the reaction

N2(g) + H2(g) NH3(g) is G°rxn = 26.9 kJ/mol at 700 K. Calculate the equilibrium constant at this temperature.

b) Calculate G at 700 K if the reaction mixture consists of 30.0 atm of H2, 20.0 atm of N2, and 0.500 atm of NH3. Use G°rxn = 26.9 kJ/mol G = kJ/mol

c) The equilibrium constant for the reaction: AgBr(s) Ag+(aq) + Br–(aq) –13 is the solubility product constant, Ksp = 7.7 x 10 at 25°C. Calculate G for the reaction when [Ag+] = 1.0 x 10–2 M and [Br–] = 1.0 x 10–3 M. Is the reaction spontaneous or nonspontaneous at these concentrations?

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