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Entropy and the Driving Force

Entropy and the Driving Force. Each substance has its own entropy value. This is an absolute scale, because a perfectly ordered substance at 0 K has an entropy of 0. Compare this to enthalpy, where only changes, Δ H, can be measured. Some entropy values. Temperature effects on entropy .

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Entropy and the Driving Force

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  1. Entropy and the Driving Force • Each substance has its own entropy value. This is an absolute scale, because a perfectly ordered substance at 0 K has an entropy of 0. • Compare this to enthalpy, where only changes, ΔH, can be measured.

  2. Some entropy values

  3. Temperature effects on entropy • S of water • (J/mol.K): • Increasing temperature increases entropy because random motion of molecules has increased

  4. Why isn’t everything a gas? • Gases have more entropy than the equivalent solids, but energy factors – such as intermolecular forces – can make it favorable to exist as a solid • Other things being equal, reactions that produce gases are favored

  5. Entropy change of reactions • ∆S can be calculated just like ΔH: • ∆So = ΣSo(prod) - ΣSo(react) • For: ½N2(g) + O2 (g) → NO2(g) • So: 191.6 205.1 240.1 J/mol.K • ΔSo = So(NO2) – [½So(N2) + So(O2)] • = 240.1 - [½(191.6) + 205.1)] • = -60.8 J J/mol.K

  6. ½N2(g) + O2 (g) → NO2(g) • ΔSo = -60.8 J/mol.K • Entropy decreases in this process, consistent with the fewer moles of gas in the products than in the reactants

  7. Entropy, enthalpy and spontaneity: Surroundings System ΔH “The Universe” An increase in temperature also causes an increase in entropy

  8. The “Second Law” of Thermodynamics • In a spontaneous process, the total entropy change of the universe must be positive (i.e., ∆S > 0) • The S of the system can decrease, but only if S of the surroundings increases to a greater extent. • Exothermic reactions always raise the entropy of the surroundings, which is why they are intrinsically more favorable.

  9. The four cases: • ΔH < 0; ΔSsys > 0: reaction is spontaneous • ΔH < 0; ΔSsys < 0: depends on values • ΔH > 0; ΔSsys > 0: depends on values • ΔH > 0; ΔSsys < 0: reaction not spontaneous • There ought to be just one function that tells us whether the reaction is spontaneous [“will tend to occur”] or not!

  10. A quick derivation • ΔSsurroundings = - ΔH/T • Negative sign, because if reaction is exothermic, ΔH is negative, but if it makes the temperature of the surroundings increase, the entropy of the surroundings should increase (positive ΔS) • 1/T because effect of increased random molecular motion is more pronounced at lower temperatures than higher

  11. A quick derivation

  12. Reactants G Products For a spontaneous process, ΔG is always negative

  13. The positive sign indicates that this reaction is not spontaneous at 25oC • The o indicates standard conditions (P = 1 atm; concentrations = 1 M) • ΔGo = 33.2 kJ/mol – T(-60.8 J/mol.K) • Because ΔH is positive and ΔS is negative, this reaction can never be spontaneous.

  14. Another way to calculate ΔGo • ΔGfo values are also tabulated • ΔGfo = Gibbs free energy change when one mole of a substance is formed from its elements in their standard states • ΔGo = ΣΔGfo(prod) – ΣΔGfo(reactants)

  15. So is ΔGo related to Keq?And what can ΔG tell us about systems not at equilibrium? • Consider: -ln(K/Q) • Recall: ln(1) = 0 • ln (>1) = positive number • ln (<1) = negative number • If Q < K, reaction goes to right, -ln(K/Q) is - • If Q> K, reaction goes to left, -ln(K/Q) is +

  16. If Q < K, reaction goes to right, -ln(K/Q) is -If Q> K, reaction goes to left, -ln(K/Q) is + • So we can say ∆G α -ln(K/Q) • Specifically: ∆G = -RTln(K/Q) • At standard conditions, Q = 1 (!) . . . so . . . • ∆Go = -RT lnK a marvelous equation! • R = 8.31 J/mol.K • This allows us to calculate an equilibrium constant from basic thermodynamic information

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