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Thermodynamics

Thermodynamics. Spontaneous Processes. Proceeds on its own without any assistance The process in the other direction will be nonspontaneous May or may not occur at a fast rate. Examples:. Some spontaneous processes are endothermic. Example: Baking soda dissolves in water

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Thermodynamics

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  1. Thermodynamics

  2. Spontaneous Processes • Proceeds on its own without any assistance • The process in the other direction will be nonspontaneous • May or may not occur at a fast rate Examples:

  3. Some spontaneous processes are endothermic • Example: Baking soda dissolves in water • Energy is absorbed • Surroundings feel cold • Why?

  4. Entropy • The degree of randomness in a system • Is a state function • S = symbol • Unit: J/K S = Sfinal - Sinitial • For isothermal processes: S = qrev T

  5. Second Law of Thermodynamics The total entropy of the universe increases in any spontaneous process

  6. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is zero S at 0 K = 0

  7. Qualitative Predictions about S An increase in the number of microstates of a system will increase entropy If one of the following increases, entropy will as well: 1. temperature 2. volume 3. number of independently moving particles

  8. Qualitative Predictions about S Entropy increases for the following processes: 1. Gases are formed from solids or liquids 2. Liquids or solutions are formed from solids 3. The number of gas molecules increase during a chemical reaction

  9. Example: is S positive or negative? H2O (l)  H2O (g) Ag+ (aq) + Cl- (aq)  AgCl (s) 4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s) N2 (g) + O2 (g)  2 NO (g)

  10. Entropy Changes in Chemical Reactions Standard molar entropy (S) – the molar entropy value of a substance in its standard state at 298 K Are NOT zero Gases  Liquids  Solids Increases with increasing molar mass Increases with an increasing number of atoms in the formula of a substance

  11. Entropy Change in a Chemical Reaction *multiply S values by coefficients from the balanced equation S =  Sproducts -  Sreactants

  12. Calculating S from Tabulated Entropies Calculate S for the synthesis of ammonia from N2 (g) and H2 (g) at 298 K.

  13. Gibbs Free Energy • Symbol = G • The maximum useful work that can be done by the system on its surroundings (at constant T and P) • Some energy is lost as heat to the environment

  14. Gibbs Free Energy G = H – TS G = H - TS (constant temperature) G = H - TS (standard conditions)

  15. Gibbs Free Energy and Spontaneity If G is: Negative The reaction is spontaneous in the forward direction Zero The reaction is at equilibrium Positive The reaction is nonspontaneous Work must be supplied from the environment

  16. Summary of Gibbs Free Energy In any spontaneous process, the free energy always decreases

  17. Calculating Free Energy Change from H , T, and S Calculate the standard free energy for the formation of NO (g) from N2 (g) and O2 (g) at 298 K N2 (g) + O2 (g)  2 NO (g) given that H = 180.7 kJ and S = 24.7 J/K. Is the reaction spontaneous under these circumstances?

  18. Standard Free Energies of Formation (Gf) *multiply Gfvalues by coefficients from the balanced equation • Each substance has its own Gf • Gf = zero for elements in their standard states Gf =  Gfproducts -  Gfreactants

  19. Calculating Standard Free Energy Change of a Reaction a) Calculate the standard free energy change for the following reaction at 298 K: P4 (g) + 6 Cl2 (g)  4 PCl3 (g) b) What is Grxn for the reverse of the above reaction?

  20. Estimating and Calculating G For the combustion of propane gas, H = -2220 kJ C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l) Estimate whether G for this reaction is more or less negative than H. Use thermodynamic data to calculate the actual value for G. Were your predictions correct?

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