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Understanding Spontaneity and Entropy in Reactions

Learn about spontaneous reactions, factors affecting entropy, calculating entropy change, and the Gibbs free energy equation.

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Understanding Spontaneity and Entropy in Reactions

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  1. Spontaniety • Spontaneous reactions are reactions, that once started, continues by itself without further input of energy from the outside. • If a reaction is spontaneous under a given set of conditions, then the reverse reaction is considered non-spontaneous.

  2. Identifying Spontaneity • Identify whether the reactions below are spontaneous or not: • H2O (s)  H2O (l) • 2H2 (g) + O2 (g)  2H2O (l) *requires a spark to begin • 2H2O (l)  2H2 (g) + O2 (g) *requires consistent electric current

  3. Enthaply and Spontaneity: • Almost all exothermic reactions are considered to be spontaneous (at 25 *C and 1 atm). • ΔH for a spontaneous reactions tends to be negative. • However, some endothermic reactions at specific temperatures may be considered spontaneous, for example the melting of ice at 1 atm above 0 *C: H2O (s)  H2O (l) ΔH = 6.0 kJ • Endothermic reactions that are non-spontaneous at room temperature often become spontaneous when the temperature is raised. • The Randomness Factor: In general, nature tends to move spontaneously from more ordered to more random states (order to disorder).

  4. Entropy • The randomness factor discussed is treated quantitatively as entropy (S). Basically the greater the disorder (more random the distribution of molecules) the greater the entropy. • -Entropy, like enthalpy is a state property so that ΔS = Sfinal - Sinitial • unit for S = J/mol K

  5. Factors that influence entropy that a system has in a particular state: • A liquid has higher entropy than the solid from which it is formed. • A gas has a higher entropy than the liquid from which it is formed. • Increasing the temperature of a substance increases its entropy. • A completely ordered pure crystalline solid has an entropy of 0 K (3rd law of thermodynamics).

  6. Standard molar entropies of elements, compounds, and aqueous ions: • Elements have nonzero standard entropies. • Standard molar entropies of pure substances are always positive quantities. • Aqueous ions may have negative entropy values. • As a group, gases tend to have higher entropies than liquids. An increase in the number of moles of a gas also leads to a higher entropy and vice versa. • As a molecule becomes more complex, the higher the entropy (more ways for the atoms to move about with respect to one another).

  7. Example 17.1 • Predict whether ΔS is positive or negative for each of the following processes: • Taking dry ice from a freezer where its temperature is -80°C and allowing it to warm to room temperature • Dissolving bromine in hexane • Condensing gaseous bromine to liquid bromine

  8. Which of the following reactions results in the largest increase in entropy? (A) CO2(s) CO2(g) (B) H2(g) + Cl2(g) 2HCl(g) (C) KNO3(s) KNO3(l) (D) C(diamond) C(graphite)

  9. Calculating Entropy • To calculate the standard entropy change, ΔS, use the following relation: ΔS = ∑ ΔS products - ∑ ΔS reactants • The 2nd Law of Thermodynamics: In a spontaneous process, there is a net increase in entropy, taking into account both the system and surroundings, ΔS > 0

  10. Example • Calculate the entropy change at 25ºC in J/K for 2SO2(g) + O2(g) 2SO3(g) given the following data: • SO2(g): 248.1 J/mol-K • O2(g): 205.3 J/mol-K • SO3(g): 256.6 J/mol-K • [2(256.6)] - [2(248.1) + 1(205.3)] = -188.3 J/K

  11. Example 17.2 • Calculate ΔSfor • (1) dissolving one mole of calcium hydroxide in water • (2) the combustion of one mole of methane, CH4, to form carbon dioxide and liquid water

  12. Gibbs Free Energy (G) • Gibbs free energy is a quantity that basically allows us to put the two quantities, enthalpy and entropy, together in such a way as to arrive at a single function whose sign will determine whether the reaction is spontaneous. The basic definition of Gibbs free energy is: ΔG = ΔH - TΔS (The Gibbs-Helmholtz Equation)  • G = Gibbs Free Energy (J) • H = Enthalpy (J/mol) • T = Temperature (K) • S = Entropy (J/mol K -- please be aware that the mole in the unit tends to cancel out from the ΔS equation)

  13. Understanding ΔG • The Gibb’s free energy equation combines all the information that we have learned thus far. But what does the Gibb’s free energy value tell us about a reaction? It tells us the following: • If ΔG is negative, the reaction is spontaneous in the forward direction. • If ΔG is equal to zero, the reaction is at equilibrium. • If ΔG is positive, then the reaction is nonspontaneous in the forward direction, but the reverse reaction will be spontaneous. • For elements at standard state (pure elements at 25ºC and 1 atm are assigned a value of zero).

  14. Example 17.3 • For the reaction CaSO4 (s) Ca2+(aq) + SO42- (aq) calculate: • ΔH° • ΔS° • ΔG° at 25°C

  15. Gibbs Free Energy of Formation • The standard free energy of formation, ΔGf, can also be used to solve for the free energy of a reaction (reference Appendix 1 in text): ΔGf rxn = ∑ ΔGfproducts - ∑ ΔGfreactants • If it is a negative quantity then the compound can be formed spontaneously from the elements, like in the formation of water: H2 (g) + ½ O2 (g)  H2O (l) ΔGf = -237.2 kJ • Elements in their elemental state will have a ΔGf = 0

  16. Example 17.4 • Using ΔG°f values from Appendix 1, calculate the standard free energy change at 25°C for the reaction CaSO4 (s) Ca2+(aq) + SO42- (aq) • Using ΔG°f values from Appendix 1, calculate the standard free energy change at 25°C for the dissolution of 1 mole of calcium chloride.

  17. Δ S + ΔG = negative spontaneous at all temperatures ΔG = ?? spontaneous at high temperatures 0 K=1 Δ H – Δ H + ΔG = ?? spontaneous at low temperatures ΔG = positive non-spontaneous at all temperatures Δ S –

  18. Example 17.6 • At what temperature does ΔG°become zero for the reaction Fe2O3 (s) + 3H2 (g)  2Fe (s) + 3H2O (g)

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