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Chapter 17

Chapter 17. Spontaneity, Entropy, and Free Energy. Spontaneous Processes Section 17.1. 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.

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Chapter 17

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  1. Chapter 17 Spontaneity, Entropy, and Free Energy

  2. Spontaneous ProcessesSection 17.1 • 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

  3. 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

  4. 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

  5. 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

  6. 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) Predicting the Sign of S

  7. Entropy Changes in Chemical Reactions • 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)

  8. Gibbs Free EnergySection 17.4 • 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

  9. 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)

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

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

  12. Effect of Temperature on Spontaneity

  13. 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 Determining Effect of Temperature on Spontaneity

  14. 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. Relating G to a Phase Change at Equilibrium

  15. Free Energy and the Equilibrium Constant Section 17.9 • 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)

  16. 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 Interactive Example 17.14 (Pg. 695) Calculating the Free Energy Change Under Nonstandard Conditions

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

  18. 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. Calculating an Equilibrium Constant from G

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