1. Thermochemistry

2. Overview Entropy & Second Law ThermodynamicsEntropy & Second Law Thermodynamics Predicting SpontaneityPredicting Spontaneity Free EnergyFree Energy

3. Entropy & Second Law Thermodynamics Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings.Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings. A spontaneous process is one that proceeds on its own without any continuous external influence.A spontaneous process is one that proceeds on its own without any continuous external influence. A nonspontaneous process takes place only in the presence of a continuous external influence.A nonspontaneous process takes place only in the presence of a continuous external influence.

4. Entropy & Second Law Thermodynamics The measure of molecular disorder in a system is called the system’s entropy; this is denoted S.The measure of molecular disorder in a system is called the system’s entropy; this is denoted S. Entropy has units of J/K (Joules per Kelvin).Entropy has units of J/K (Joules per Kelvin). –  S = S final – S initial –Positive value of  S indicates increased disorder. –Negative value of  S indicates decreased disorder.

5. Entropy & Second Law Thermodynamics

6. Predicting Spontaneity To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered:To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered: Spontaneous process:Decrease in enthalpy (–  H). Increase in entropy (+  S).Spontaneous process:Decrease in enthalpy (–  H). Increase in entropy (+  S). Nonspontaneous process:Increase in enthalpy (+  H). Decrease in entropy (–  S).Nonspontaneous process:Increase in enthalpy (+  H). Decrease in entropy (–  S).

7. Predicting Spontaneity Predict whether  S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate  S° for each:Predict whether  S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate  S° for each: –a. 2 CO(g) + O 2 (g)  2 CO 2 (g) b. 2 NaHCO 3 (s)  Na 2 CO 3 (s) + H 2 O(l) + CO 2 (g) c. C 2 H 4 (g) + Br 2 (g)  CH 2 BrCH 2 Br(l) d. 2 C 2 H 6 (g) + 7 O 2 (g)  4 CO 2 (g) + 6 H 2 O(g)

8. Free Energy Gibbs Free Energy Change (  G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.Gibbs Free Energy Change (  G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process. –  G =  H – T  S –  G < 0Process is spontaneous –  G = 0Process is at equilibrium –  G > 0Process is nonspontaneous

9. Free Energy Situations leading to  G < 0:Situations leading to  G < 0: –  H is negative and T  S is positive –  H is very negative and T  S is slightly negative –  H is slightly positive and T  S is very positive Situations leading to  G = 0:Situations leading to  G = 0: –  H and T  S are equally negative –  H and T  S are equally positive Situations leading to  G > 0:Situations leading to  G > 0: –  H is positive and T  S is negative –  H is slightly negative and T  S is very negative –  H is very positive and T  S is slightly positive

10. Free Energy Which of the following reactions are spontaneous under standard conditions at 25°C?Which of the following reactions are spontaneous under standard conditions at 25°C? –a. AgNO 3 (aq) + NaCl(aq)  AgCl(s) + NaNO 3 (aq)  G° = –55.7 kJ –b. 2 C(s) + 2 H 2 (g)  C 2 H 4 (g)  G° = 68.1 kJ –c. N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92 kJ;  S° = –199 J/K

11. Free Energy Equilibrium (  G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature?Equilibrium (  G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature? – N 2 (g) + 3 H 2 (g)  2 NH 3 (g)  H° = –92.0 kJ  S° = –199 J/K –Equilibrium is the point where  G° =  H° – T  S° = 0