Chapter 17.3 Entropy and Spontaneity Objectives Define entropy and examine its statistical nature Predict the sign of entropy changes for phase changes Apply the second law of thermodynamics to chemical systems Recognize that absolute entropies can be measured because the third law of thermodynamics defines a zero point
Entropy is a measure of randomness An increase in randomness (also known as disorder) is an important driving force for many changes. Two different gases, initially separated by a partition, will mix spontaneously when the partition is removed, increasing the randomness of the system.
Entropy and Spontaneity Thermodynamics is able to relate the spontaneity to a state function called entropy. Entropy (S) is the thermodynamic state function that describes the amount of disorder or randomness. A large value for entropy means a high degree of disorder or randomness.
Second Law of Thermodynamics The second law of thermodynamics states that total entropy will always increase. In any spontaneous process, the entropy of the universe increases. S univ = S sys + S surr > 0 If S sys < 0 for a spontaneous process, then a larger positive change in S surr must occur.
Spontaneity and DS univ S univ = S sys + S surr When S univ > 0, the change occurs spontaneously. When S univ < 0, the reverse change occurs spontaneously. When S univ = 0, the system is at equilibrium.
Entropy Change (DS) An increase in randomness results in an increase in entropy. Some general guides are: The entropy of a substance increases when solid becomes liquid, and when liquid becomes gas. The entropy generally increases when a solute dissolves. The entropy decreases when a gas dissolves in a solvent. The entropy increases as temperature increases.
Phase Change (DS) The entropy of a system generally increases when a solid dissolves in a liquid. Low entropy, very ordered or not very random. S = n q T High entropy, disordered or more random.
The Third Law of Thermodynamics The third law of thermodynamics states that the entropy of a perfect crystal of a substance at absolute zero is equal to zero (0). A perfect crystal at 0 K has no disorder. Unlike enthalpy and internal energy, absolute values of entropy can be determined.
Absolute Entropies In Appendix G of your text book, absolute entropies are given for substances in their standard state 298 K. The entropy change for a reaction is given by: S o = ns rxn o ms products o reactants where n and m are the coefficients of the products and reactants in the reaction.
In Class Example Calculate the standard entropy change for the reaction 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) Substance S o, J/ mol-k C 2 H 6 (g) 229.49 CO 2 (g) 213.63 O 2 (g) 205.03 H 2 O(l) 69.91
In Class Example The combustion of methane is spontaneous, but S o is -620.21 J/K. Does this violate the second law of thermodynamics? Why or why not?
Chapter 17.4 Gibbs Free Energy Objectives Define Gibbs free energy and relate the sign of a Gibbs free energy change to the direction of spontaneous reaction. Predict the influence of temperature on Gibbs free energy. November 10 th, 2016
The Gibbs Energy Equation J.W. Gibbs defined a state function called the Gibbs free energy, G, and directly related it to enthalpy and entropy: G = H - TS At constant standard temperature and pressure, this becomes G o = H o - T S o
Free Energy and Spontaneity G is a state function of the system. For any spontaneous change G < 0 S univ > 0 From the equation G = H - T S, a negative H and a positive S favor spontaneity. A minimum free energy occurs when the system is a equilibrium, and G = 0.
Standard Free Energy of Formation The standard free energy of formation is the free energy change to form one mole of a compound from its elements in their standard state. G fo = H fo - T S fo, can be calculated from tabulated data. G rxn = n G o f products m G o f reactants where n and m are the coefficients of the products and reactants in the reaction.
In Class Example Calculate the standard free energy change for the reaction 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6H 2 O(l) Substance DG o, kj/mol C 2 H 6 (g) -32.89 CO 2 (g) -394.4 O 2 (g) 0 H 2 O(l) -237.2
Temperature Dependence of DG The dependence of G on temperature arises mainly from the T S term in the definition of G. The sign of G is dominated by the sign of H at low temperatures and by the sign of S at high temperatures. Negative values of H and positive values of S favor spontaneity.
Direction of Spontaneous Reaction H S Temperature G Spontaneous Direction - + All - Forward - - + + Low High Low High - + + - Forward Reverse Reverse Forward + - All + Reverse
In Class Example CS 2 (l) CS 2 (g) At 298 K, H o = 27.66 kj/mol and S o = 86.39 J/mol*K. Calculate DG o Assuming H o and S o do not change with temperature, calculate the temperature for which G is 0 for the reaction.
Student Example Given the following chemical reaction and data at 298 K calculated DG o : N 2 (g) + 3H 2 (g) 2NH 3 (g) H fo, kj/mol 0 0-46.1 S o, J/mol-K 191.5 130.6 192.3 Assuming that H o and S o do not change with temperature, calculate G o at 1000 K.
Concentration and Free Energy Concentrations of reactant and products in an equilibrium reaction influence the free energy change of a reaction according to the equation G = G o + RTlnQ where Q is the reaction quotient (see chapter 14).
In Class Example For the reaction 2NO 2 (g) N 2 O 4 (g) Calculate G o at 298 K. Substance G fo kj/mol NO 2 (g) 51.29 N 2 O 4 (g) 97.82 Calculate G when P NO2 = 0.12 atm and P N2 O 4 = 0.98 atm at 298 K.
Thermodynamics Summary Exothermic Reaction DH is negative Endothermic Reaction DH is positive Disorder Increases (more gases formed) DS is positive Disorder Decreases (less gases formed) DS is negative Spontaneous Reaction DG is negative Non-spontaneous Reaction DG is positive Look to table on slide 18 for how DH and DS can influence DG
Chapter 17.5 DG o and K eq Objectives Determine the effect of concentration on Gibbs free energy Calculate the standard Gibbs free energy change from the equilibrium constant, and vice versa Determine the effect of temperature on the equilibrium constant Understand the connection between Gibbs free energy and work November 10 th, 2016
Free Energy and Equilibrium Constant G = G o + RTlnQ For a system at equilibrium, G = 0, and Q = K eq, so G o = -RTlnK eq G o, calculated from the data in Appendix G, can be used to calculate the value of the equilibrium constant.
In Class Example Given the following equation and the data below calculate the equilibrium constant (K eq ) at 298 K. 2NO(g) + Br 2 (g) 2NOBr(g) G fo, kj/mol 86.55 3.14 82.4
Student Example Given the following equation and that the concentrations at equilibrium at NH 3 = 0.1 M, N 2 = 0.1 M and H 2 = 0.005 M calculate Gibbs Free Energy (DG o ) of the reaction at 298 K. N 2 (g) + 3H 2 (g) 2NH 3 (g)
Temperature and K eq The temperature dependence of K eq is derived from two equations given earlier: G o = -RTlnK eq = H o T S o lnk eq = S o /R - H o /RT A graph of lnk eq vs. 1/T gives a straight line with slope = - H o /R and an intercept of S o /R. ln K 1 K 2 = Ho R 1 T 2 1 T 1
Bringing it all together The temperature dependence of G comes from the sign of S. G = H - T S The temperature dependence of K eq comes from the sign of H. lnk eq = S o /R - H o /RT Increasing T decreases the value of K eq if the reaction is exothermic. Exactly as predicted by Le Chatelier s principle.
Bringing it all together still The change in Free Energy is the maximum work that can be performed by a spontaneous chemical reaction at constant temperature and pressure. w max = G When G > 0 (spontaneous in the reverse direction), it represents the minimum work that must be provided to cause a change.