Thermodynamics I. Introduction.

1. In chemistry, thermodynamics is involved with the spontaneity and equilibrium position of chemical reactions.

2a. "System": the little bit of the universe we are interested in.

2b. a "closed system" is a system that allows only the passage of energy across its boundary;

an "open system" allows the passage of matter and energy across its boundary;

an "isolated system" allows neither matter nor energy to cross its boundary.

2c. The "closed system" is the only one dealt with in this course. Measuring the energy that crosses the boundary of a closed system allows the calculation of entropy changes which would not be possible if matter was also crossing the boundary. Nothing can be measured in an isolated system because nothing goes in or out.

3a. A state function is a measurable property of a system which is independent of how the state of the system was reached, that is it's value does not depend upon what has happened to the system before it is measured..

3b. E, P, T, mass, n, S, V, d , [A] are possible state functions; t, reaction rate, k are not. K in the form of ln K is a state function.

3c. Extensive: P, T, d, [A]

Intensive: E, mass, n, S, V.

4. D means difference between final function value and initial function value.

DV means Vproductts - Vreactants, DE means Eproducts- Ereactants and DS means Sproducts - Sreactants.

5 Entropy, symbol S, is a measure of the "likeliness" of the state of a system. It is also associated with "randomness" and "chaos" in a system.

6a. The Second Law: The entropy of the universe increases in any spontaneous process.

In chemistry this law allows us to calculate from tabulated data whether a reaction will occur as written or not.

6b. To this point in time, no system changes have been found that contravene the second law. Living systems are usually open systems, so care must be taken when dealing with them. If we choose the whole solar system as a system, and assume that it is closed (i.e. that it is not losing or gaining matter, which is not true), then the drop in entropy as life orders matter is matched by a greater, positive change in entropy in the sun to provide life with the energy it needs to arrange things. The net entropy change in the solar system is therefore positive, as required by the second law.

6c. No exceptions to the second law are known.

7a For one way of organizing a system, S = k*log 1 = 0.

For 2 ways, S = k*0.693; anf for 1000 ways, S = k*6.91.

7b. Because chemical systems are comprised of many, many particles, the number of ways in which such a system can be organized is incalculable.

8. DS(universe) = DS(system) + DS(surrounds). Entropy changes are intensive and so addable.

9. DSsurrounds for a closed system is only affected by the energy passing across the system's boundary.

10a. Because heat always flows from the hotter to the colder body, this must mean that the change in entropy for the cold body is greater than the change in entropy for the hot body. Since the amount of energy flowing between the two is the same for both the hot and cold body, this means the a given energy transfer must cause a greater entropy change the colder the initial temperature. Thus a measure of entropy change in terms of energy passing across a boundary must also include a factor which decreases the entropy change as the initial energy in the system increases. The initial energy in the system depends on its temperature, so the entropy change due to a given amount of energy change will depend on that energy change, and inversely on the temperature at which the energy transfer takes place.

10b. Although energy flows out of the refrigerator, thereby reducing the entropy, that energy, plus energy needed to do the work of moving gases , is transferred to the surrounds. When summed, the entropy change in the surrounds is greater than the entropy change within the refrigerator, and the second law is upheld.

11b. The final temperature will be the average for the two blocks: 20ļC.

For the cold block DS = DE/T = 500 kJ/ 273 K = 1.83 kJ/K.

For the hot block, DS = DE/T = -500 kJ/313 K = - 1.60 kJ/K

Thus DSuniverse = 1.83 kJ/K - 1.60 kJ/K = 0.23 kJ.K (a positive value for this spontaneous process).

Thermodynamics II. Thermochemistry.

1a. Energy is conserved.

1b. DE is the energy change, q is the heat put into the system, w is the work done on the system.

1c. DE = -1000 J + 500 J = -500 J.

1d. DE = -20.0 kJ - 15.0 kJ = - 35.0 kJ.

2a. Work is force * distance moved.

2b. In chemistry we can neglect all work terms except expansion/contraction work.

3a. Work is equal to {force x distance moved}.

Pressure is force per unit area, so force is P x area.

Work is then P x area x distance moved = P x volume change = PDV.

3b. DE = q - PDV.

DE: E(final) - E(initial), the change in internal energy of a system

q: the heat change in the process

- PDV: the work change in the process. The negative sign is due to the work being measured by a change in volume of the system, a negative volume change indicates that work is being put into the system to compress it.

4a,b. Terms are defined in your notes.

5. Measurements at constant pressure can be done in an open calorimeter, whereas those at constant volume must be carried out in a suitably designed (and so expensive) pressurized reaction vessel.

6a. Standard pressure is 100 kPa. (It used to be 1 atm = 101.3 kPa.)

Standard concentration is 1 mol/L.

6b. DEļ is the internal energy change at a constant 100 kPa pressure.

DHļ is the enthalpy change at a constant 100 kPa pressure.

6c/a. The molar mass of thiourea is 76 g/mol, so this many grams of thiourea would evolve:

15.37 kJ/g * 76 g/mol = 1168 kJ / mol.

DE = 1168 kJ/mol.

6c/b. CS(NH2)2(s) + 3 O2(g) --> CO2(g) + SO2(g) + N2(g) + 2 H2O(l)

Dn (for gases only) = 3 - 3 = 0, so PDV = RTDn = 0

So DH = DE + PDV = DE = 1168 kJ/mol.

7. No known zero of energy means that we can only tabulate energy differences. Sometimes it is possible to define an arbitrary energy as the zero. In thermodynamics the energies of the elements in their standard states (at 298 K and 100 kPa) are designated O.

8a. Hess's law is a realization that energy is a state function; that is it does not matter how a change takes place, the sum of all energies for the steps in the change will equal the energy change no matter what or how many steps there are.

8b. By suggesting that all reactions pass through the elements in their standard states, we can tabulate the energy or enthalpy changes when compounds are made from the elements in their standard states.

8c. Standard state refers to the lowest energy form of an element at 100 kPa pressure and 298 K.

8d. Enthalpy of formation is defines as the enthalpy change when 1 mole of a compound is formed at 298 K and 100 kPa from its elements in their standard states.

9a. DHļ = S miDf(i).

>DHļ: the enthalpy change for the reaction shown by a balanced chemical equation.

S: sum the following terms.

mi : the coefficient of component i in the balanced chemical equation with the sign + for products and - for reactants.

Df(i): the enthalpy of formation of component i in the balanced chemical equation.

9b. SO3(g) + H2O(l) é H2SO4(l).

DHļ = S miDf(i) = (-1)Df(SO3) + (-1)Df(H2O) + (+1)Df(H2SO4).

From tables:

D = {(-1)(-396) + (-1)(-286) + (+1)(-814)} kJ = -132 kJ.

9c. CS(NH2)2(s) + 3 O2(g) --> CO2(g) + SO2(g) + N2(g) + 2 H2O(l)

DHļ = S miDf(i) = (-1)Df(CS(NH2)2) + (-3)Df(O2) + (+1)Df(CO2) + (+1)Df(SO2) + (+1)Df(N2) + (+1)Df(H2O) .

From tables:

D = 1168 kJ

= (-1)Df(CS(NH2)2) + {(-3)(0) + (+1)(-393.5) + (+1)(-296.1) + (+1)(0) + (+2)(-285.8)} kJ

Df(CS(NH2)2) = -249.3 kJ/mol.

9d. DHļ = S miDf(i) = (-1)Df(Fe2O3) + (-3)Df(CO) + (+2)Df(Fe) + (+3)Df(CO2).

From tables:

D298 = {(-1)(-822.2) + (-3)(-110.5) + (+2)(0) + (+3)(-393.5)} kJ = -26.8 kJ.

9e. The reaction for the enthalpy of formation of N2O5(g) is:

N2(g) + 5/2 O2(g) --> N2O5(g)

This reaction is the following reactions summed:

N2(g) + O2(g) --> NO(g)

2 NO(g) + O2(g) --> 2 NO2(g)

2 NO2(g) + Ĺ O2(g) --> N2O5(g)

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N2(g) + 5/2 O2(g) --> N2O5(g)

So add up the enthalpy changes for these reactions to get the enthalpy of formation:

180.5 kJ + -114.1 kJ + -55.1 kJ =+11.3 kJ.

10. Heat of combustion is for the burning of 1 mole of the compound in oxygen to give CO2(g) and H2O(l).

For propanone:

CH3COCH3 + 4 O2 ----> 3 CO2 + 3 H2O(l) DH = -1000kJ

For propan-2-ol:

CH3CH(OH)CH3 + 4.5 O2 ----> 3 CO2 + 4 H2O(l) DH = -1400 kJ

The reaction for which DH is required is:

CH3CH(OH)CH3 + 0.5 O2 ----> CH3COCH3 + H2O(l) DH = ?

The required reaction is obtained by adding the reverse of reaction I to reaction II.

Do the same with the DH values to obtain the required DH:

DHļ = +1000 kJ + (-1400) kJ = -400 kJ.

End of Assignment for quiz 5

2/10a. "Bond energy" is the amount of energy needed to break a bond type. It is an average value rather than a value for a specific bond.

2/10b. Because bond energies are averages, the value calculated for DHļ is only approximate. Additionally bond energies apply strictly to the gas phase and can only be used when all reactants and products are in the gas phase.

2/10c. CH3-CH3(g) + Cl2(g) é CH3-CH2Cl(g) + HCl(g)

DHļ = S BE(bonds broken) - S BE(bonds made)

= (C-H) - (Cl-Cl) - (C-Cl) + (H-Cl)

= {415 + 240 - 330 - 430} kJ

= -105 kJ.

2/11. Combustion is the complete oxidation of an organic compound to carbon dioxide and water.

CH3COCH3(l) + 4 O2(g) é 3 CO2(g) + 3 H2O(l) DHļ = -1000 kJ

CH3CH(OH)CH3(l) + 5 O2(g) é 3 CO2(g) + 4 H2O(l) DHļ = -1400 kJ

If equation 1 is reversed and added to equation 2, the result is:

CH3CH(OH)CH3(l) + O2(g) é CH3COCH3(l) + H2O(l) DHļ = +(-1400) - (-1000) = -400 kJ

3/4a. For a change in phase, DS = DH/T.

In this case: DH = 1 mol x (6.01 / 273) kJ/mol.K = 22.0 J/K.

3/4b. For a change in phase, DS = DH/T.

In this case: DH = 62.4 kJ/mol; DS = 145 J/mol.K; T = {62 400 / 145} K = 430 K.

3/5. For heating without a phase change and with a constant value for Cp: DS = Cp ln {T2/T1}.

In this case, DS = 54.4 J/mol.K * ln {293/253} = 8.00 J/mol.K.

3/8. The process is: Hg(l) ĺ Hg(g), at equilibrium at 100 kPa.

DT = DHļ - TDSļ = 0

So T = DHļ / DSļ = {(61 320 J/mol) / (98.94 J/K.mol) = 620 K

3/9a. 2 MnO2(s) é 2 MnO(s) + O2(g)

DHļ = (-2)(-520) + (+2)(-385) + (+1)(0) = 270 kJ

DSļ = (-2)(53.1) + (+2)(59.7) + (+1)(205) = 217 J/K

D(univ) = DSļ - DHļ/T = {217 - 270 000/298} = -689 J/K

The process is predicted to be non-spontaneous.

3/9b. SiH4(g) + 4 Br2(l) é SiBr4(l) + 4 HBr(g)

DHļ = (-1)(34.3) + (-4)(0) + (+1)(-457) + (+4)(-36.4) = -636.9 kJ

DSļ = (-1)(205) + (-4)(152) + (+1)(278) + (+4)(199) = 261 J/K

D(univ) = DSļ - DHļ/T = {261 - (-637 000)/298} = 2400 J/K

The process is predicted to be spontaneous.

3/9c. 2 CuCl(s) é CuCl2(s) + Cu(s)

DHļ = (-2)(-137) + (+1)(-220) + (+1)(0) = 54 kJ

DSļ = (-2)(86.2) + (+1)(108) + (+1)(33.2) = -31.2 J/K

D(univ) = DSļ - DHļ/T = {(-31.2) - (54 000)/298} = -212 J/K

The process is predicted to be non-spontaneous.

4. The combustion of 1.000g of thiourea, CS(NH2)2 (s), in a bomb calorimeter (at constant volume) evolves 15.37 kJ of heat at 25ļC. The products of the combustion are CO2 (g),SO2(g), N2(g) and H2O(l).

 a) What is DE for the combustion of 1 mol of thiourea?

DE = qp = -15.37 kJ/g * 76 g/mol = - 1168 kJ/mol

 b) Write a balanced chemical equation and calculate DH the heat of combustion.

CS(NH2)2(s) + 3 O2(g) é CO2(g) + SO2(g) + N2(g) + 2 H2O(l)

DH = DE + PDV = DE + RTDn.

But, in this case, Dn for gases is 0, so DH = DE = - 1168 kJ.

 c) Use values from the appendix of your text to calculate the enthalpy of formation of thiourea.

CS(NH2)2(s) + 3 O2(g) é CO2(g) + SO2(g) + N2(g) + 2 H2O(l)

DH = (-1)Df(CS(NH2)2) + (-3)(0) + (1)(-394) + (1)(-297) + (1)(0) + (2)(-286)

-1168 kJ = - Df(CS(NH2)2) - 1263

Df(CS(NH2)2) = - 95 kJ/mol

5. The reduction of iron(III) oxide by carbon monoxide in the reaction :

Fe2O3(s) + 3 CO(g) é 2 Fe(s) + 3 CO2(g)

is one of the steps in the production of iron metal. What is D298 for this reaction ?

D298 = (-1)(-824) + (-3)(-111) + (+2)(0) + (+3)(-394)

D298 = - 25 kJ

6. Determine the standard enthalpy of formation of gaseous N2O5 on the basis of the following data ONLY:

N2(g) + O2(g) é 2 NO(g), DHļ = 180.5kJ

2 NO(g) + O2(g) é 2 NO2(g) DHļ = -114.1kJ

4 NO2(g) + O2(g) é 2 N2O5(g) DHļ = -110.2kJ

If reaction 3 is halved and then all three equations added, the result is:

N2(g) + 2Ĺ O2(g) é N2O5(g) DHļ = 11.3 kJ

7. A large shipment of salicylic acid (C7H6O3), used in the manufacture of aspirin, is found to be contaminated with boric oxide which, like salicylic acid, is a white powder. The heat of combustion of salicylic acid is known to be -3.00 x 103 kJ/mol. Boric oxide is fully oxidized and so will not burn. When a 3.556 g sample of the contaminated salicylic acid is burned in a bomb calorimeter, the temperature increases 2.556 ļ. From previous measurements, the heat capacity of the calorimeter is known to be 13.62 kJ/K. What is the amount of boric oxide in the sample in terms of percent by weight?

qv = -13.62 kJ/K * 2.556 K = -34.81 kJ for this sample.

If the sample were pure salicylic acid, qv for 3.556 g would be:

3.556 g * (1 mol/138 g) * -3 000 kJ/mol = -77.30 kJ.

Thus the sample is (34.81 * 100%)/77.30 = 45.0% salicylic acid, and therefore 55.0% boric oxide.

8. Use the reaction :

SO2(g) + Cl2(g) é SO2Cl2(l)

to find the absolute entropy, S298, of SO2Cl2(l).

Df for SO2Cl2(l) is -389 kJ/mol; Df for SO2Cl2(l) is -314 kJ/mol.

Data for SO2 (g) and Cl2(g) are in your text.

D298 = DHļ - 298DSļ, so DSļ = (DHļ - D298) / T

D298 = (-1)(-300) + (-1)(0) + (+1)(-314) = -14 kJ

DHļ = (-1)(-297) + (-1)(0) + (+1)(-389) = -92 kJ

DSļ = (DHļ - D298) / T = (-92 000 - (-14 000)) / 298 = -262 J/K

DSļ = (-1)(248) + (-1)(223) + (+1)(Sļ(SO2Cl2)) = -262 J/K

Sļ(SO2Cl2) = 209 J/K

9. For the reaction

PCl5(g) é PCl3(g) + Cl2(g) at 25ļC

DHļ = + 92.5 kJ and DSļ = + 182 J/K

 a) Is the reaction spontaneous under standard conditions at 25ļC?

D298 = DHļ - TD

D298 = 92.5 - 298 * 182/1000 = 38.3 kJ

The reaction is not spontaneous at 298 K

b) Assuming that the values of DHļ and DSļ do not change with a change in temperature

 i) Is the reaction spontaneous at 300ļC?

D573 = DHļ - TD

D573 = 92.5 - 573 * 182/1000 =- 11.8 kJ

The reaction is spontaneous at 573 K

 ii) At what temperature (if one exists) does equilibrium occur with all three gases at standard pressure?

If all three gases are at standard pressure (100 kPa) then the equilibrium is under standard conditions and DT = 0.

DT = DHļ - TDSļ = 0

DT = 92.5 - T * 182/1000 = 0 kJ

T = 508 K