Thermodynamics

The Meaning of ΔG for a Chemical Reaction

Even when value of ΔG provides information regarding whether the system is favored under a given set of conditions: System may not proceed to pure products (if ΔG is negative) System may not remain at pure reactants (if ΔG is positive) A system will spontaneously seek equilibrium System can achieve the lowest possible free energy by going to equilibrium, not by going to completion

G= G°+RTln(P)

G° - Free energy of a gas at 1 atm G - Free energy of the gas at a pressure of P atm R - Universal gas constant T - Temperature in Kelvin

The Temperature Dependence of K

If the values of K are determined at various levels of T: Plot of ln(K) versus 1/T will be linear Slope = - ΔH°/R Intercept = ΔS°/R Assume that ΔH°and ΔS°are independent of T

K values and direction of reaction

K is large- mostly products K is small- mostly reactants K is between .001 and 1000 we will have a significant concentration of both reactant and product species present at equilibrium.

Calculate K for the first step of the process at 825 K, assuming delta Ho and delta So don't depend on temperature ΔH° = 6(−242) + 4(90.) − [4(−46)] = −908 kJ ΔS° = 4(211) + 6(189) − [4(193) + 5(205)] = 181 J/K ΔG° = −908 kJ - (298 K)(0.181 kJ/K) = −962 kJ

4NH3 (g) + 5O2 (g) = 4NO (g) + 6H2O (g) New Delta Go at 1098K = ΔH° − TΔS° = −908 kJ − (1098 K)(0.181 kJ/K) = −1107 kJ Delta Go = -RTlnk -1107 = -(.008315)(1098)lnK k= e^ -1107/-.008315*1098 = 4.589 × 10^52 VERY BIG! (these are sometimes very big, too big to calcuate, that's okay!)

specific heat capacity

C how much energy is required to raise 1 g of substance 1 degree C or K J/g.K

Standard Free Energy Change (ΔG°)

Change in G that will occur if the reactants in their standard states are converted to the products in their standard states More negative the value of ΔG°, the further the reaction shifts to the right to attain equilibrium

Standard Free Energy of Formation (Delta G. f)

Change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements All reactants and products are in their standard states Used to calculate the free energy change for a reaction of an element in its standard state = 0

molar heat capacity

Cmol how much energy is required to raise 1mol if a substance 1 degree C or K J/mol.K

Gibbs Free Energy

Negative delta G means rx'n is spontaneous ΔG = ΔH - TΔS at constant temp Energy available to do work Negative ΔG means positive ΔSuniv In any spontaneous process there is always an increase in entropy of the universe

Equilibrium Point

Occurs at the lowest value of free energy available to the system When substances undergo a chemical reaction, the reaction proceeds to the minimum free energy (equilibrium) This corresponds to the point where: Delta G = 0

To calculate ΔG for this process, use the following equation: ΔG =ΔG .+RTlnQ First compute ΔGₒ from standard free energies of formation= -2.9x10^4 j/mol Use this value to calculate the value of ΔG ΔG°= - 2.9×104 J/mol rxn R = 8.3145 J/K·mol T = 273 + 25 = 298 K Q= 1/(PCO)(PH2^2) =-3.8x10^4 j/mol *Note that the pure liquid methanol is not included in the calculation of Q *Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm This result can be expected from Le Châtelier's principle

One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol

Entropy changes in chemical reactions

Positional probability determines the changes that occur in a chemical system Fewer the molecules, fewer the possible configurations Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products If the number of product molecules is greater than the number of reactant molecules: Positional entropy increases ΔS is positive mixing involves an increase in disorder (an increase in positional probability

Since in this reaction a gas is produced from a solid reactant, the positional entropy increases, and ΔS°is positive

Predict the sign of ΔS°for each of the following reactions:

Negative ΔS°= -186 J/K

Predict the sign of ΔSₒ and then calculate ΔS°for each of the following reactions:

ΔSsurr

Reaction takes place under conditions of constant temperature (in Kelvins) and pressure If the reaction is exothermic: ΔH has a negative sign ΔSsurr is positive since heat flows into the surroundings ΔSsurr=-ΔH/T

For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature Solid CO2 and gaseous CO2 N2 gas at 1 atm and N2 gas at 1.0×10-2 atm

Since a mole of gaseous CO2 has the greater volume by far, the molecules have many more available positions than in a mole of solid CO2 Thus, gaseous CO2 has the higher positional entropy A mole of N2 gas at 1.0×10-2 atm has a volume 100 times that (at a given temperature) of a mole of N2 gas at 1 atm Thus, N2 gas at 1.0×10-2 atm has the higher positional entropy

S°

Standard entropy values (S°) represent increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm More complex the molecule, the higher the standard entropy value

Predict the sign of the entropy change for each of the following processes: Solid sugar is added to water to form a solution

Sugar molecules become randomly dispersed in the water when the solution forms and thus have access to a larger volume and a larger number of possible positions Positional disorder is increased, and there will be an increase in entropy ΔS is positive, since the final state has a larger entropy than the initial state, and ΔS = Sfinal - Sinitial

Free Energy and Pressure

System under constant P and T proceeds spontaneously in the direction that lowers its free energy Free energy of a reaction system changes as the reaction proceeds Dependent on the pressure of a gas or on the concentration of species in solution Equilibrium - Point where free energy value is at its lowest For ideal gases: Enthalpy is not pressure-dependent Entropy depends on pressure due to its dependence on volume At a given temperature for 1 mole of ideal gas: Slarge volume > Ssmall volume Or, Slow pressure > Shigh pressure

Third law of thermodynamics

The entropy of a perfect crystal at absolute zero is exactly equal to zero. At absolute zero (zero kelvin), the system must be in a state with the minimum possible energy, and the above statement of the third law holds true provided that the perfect crystal has only one minimum energy state. No mircrostates, no movement

Enthalpy (H)

The heat content of a system at constant pressure H

Calculate ΔHₒ, ΔSₒ, and ΔGₒ 2 SO2 (g) + O2 (g) = 2SO3 (g)

The value of ΔH°can be calculated from the enthalpies of formation using the following formula n ΔH°products -n ΔH°reactants The value of ΔS°can be calculated using the following formula nΔS°products -nΔS°reactants We would expect ΔS°to be negative because three molecules of gaseous reactants give two molecules of gaseous products THEN: Delta G . = Delta H. - T Delta S.

At what temperatures is the following process spontaneous at 1 atm? What is the normal boiling point of liquid Br2? Br2 (l) -> Br2 (g) delta H= 31.0 kj/mol delta S = 93.0j/mol

The vaporization process will be spontaneous at all temperatures where ΔG°is negative Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium (ΔG°= 0) ΔG°=ΔH°-TΔS° ΔH°=TΔS° T= delta H dot/delta s dot =3.1 x 10^4/93 = 333K At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative Above 333 K, the vaporization process is spontaneous The opposite process occurs spontaneously below this temperature At 333 K, liquid and gaseous Br2 coexist in equilibrium T = 333 K The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist This is the normal boiling point

Methods for Calculating ΔG°

Use the following formula Treat free energy as a state function and use Hess's law Use standard free energy of formation

temperature

a measure of the average kinetic energy of the particles in a sample of matter A measure of how hot or cold something is.

spontaneous reactions

entropy of the universe (delta s) always increases occurs without addition of external energy reactions that release energy reactions that favor formation of products under the specified conditions

positional entropy

for GASEOUS molecules: change in positional entropy is dominated by relative numbers of molecules of products vs reactants ex: H2 -> 2H delta S is positive

Heat (q)

heat absorbed or released by material mass x specific heat capacity x tfinal-tinitial if +, system increases in temp, system's energy increases if delta is negative, q is negative The energy transferred between objects that are at different temperatures Not necessarily accompanied by T change (ex, phase change)

heat capacity

how much energy is needed to change the temperature of a substance (assuming no phase changes are occurring) the amount of heat needed to increase the temperature of an object exactly 1°C

Entropy (S)

increases as phases change solid to liquid to gas (more micro states) perfect crystal entropy =0 at 0k dissolving pure substance in solvent increases entropy reactions increasing # of mols often increase entropy units= J/K.mol delta s rxn= delta s of products - delta s of reactants increases with #of atoms in compound and higher molecular mass Randomness or disorder of the components of a chemical system. ΔS = + MORE DISORDER (FAVORED CONDITION) ΔS = - MORE ORDER

Temperature dependence of K

ln(K) = -deltaH/R x 1/T + deltaS/R

density

m/v can be used to calculate mass ex: m=dv 250 ml x 1g/ml = 250 g

Dependence of free energy on pressure

ΔG= ΔG• + RT ln(Q) Q - Reaction quotient T - Temperature in Kelvin R - Universal gas constant (8.3145 J/K·mol) ΔG°- Free energy change at 1 atm ΔG - Free energy change at specified pressures

Entropy Changes in the Surroundings (ΔSsurr)

ΔSsurr is determined by flow of energy as heat Exothermic process increases ΔSsurr Important driving force for spontaneity Endothermic process decreases ΔSsurr Impact of transfer of energy as heat to or from the surroundings is greater at lower temperature Sign of ΔSsurr depends on the direction of the heat flow At constant temperature: ΔSsurr for exothermic processes is positive ΔSsurr for endothermic processes is negative

Entropy of surroundings

ΔSsurr=-H/ΔT If the reaction is exothermic: ΔH has a negative sign ΔSsurr is positive since heat flows into the surroundings

ΔSuniv

ΔSuniv= -ΔG/T Processes that occur at constant T and P are spontaneous in the direction in which the free energy decreases Negative ΔG means positive ΔSuniv