CMY 282 Sem Test 2019 Memo PDF

Preview

Summary

UNIVERSITY OF PRETORIA: Department of Chemistry CMY 282- Physical ChemistrySemestertest 201 9-04- 30 Time: 120 min Marks: 55 Examiner: Prof. E. RodunerInstructions:  This paper contains 8 pages, please check that you have them all.  Answer all questions in the spaces provided; use space overleaf i...

Description

UNIVERSITY OF PRETORIA: Department of Chemistry CMY 282- Physical Chemistry Semestertest Time: 120 min Examiner: Prof. E. Roduner

2019-04-30 Marks: 55

Instructions:  This paper contains 8 pages, please check that you have them all.  Answer all questions in the spaces provided; use space overleaf if required, and indicate clearly that you have done so.  Please write neatly.  Answer all questions.  Show all calculation steps.  When deriving an equation, all steps should be shown in a logical manner, with explanations where necessary. Don’t just write the equations.  A data and formula sheet is provided separately.  The use of non-programmable scientific calculators is allowed.

Surname and initials: Registration number:

Total / 55

Question number

1

2

3

4

5

6

7

Max mark

9

7

8

4

4

14

9

Mark obtained

1

Question 1

[9]

a) An ice cube of 5 g at an initial temperature of 8°C melts and warms up further until equilibration with ambient temperature at 20°C. Calculate the amount of heat that this system takes up during this process, given the heat capacities of ice and water at constant pressure are Cp = 2.108 kJ kg1 K1 and 4.187 kJ kg1 K1, respectively, and the heat of melting of ice is 333.6 kJ kg1. b) Explain why the heat capacities of ice and water at constant volume (not given) are not much different from the values at constant pressure. c) Calculate the molar heat capacities of ice and water and compare them with the value of gas phase water, Cp,m = 33.58 J mol1 K1. Justify the gas phase value and rationalise why Cp of liquid water is so much larger than that of ice.

Answers: a) Cp assumed constant over the limited temperature range q = Cp(ice) T + meltH + Cp(water) T = (10.54  8) + 1668 + (20.94  20) = 2171 J = 2.171 kJ (values for 5 g)

(2)

b) qp = Cp dT = CV dT + pdV  qV since volume expansion is small (pdV  0)

(2)

c) Cp,m(ice) = 2.108 J g1 K1  18 g mol1 = 37.99 J mol1 K1 Cp,m(liquid water) = 4.187 J g1 K1  18 g mol1 = 75.45 J mol1 K1 Cp,m(gas) = 33.58 J mol1 K1 = 4.04 R  4 R H2O is a non-linear molecule that has 3 translational and 3 rotational degrees of freedom. If vibrations do not contribute, this results in CV,m(gas)  6/2 R = 3 R, and if water is a good ideal gas then Cp,m(gas) = CV,m(gas) + R = 4 R, which agrees well with the tabulated value. Cp,m(liquid water) = 9.07 R >> 4 R (gas), which must be due to the soft vibrations in the hydrogen bonds. (5)

2

Question 2

[7] The Linde refrigerator contains a cooling agent in a closed cycle. Discuss the following aspects of its functionality: a) Should the system be classified as open, closed or isolated? Please justify. b) Explain the purpose of the heat exchanger. c) The pressurised gas expands through a throttle and liquefies. What is the effect called that operates here, and why does the gas cool on expansion? d) Do you expect that you can use helium as a cooling agent?

Answers: a) The system exchanges energy but no matter with the surrounding. It is therefore a closed system. (1) b) The gas heats up when compressed in the compressor. The heat exchanger dissipates this heat to the surroundings (the kitchen). The gas is then still compressed but no longer at so high temperature. (2) c) Liquefaction of a gas on expansion through a throttle is the Joule-Thomson effect. On expansion, the distance between the molecules increases. If this is done at the proper density (intermolecular distance), the molecules spend energy to escape the attractive short-range interactions. In this approximately adiabatic process, the necessary energy is taken from the kinetic energy of the molecules themselves, so that the molecules cool down (transformation of kinetic into potential energy). (3) d) Helium is a good ideal gas that has a zero Joule-Thomson coefficient over much of the temperature range (except at very low temperatures). (1)

3

Question 3 [8] 1.00 g of glucose (C6H12O6) is incinerated in a bomb calorimeter under an excess of oxygen at nearambient temperature according to the reaction C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l) The measured heat of combustion is 15.6 kJ. a) Is the heat of combustion an enthalpy H or an internal energy U, and what is the expected difference between these two properties? b) Given the standard heats of formation of CO2(g) (393.51 kJ mol1) and of H2O(l) (285.83 kJ mol1), calculate the standard molar heat of formation of glucose. c) What should be the nutritional value of sugar as printed on food packages? Answers:

a) A bomb calorimeter works at constant volume, therefore the measured heat is qV = U, an internal energy. qp = H is larger by the volume work on expansion, (pV) = (nRT) = nRT, where n is the change in number of gas molecules. For the above reaction, 6 mol of O2(g) react to 6 mol of CO2(g), the rest does not count. Therefore, n = 0, and U = H for this particular reaction. The measured heat of combustion, 15.6 kJ g1, corresponds to 1.00 g = 1/180 mol The molar heat of combustion is 180 g  15.6 kJ g1 = 2810 kJ mol1 = CH° (4) b) The heat of formation of glucose is: 6 fH°(CO2(g)) + 6 fH°(H2O(l))  6 fH°(O2(g))  CH°(C6H12O6(s)) = fH°(C6H12O6(s)) 2361.1 1715.0 +0 + 2808 = 1266 kJ mol1

(3)

c) The nutritional value is the heat of combustion, 15.6 kJ g1

(1)

4

Question 4

[4]

Given are the isobaric expansion coefficient, 𝛼 = 1 𝜕𝑉

1 𝜕𝑉 ( 𝜕𝑇)𝑝 ,

𝑉

and the isothermal compressibility,

𝜅 = − 𝑉 ( 𝜕𝑝) . Evaluate these two expressions for the ideal gas. 𝑇

Answers: 𝛼=

1 𝜕𝑉

𝑉

( 𝜕𝑇) For the ideal gas, 𝑉 = 𝑝

1 𝜕𝑉

𝜕𝑉

1

𝑛𝑅𝑇 𝑝

𝜕𝑉

, and (𝜕𝑇) =

𝑛𝑅

𝑉

1

𝑝

𝑝

𝑉

= 𝑇, thus we have 𝛼 = 𝑉

𝜅 = − 𝑉 ( 𝜕𝑝) , ( 𝜕𝑝) = −𝑛𝑅𝑇 𝑝2 = − 𝑝, and 𝜅 = − 𝑉 (− 𝑝) = 𝑇

𝑇

1

𝑝

1𝑉 𝑉𝑇

=

1 𝑇

(2)

(2)

5

Question 5 [4] Draw schematically a potential energy curve as a function of distance for two atoms which encounter in a head-on collision, and describe potential and kinetic energy as well as the types of interaction along this curve.

Answer: a)

The kinetic energy is the energy of motion of the atoms. At large distances this is the only contribution to energy (potential energy is zero). As the atoms approach each other to a distance of a few atomic diameters, they will first attract each other. The total energy remains constant, but the kinetic energy increases at cost of a negative contribution of potential energy. At even closer distance, the potential energy reaches a minimum since repulsion of the atoms sets in. The relative velocity slows down, and after the collision the two atoms separate again (following the same potential energy curve backwards). (4)

6

Question 6

[14]

One mole of a monatomic perfect gas initially at 298 K and a pressure of 1.00 bar is expanded to a final pressure of 0.50 bar, according to the following processes: (a) Isothermally against a constant external pressure of 0.50 bar. Calculate q, w, ∆U and ∆H. (b) Isothermal reversibly. Calculate q, w, ∆U and ∆H. (c) Reversibly, adiabatic. Calculate the final temperature, the final volume and q. (d) State and justify without calculation whether the magnitude of w of the adiabatic process (c) is smaller or larger than for the isothermal reversible process. Justify why the change in internal energy can be calculated from, U = CV,mT, using the specific heat at constant volume even though the volume is not constant in this process.

Answers: pV = nRT, Vi = nRT/p = 1.00 mol  0.0831 dm3 bar K-1 mol-1  298 K / 1.00 bar = 24.8 dm3 a) isothermal: Tf = 298 K, U = 0 for ideal gas, Vf = 2Vi, w =  p dV = p V = 0.5 bar  24.8 dm3 = 12.4 dm3 bar = 1.24 kJ U = q + w = 0, q = +1.24 kJ H = U + (pV) = U + (nRT) = U + 0 = 0 (3) b) isothermal reversibly: Tf = 298 K, U = 0, p = pext , w =  p dV = nRT (1/V) dV = nRT ln(Vf/Vi) = 1.00 mol  0.0831 dm3 bar K-1 mol-1  298 K  0.69 = 1.71 kJ q = w = 1.71 kJ H = U + (pV) = U + (nRT) = U + 0 = 0 (3) c) reversibly adiabatic: q = 0, pV  = constant. Vf = Vi pi /pf,  = Cp/CV = 5/3 = 1.67  ln(Vf/Vi) = ln(pi /pf) = ln(2) = 0.693 ln(Vf/Vi) = 0.693/1.67 = 0.415 Vf/Vi = exp(0.415) = 1.51 Vf = 1.51  24.76 dm3 = 37.5 dm3, Tf = Ti (Vi/Vf)1/c, c = CV/R = 3/2, Tf = 298 K (1/1.51)2/3 Tf = 0.760  298 K = 226 K

(5)

d) In a pV diagram, the adiabatic curve decays more steeply than the isothermal curve, therefore, the area under the curve, which is identical to w, is smaller. U is a state function, and the adiabatic process can be seen as occurring in two steps, first an isothermic step to the final volume for which U = 0, and a second step that is isochoric and for which

U = CVT (= 1.50 R  (226-298) K  1 mol = 0.09 kJ. No numerical value required)

(3)

7

Question 7 a) Calculate the standard enthalpy of reaction of the Haber-Bosch process:

[9]

3 H2(g) + N2(g) → 2 NH3(g) at 100C, from its value at room temperature. Given: ΔfHo (NH3, g) = 46.11 kJ mol-1 at 298 K Cp(H2,g) = 28.824 J K-1 mol-1 Cp(N2,g) = 29.125 J K-1 mol-1 Cp(NH3,g) = 35.06 J K-1 mol-1 b) Comment on the values of the three heat capacities and provide possible reasons for deviations from expectation if any. Answers: a) At 298 K: ΔRH = 2 ΔfHo (NH3,g)  ΔfHo (N2,g)  3 ΔfHo (H2,g) = 2  (46.11 kJ mol-1) = 92.22 kJ mol-1 (the standard heats of formation of the elements N2,g and H2,g is zero by definition) irCp = 45.49 J K1 irCp 75 K = 3.41 kJ o ΔfH(NH3,g,372K) = ΔfH (NH3,g) + Cp(NH3,g)T= 46.11 kJ mol-1 + 35.06 J K-1 mol-1  75 K = 43.48 kJ mol-1 ΔfH(N2,g,372K) = 0 + 29.125 J K1 mol-1  75 K = +2.15 kJ mol-1 ΔfH(H2,g,372K) = 0 + 28.824 J K1 mol-1  75 K = +2.13 kJ mol-1 At 372 K: ΔRH = 2 ΔfH(NH3,g)  ΔfH(N2,g)  3 ΔfH(H2,g) = 2  (43.52) 2.15  32.13 = 95.54 kJ (6) -1 -1 b) Diatomic ideal gases should have Cp,m = CV,m + R = 5/2 R + R = 3.5 R = 29.10 J K mol , which is very close to the observed values for N2 and H2. (1) -1 -1 NH3 is non-linear and should have Cp,m = CV,m + R = 6/2 R + R = 4.0 R = 33.26 J K mol The inversion vibration is also slightly excited, which increases the observed value of C p,m. (2)

8...