w6 Flashcards

Question

Entropy: Dispersal of Energy and Matter

Answer 1

Entropy: using a thermodynamic property to predict whether a process is spontaneous - built on the idea that spontaneous change results in dispersal of energy - The Second Law of Thermodynamics: Any spontaneous process is accompanied by an increase in the entropy of the universe. - also observed in solitions. - formation of a mixture does not always lead to greater disorder

Answer 2

S = k log W where k is the Boltzmann constant, and W represents the number of different ways that the energy can be distributed over the available energy levels.

Answer 3

The Third Law of Thermodynamics: There is no disorder in a perfect crystal at 0 K, or S = 0. All substances have positive entropy values at temperatures above 0 K.

Answer 4

the entropy of 1 mol of a substance in its standard state at the specific temperature Large molecules >> smaller molecules. Substances whose molecules have complex structures have more entropy than those with simpler molecules. increases as thetemperature is raised. There are large increases in entropy when the state changes. At a temperature T above the boiling point, the standard molar entropy includes contributions from: 1.The heat capacity of the solid between 0 K and M.P. 2.The heat of fusion of the solid 3.The heat capacity of the liquid from M.P. to B.P. 4.The heat of vaporization of the liquid 5.The heat capacity of the gas from the boiling point to T K.

Answer 5

∆S° (univ) = ∆S°(sys) + ∆S° (surr) (∆(r)S°) = (sum of the entropies of the products) - (sum of the entropies of the reactants) = ∑n(i) S°(products) - ∑n (i) S°(reactants) S is the entropy per mole of each product and reactant N is the number of moles of each reactant or product in the equation

Answer 6

The entropy change of the surroundings is ∆S° (surr) = q (surroundings) / T = -∆H (sys) / T There are 4 categories of reaction, depending on the signs of ∆S° (sys) and ∆S° (surr)

Answer 7

e used determine whether reaction is spontaneous from just the system. At a specified temperature T, the free energy is defined mathematically as: G = H – TS The free energy of a substance is usually not important. ∆rG = ∆rH ̶T∆(r) S

Answer 8

The second law states that a reaction is spontaneous if ΔSuniv > 0. Reaction spontaneity can be worked out more conveniently from: ∆S (univ) = -∆(r) G / T If ∆(r)G < 0, the reaction is spontaneous as written. If ∆(r)G = 0, the reaction is at equilibrium. If ∆(r)G > 0, the reaction is not spontaneous (but the reverse reaction is).

Answer 9

∆rG° <0, the spontaneous direction of reaction is the direction of the chemical equation. ∆rG° is directly related to the value of the equilibrium constant, thus whether reactions are reactant-favoured or product-favoured. The free energy change is the maximum energy available for work: ∆rG = w(max)

Answer 10

1. Determine ∆Ho (rxn) and ∆So (rxn) and use Gibbs equation: ∆G = ∆H - T∆S ΔHrxn = ΔHprod – ΔHreact ΔSrxn = ΔSprod – ΔSreact Values for H and S can be found in SI data book. 2. Use tabulated values of free energies of formation, ∆Gfo ∆Gfo is the standard free energy per mole for the formation of a compound from its elements in their most stable form under standard conditions (298 K). ∆G0 (rxn) = Σ∆G0 f(products) - Σ∆G0 f(reactants)

Answer 11

reaction is always spontaneous, reverse is always non-spontaneous ∆S positive favours reaction ∆H negative favours reaction

Answer 12

reaction is spontaneous at low T, becomes non-spontaneous as T increases ∆H negative favours reaction Have to do the calculation to determine spontaneity.

Answer 13

non-spontaneous at low T, spontaneous at high T ∆S positive favours reaction Have to do the calculation to determine spontaneity.

Answer 14

reaction is non-spontaneous at all temperatures, reverse is spontaneous

Answer 15

∆rG = ∆rG° + RT ln Q where R is the gas constant, 8.314 J K-1 mol-1, and T is the temperature (in K).

Answer 16

When equilibrium is reached Q = K and ∆rG = 0. If we substitute into: ∆rG = ∆rG° + RT lnQ 0 = ∆rG° + RT ln K ∆rG° = ̶RT ln K This equation allows calculation of: * an equilibrium constant from thermochemical data in tables. * the standard free energy change of a reaction an experimentally equilibrium constants.

Answer 17

Recall the cell emf (E° cell) is related to K by: ln K = + (nFE°cell) / RT We can rearrange ∆rG° = ̶RT ln K to get: ln K = -∆rG° / RT Equate these two requations equations (at constant T): ∆rG° = ̶nFE° cell Which relates the standard free energy of reaction to emf.

Answer 18

(Q = 1): K > 1, E°cell > 0, and ∆rG° < 0

Answer 19

The dependence of the equilibrium constant on temperature is given by the following relationship: ln K = (-∆rH° / RT) + (∆rS°/R) R is independent of temperature changes and enthalpy and entropy are approximately constant over small temperature ranges. The equation is linear

Answer 20

used to estimate K at one temperature from an experimentally measured value at another temperature: ln K {1} - lnK {2} = -∆rH°/R (1/T{1} - 1/T {2})