Química Geral · Entropy, free energy and electrochemistry
Entropia e energia livre
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O que você vai aprender
- Explain spontaneity in terms of microstates and predict the sign of ΔS for physical and chemical changes.
- Calculate ΔS°rxn from standard molar entropies, ΔSsurr = −ΔH/T and ΔSuniv, and apply the second and third laws.
- Use ΔG = ΔH − TΔS to classify the four sign cases and find the crossover temperature.
- Calculate ΔG° from free energies of formation and relate ΔG, ΔG°, Q and K through ΔG = ΔG° + RT ln Q and ΔG° = −RT ln K.
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Spontaneity and microstates
A spontaneous process proceeds on its own once started, without continuous outside help. Spontaneous says nothing about speed: diamond turning into graphite is spontaneous at 25 °C but immeasurably slow. Entropy, S, counts the microstates of a system, the arrangements of positions and energies that are consistent with its state: S = k ln W, where k is the Boltzmann constant and W the number of microstates. A gas released into an evacuated flask spreads into both flasks even though, for an ideal gas, ΔH = 0. With three labelled molecules and two equal flasks there are 2³ = 8 arrangements, and only one of them puts all three in the left flask. Each added molecule doubles the count, so for a mole of molecules the spread-out state is so overwhelmingly probable that the gas never gathers back into one flask. Entropy is a state function measured in J/K, and molar entropies in J mol⁻¹ K⁻¹.
Predicting the sign of ΔS
Entropy rises from solid to liquid to gas, with more moles of gas, on heating, on expansion to a larger volume (lower pressure), on mixing, and for larger, heavier and more flexible molecules. The noble gases show the mass effect: S° = 126.2 (He), 146.3 (Ne) and 154.8 (Ar) J mol⁻¹ K⁻¹. In a reaction the change in moles of gas usually dominates: CaCO₃(s) → CaO(s) + CO₂(g) has ΔS > 0, while 2 CO(g) + O₂(g) → 2 CO₂(g) has ΔS° = 2(213.8) − 2(197.7) − 205.2 = −173.0 J/K. A reaction with no change in moles of gas, such as N₂(g) + O₂(g) → 2 NO(g), has a small ΔS°. Dissolving a solid salt usually raises the entropy, because the lattice breaks into mobile ions, although the ions also order the water molecules around them. Dissolving a gas lowers the entropy: CO₂(g) → CO₂(aq) has ΔS° = 117.6 − 213.8 = −96.2 J/K.
Second and third laws; ΔS from tables
Second law: in every spontaneous process the entropy of the universe increases, ΔSuniv = ΔSsys + ΔSsurr > 0; at equilibrium ΔSuniv = 0. The entropy of the system alone may fall, as when water freezes. Third law: a perfect crystal at 0 K has S = 0. Entropies are therefore absolute, and every substance at 25 °C has S° > 0, elements included (S° of O₂(g) is 205.2 J mol⁻¹ K⁻¹), unlike ΔH°f. Some aqueous ions have negative S° because ion values are referenced to S°(H⁺, aq) = 0. For a reaction, ΔS° = ΣnS°(products) − ΣnS°(reactants). For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l): ΔS° = 213.8 + 2(69.9) − 186.3 − 2(205.2) = −243.1 J/K. Heat released at constant pressure raises the entropy of the surroundings: ΔSsurr = −ΔH/T. This combustion has ΔH° = −890.3 kJ, so at 298 K ΔSsurr = +890 300/298 = +2988 J/K, and ΔSuniv ≈ +2.74 × 10³ J/K: spontaneous although ΔSsys < 0. The same heat produces a larger ΔSsurr at a lower temperature. At a phase-transition temperature the two phases are in equilibrium and ΔS = ΔH/T: melting ice at 273.15 K has ΔSfus = 6010/273.15 = 22.0 J mol⁻¹ K⁻¹.
Gibbs free energy and temperature
At constant temperature and pressure ΔG = ΔH − TΔS, which equals −TΔSuniv. ΔG < 0: spontaneous as written; ΔG > 0: nonspontaneous (the reverse is spontaneous); ΔG = 0: equilibrium. Tables give ΔS in J/K and ΔH in kJ, so convert ΔS to kJ/K before combining them. The signs give four cases. ΔH < 0, ΔS > 0: spontaneous at every temperature. ΔH > 0, ΔS < 0: spontaneous at no temperature. ΔH < 0, ΔS < 0: spontaneous below the crossover temperature. ΔH > 0, ΔS > 0: spontaneous above it. Treating ΔH° and ΔS° as constant, ΔG° = 0 at the crossover T = ΔH°/ΔS°. Example: the dissociation of colourless N₂O₄ (drawn) into brown NO₂, N₂O₄(g) → 2 NO₂(g), has ΔH° = +57.2 kJ and ΔS° = +175.8 J/K, so T = 57 200/175.8 = 325 K (52 °C). Below about 325 K, ΔG° > 0 (K < 1); above it, ΔG° < 0 (K > 1), which is why a sealed flask of the gas turns darker brown when warmed. A plot of ΔG° against T is a straight line with intercept ΔH° and slope −ΔS°.
ΔG° from free energies of formation
Standard free energies combine like enthalpies: ΔG°rxn = ΣnΔG°f(products) − ΣnΔG°f(reactants). ΔG°f = 0 for an element in its standard state, such as O₂(g), C(graphite) or Br₂(l), but not for other forms: O₃(g) is +163.2, C(diamond) +2.9 and Br₂(g) +3.1 kJ/mol. Example: CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(l) has ΔG° = −394.4 + 2(−237.1) − (−50.8) = −817.8 kJ. ΔG° refers to standard states (gases at 1 bar, solutes at 1 M) and to the temperature of the table, usually 298 K. At another temperature estimate ΔG° ≈ ΔH° − TΔS°; ΔG° itself changes strongly with T.
Free energy, Q and K
Away from standard conditions ΔG = ΔG° + RT ln Q, with R = 8.314 J mol⁻¹ K⁻¹; express ΔG° in J/mol, or RT in kJ/mol, before adding. Gas pressures in Q are in bar and solute concentrations in mol/L. If Q < K, ΔG < 0 and the reaction runs forward; if Q > K, ΔG > 0 and it runs in reverse; if Q = K, ΔG = 0 and the system is at equilibrium. Setting ΔG = 0 and Q = K gives ΔG° = −RT ln K, so K = e^(−ΔG°/RT). Example: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) at 298 K has ΔG° = −267.8 − (−305.0) = +37.2 kJ/mol, so K = e^(−37 200/2477.6) = e^(−15.01) = 3.0 × 10⁻⁷. A positive ΔG° means K < 1, not that no product forms. In a mixture with Q = 1.0 × 10⁻⁹, ΔG = 37.2 + (2.4776)(ln 1.0 × 10⁻⁹) = 37.2 − 51.3 = −14.1 kJ/mol, so this mixture reacts forward. At 298 K every 5.7 kJ/mol of ΔG° (2.303RT) changes K by a factor of 10.
What to carry forward
The entropy of the universe decides spontaneity; at constant T and P that becomes ΔG = ΔH − TΔS < 0. Predict ΔS from phases and moles of gas. Compute ΔS° and ΔG° as products minus reactants, remembering that S° of an element is not zero but ΔG°f of an element in its standard state is. Put ΔS in kJ/K beside ΔH in kJ, find the crossover at T = ΔH°/ΔS°, and connect free energy to composition with ΔG = ΔG° + RT ln Q and ΔG° = −RT ln K.
Problemas de prática (40)
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Preveja o produto 2
Entropy and spontaneity
- Which change lowers the entropy of a gas?Predict how volume, temperature and mixing change the entropy of an ideal gas sample.
- Which process lowers the entropy?Predict which of several dissolution and gas-phase processes has a negative ΔS° by weighing loss of gas against ion formation.
Classificar 5
Free energy and equilibrium
- What is zero at equilibrium?Classify statements about ΔG, ΔG°, Q and K for a system at equilibrium.
- Which species has ΔG°f = 0?Identify which form of an element has a standard free energy of formation of zero.
Free energy and temperature
- Which reaction is never spontaneous?Classify reactions into the four ΔH/ΔS sign cases and pick the one that is nonspontaneous at every temperature.
Entropy and spontaneity
- Which process increases the entropy?Classify processes by the sign of ΔS using phase changes and the change in moles of gas.
- Which statement is the second law?Identify the second law of thermodynamics among the statements of the three laws.
Comparar 4
Free energy and temperature
- Rank crossover temperaturesRank reactions with positive ΔH° and ΔS° by the temperature at which ΔG° becomes zero.
Free energy and equilibrium
- Rank equilibrium constants at different temperaturesRank equilibrium constants from ΔG° values given at different temperatures using ln K = −ΔG°/RT.
Entropy and spontaneity
- Rank gaseous halogens by S°Rank the standard molar entropies of gaseous diatomic molecules of the same shape by molar mass.
Entropy changes from data
- Rank reactions by ΔS°Rank reactions by ΔS° from the change in moles of gas.
Calcular 16
Free energy and equilibrium
- ΔG for an ammonia mixtureCalculate ΔG under nonstandard partial pressures with ΔG = ΔG° + RT ln Q, using the coefficients as exponents in Q.
- K for N₂O₄ dissociation from ΔG°fCalculate an equilibrium constant from standard free energies of formation through ΔG° = −RT ln K.
- ΔG° for oxidizing glucoseCalculate ΔG°rxn for glucose oxidation from standard free energies of formation.
- K for methanol synthesis at 500 KEstimate an equilibrium constant at a temperature other than 298 K from ΔH° and ΔS°.
- log K for oxidizing NO to NO₂Calculate log K from ΔG°f data for a reaction with stoichiometric coefficients, converting ln K to log K.
- ΔG° for dissolving silver chlorideCalculate ΔG° for a dissolution equilibrium from its solubility product.
Free energy and temperature
- Estimate the boiling point of bromineEstimate a normal boiling point as the temperature where ΔG° of vaporization is zero, using formation and entropy data.
- Highest temperature for favourable methanol synthesisFind the upper temperature limit for a reaction with negative ΔH° and negative ΔS°.
- ΔG° for oxidizing SO₂Calculate ΔG° from ΔH° and ΔS° at 298 K, converting ΔS° to kJ/K.
Entropy changes from data
- ΔS° for precipitating calcium carbonateCalculate ΔS° for a precipitation using aqueous ions with negative standard entropies.
- ΔS° for burning liquid ethanolCalculate ΔS°rxn for a combustion with a liquid fuel and liquid water from tabulated entropies.
- Entropy of vaporization of ethanolCalculate the entropy of vaporization of a liquid from its enthalpy of vaporization at the normal boiling point.
- Entropy gained by the surroundingsCalculate ΔSsurr from the enthalpy change of a reaction and the temperature.
- ΔS° for burning propaneCalculate ΔS°rxn for propane combustion from standard molar entropies, including the stoichiometric coefficients.
- Entropy of the universe when iron rustsCombine ΔSsys and ΔSsurr to show that a reaction with a negative ΔSsys can increase the entropy of the universe.
Entropy and spontaneity
- Counting arrangements of four moleculesCount the positional microstates that correspond to an even distribution of four gas molecules between two flasks.
Encontre o erro 2
Entropy changes from data
- Find the error in an entropy calculationDiagnose the error of setting S° of elements to zero in a ΔS°rxn calculation.
Free energy and temperature
- Find the error in a ΔG° calculationDiagnose a unit mismatch between kJ and J/K that reverses the sign of a calculated ΔG°.
Interprete as evidências 3
Entropy and spontaneity
- Reading a plot of entropy against temperatureInterpret the vertical steps in a plot of molar entropy against temperature using ΔS = ΔH/T for phase transitions.
Free energy and temperature
- Read ΔH° and ΔS° from a ΔG° plotExtract the signs and values of ΔH° and ΔS° from the slope and zero crossing of a ΔG° versus T line.
Free energy and equilibrium
- A favourable ΔG° but a positive ΔGInterpret a positive ΔG for a mixture of a reaction with negative ΔG° in terms of Q, K and the direction of change.
Trabalhe de trás para frente 2
Free energy and temperature
- ΔH° from ΔG° at two temperaturesWork backwards from ΔG° at two temperatures to ΔS° and ΔH°.
Entropy changes from data
- Work back to S° of nitric oxideWork backwards from ΔS°rxn and known entropies to the standard molar entropy of one species.
Planeje uma síntese 1
- Choose conditions for the water-gas reactionPlan the temperature needed to make an endothermic, entropy-increasing reaction thermodynamically favourable.
Explique por quê 5
Entropy changes from data
- Why does an endothermic salt dissolve?Explain the spontaneity of an endothermic dissolution by comparing ΔSsys with ΔSsurr.
Entropy and spontaneity
- How can freezing be spontaneous?Explain why a process that lowers the entropy of the system can be spontaneous by including the entropy change of the surroundings.
- Why a gas spreads into an empty flaskExplain the free expansion of an ideal gas in terms of the number of microstates.
- Why S° of an element is not zeroExplain, using the third law, why standard molar entropies of elements are positive while their standard enthalpies of formation are zero.
Free energy and equilibrium
- Making lime when ΔG° is positiveExplain how keeping a product's pressure low makes a reaction with positive ΔG° proceed, using ΔG = ΔG° + RT ln Q.
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