Chemical Thermodynamics Chapter 19 Chemical Thermodynamics

Spontaneous Processes Section 19.1 A spontaneous process is one that occurs without the influence of an outside force Occurs in a single direction and the opposite direction is nonspontaneous

Spontaneity and the Relationship Between Thermodynamics and Kinetics A chemical or physical process may be highly spontaneous, but at the same time may take place slowly Thermodynamics offers information about the direction and extent that a process occurs but yields no information as to the rate of reaction

The Link Between Spontaneity and Enthalpy The vast majority of spontaneous chemical reactions are exothermic in nature; however, there are examples of spontaneous processes that are endothermic For an endothermic process to be spontaneous there must be some type of “compensation” to offset the increase in energy between initial and final states This “compensation” is referred to as entropy

Making Qualitative Predictions About S While entropy itself is an abstract concept, making predictions about the change in entropy for a chemical process is simple Some simple guidelines: Temperature Increase in T leads to higher entropy Change of phase Increasing entropy: solid<liquid<gas Number of particles More particles in solution results in higher entropy Dissolution Particles in solution have more entropy than when undissolved

Predicting the Sign of S Predict whether S is positive or negative for the following processes: CO2(s)  CO2(g) CaO(s) + CO2(g)  CaCO3(s) HCl(g) + NH3(g)  NH4Cl(s) 2 SO2(g) + O2(g)  2 SO3(g) See Sample Exercise 19.3 (Pg. 814)

Third Law of Thermodynamics The Third Law states that at 0 K, all motion completely stops (S = 0)

Entropy Changes in Chemical Reactions Section 19.4 Just like enthalpy, entropy change for a reaction can be calculated using tabulated standard molar enthalpies: S = S(products) - S(reactants) Example: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

Gibbs Free Energy Section 19.5 Because there are endothermic processes that take place spontaneously and exothermic processes that take place with a decrease in entropy, there must be a dependence upon both these quantities on reaction spontaneity These terms have been combined along with a new term, Gibbs free energy: G = H – TS or G = H - T S Std. Conditions: G = H - T S

Gibbs Free Energy and Spontaneity Since Gibbs free energy combines both enthalpy and entropy, this is what ultimately determines reaction spontaneity: If G is negative, the reaction is spontaneous If G is positive, the reaction is nonspontaneous If G = 0, the reaction is at equilibrium (both directions equally spontaneous)

Calculating Grxn Calculate Grxn and determine whether the following reaction will take place spontaneously under standard state conditions H2S(g) + 2H2O(l) → SO2(g) + 3H2(g) See Sample Exercise 19.7 (Pg. 823)

Estimating G Predict whether G for this reaction is more or less negative than H: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Hrxn = -2220 kJ See Sample Exercise 19.8 (Pg. 823)

Free Energy and Temperature Section 19.6 G = H - T S G = H + (-T S) Enthalpy term Entropy Term Ex: H2O(s)  H2O(l) H > 0, S > 0

Effect of Temperature on Spontaneity

Determining Effect of Temperature on Spontaneity Consider the production of carbon dioxide and methane from carbon monoxide and hydrogen. 2CO(g) + 2H2(g)  CO2(g) + CH4(g) a.) Calculate G, S, and H for the reaction at 298 K b.) Calculate G for the same reaction at 1000 K. Assume H and S do not change much with temperature See Sample Exercise 19.9 (Pg. 826)

Relating G to a Phase Change at Equilibrium a.) Write the chemical equation that defines the normal boiling point of liquid methanol (CH3OH). b.) Determine the value of G for the equilibrium in part a.) c.) Use thermodynamic data from Appendix C to estimate the normal boiling point of CH3OH. See Sample Exercise 19.10 (Pg. 827)

Free Energy and the Equilibrium Constant Section 19.7 G determines spontaneity of a process, but only under standard conditions: [ ] = 1 M; P = 1 atm; T = 25 C (298 K) If conditions are other than standard: G = G + RT ln(Q)

Calculating the Free Energy Change Under Nonstandard Conditions Calculate the Gibbs free energy change for the reaction shown below at 298 K. PNO = 0.1 atm and PNOBr = 2.0 atm. G(NOBr) = 82.4 kJ/mol 2NO(g) + Br2(l)  2NOBr(g) See Sample Exercise 19.11 (Pg. 828)

Calculating Equilibrium Constants Using Gibbs Free Energy At equilibrium G = 0; therefore: G = -RT ln(K) or rearranging:

Calculating an Equilibrium Constant from G The Gibbs free energy change for the following reaction is +55.69 kJ. AgCl(s)  Ag+(aq) + Cl-(aq) Calculate the equilibrium constant for this reaction at 350 K. See Sample Exercise 19.12 (Pg. 829)