TUGAS KIMIA FISIKA LANJUT Dosen Pengampu : Dr. Kasmadi IS, MS

Disusun oleh :

SUPRIYANTO

NIM : 0402510082

PROGRAM PASCA SARJANA

PROGRAM STUDI PENDIDIKAN IPA (KIMIA)

UNIVERSITAS NEGERI SEMARANG (UNNES)

Tahun Akademik 2010/2011

7.37 At 25 °C and 1 atm pressure, the data are :

Substance H2 (g) C (graphite) C6H6 (l) C2H2(g)

∆Hocombustion (kJ /mol) –285.83 –393.51 –3267.62 –1299.58

a) Calculate the ∆H° of formation of liquid benzene.

b) Calculate ∆H° for the reaction 3 C2H2(g) C6H6(l).

ANSWER :

a. {H2(g) + ½ O2 (g) H2O (g) ∆H°= –285.83 kJ/mol } x 3

{C (s) + O2 (g) CO2 (g) ∆H°= – 393.51 kJ/mol } x 6

6CO2 (g) + 3H2O (g) C6H6(l) + 15

/2O2 (g) ∆H°= +3267.62 kJ/mol

3H2(g) + 6C(s) C6H6 (l)

∆Hf° C6H6 = (3)(–285.83) +(6)(–393.51) + 3267.62

= –857.49 – 2361.06 + 3267.62

= 49.07 kJ/mol

b. {C2H2(g) + 5/2O2 2CO2(g) + H2O ∆H°= –1299.58 kJ/mol } x 3

6CO2 (g) + 3H2O C6H6 (l) + 15

/2O2 ∆H°= +3267.62 kJ/mol

3C2H2 (g) C6H6(l)

∆Hreaksi = (3)(–1299.58) + 3267.62

= –3898.74 + 3267.62

= –631.12 kJ/mol

7.38 For the following reactions at 25 °C

∆H° (kJ/mol)

CaC2(s) + 2H2O (l) Ca(OH)2(s) + C2H2(g) –127.9

Ca(s) + ½ O2(g) CaO –635.1

CaO(s) + H2O(l) Ca(OH)2(s) –65.2

The heat of combustion of graphite is –393.51 kJ/mol, and that of C2H2(g) is –1299.58 kJ/mol.

Calculate the heat of formation of CaC2(s) at 25°C.

ANSWER:

∆H° (kJ/mol)

Ca(OH)2(s) + C2H2(g) CaC2(s) + 2H2O (l) +127.9

Ca(s) + ½ O2(g) CaO –635.1

CaO(s) + H2O(l) Ca(OH)2(s) –65.2

{C(s) + O2 CO2(g) –393.51} x 2

2CO2(g) + H2O(l) C2H2(g) + 5/2O2(g) +1299.58

Ca(s) + C(s) CaC2(s)

∆Hf° CaC2(s) = (127.9) +(–635.1) + (–65.2) + (2)(–3

93.51) + (1299.58)

= –59.84 kJ/mol

7.39 A sample of sucrose, C12H22O11, weighing 0.1265 g is burned in a bomb calorimeter. After the

reaction is over, it is found that to produce an equal temperature increment electrically, 2082.3

joules must be expended.

a) Calculate the heat of combustion of sucrose.

b) From the heat of combustion and appropriate data in Table A-V calculate the heat of

formation of sucrose.

c) If the temperature increment in the experiment is 1.743°C, what is the heat capacity of the

calorimeter and contents ?

ANSWER:

a. Mol sukrosa = 0.1265 𝑔𝑟

342 = 3.699 x 10

-4 mol

ΔHcombustion C12H22O11 = – 2082.3

3.669 𝑥 10−4 = – 5676.4 kJ/mol

b. Reaksi pembakaran: C12H22O11(s) + 12O2(g) 12CO2(g) + 11H2O(g)

ΔHcombustion C12H22O11 = (12ΔHf CO2 + 11ΔHf H2O) – (ΔHf C12H22O11 + 12ΔHf O2)

–5676.4 = {(12)(–393.51) + (11)(–285.83)} – (ΔHf C12H22O11 + 0)

–5676.4 = (–4722.12 – 3144.13) – ΔHf C12H22O11

–5676.4 = –7866.25 – ΔHf C12H22O11

ΔHf C12H22O11 = 5676.4 –7866.25

= – 2188.85 kJ/mol

c. ∆𝐻 = 𝐶𝑝𝑑𝑇𝑇2

𝑇1

ΔH = Cp ΔT

2082.3 = Cp (1.743)

Cp = 2082.3

1.743

= 1194.66 joule/K

7.40 If 3.0539 g of liquid ethyl alcohol, C2H5OH, is burned completely at 25°C in a bomb calorimeter,

the heat evolved is 90.447 kJ.

a) Calculate the molar ∆H° of combustion for ethyl alcohol at 25°C.

b) If the ∆Hf° of CO2(g) and H2O(l) are –393.51 kJ/mol and –285.83 kJ/mol, calculate the ∆Hf°

of ethyl alcohol.

ANSWER:

Q = – 90.447 kJoule Mol C2H5OH = 3.0539 𝑔𝑟

46= 0.0664 𝑚𝑜𝑙

a. Harga ΔHcombustion C2H5OH = −90.447

0.0664 = –1362.2 kJ/mol

b. C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)

ΔHcombustion = {2ΔHf CO2 + 3ΔHf H2O} – ΔHf C2H5OH

–1362.2 = 2(–393.51) + 3(–285.83) – ΔHf C2H5OH

= (–787.02 – 857.49) – ΔHf C2H5OH

ΔHf C2H5OH = (–787.02 – 857.49) + 1362.2

= – 1644.51 + 1362.2

= – 282.31 kJ/mol

7.41 From the data at 25 °C :

Fe2O3(s) + 3C(graphite) 2Fe(s) + 3CO(g), ∆H° = 492.6 kJ/mol ;

FeO(s) + C(graphite) Fe(s) + CO(g), ∆H° = 155.8 kJ/mol ;

C(graphite) + O2(g) CO2(g), ∆H° = –393.51 kJ/mol ;

CO(g) + ½ O2(g) CO2(g), ∆H° = –282.98 kJ/mol.

Compute the standard heat of formation of FeO(s) and of Fe2O3(s).

ANSWER:

Fe(s) + CO(g) FeO(s) + C(graphite) ∆H° = –155.8 kJ/mol

C(graphite) + O2(g) CO2(g) ∆H° = –393.51 kJ/mol

CO2(g) CO(g) + ½ O2(g) ∆H° = +282.98 kJ/mol

Fe(s) + ½ O2(g) FeO(s)

∆Hf° FeO(s) = (–155.8) + (–393.51) + (282.98)

= –266.33 kJ/mol

2Fe(s) + 3CO(g) Fe2O3(s) + 3 C(graphite) ∆H° = – 492.6 kJ/mol

{C(graphite) + O2(g) CO2(g) ∆H° = –393.51 kJ/mol } x 3

{CO2(g) CO(g) + ½ O2(g) ∆H° = +282.98 kJ/mol } x 3

2Fe(s) + 3/2 O2(g) Fe2O3(s)

∆Hf° Fe2O3(s) = (– 492.6) + (3)(–393.51) + (3)(282.98)

= –824.19 kJ/mol

9.12 Between O°C and 100°C liquid mercury has Cp/(J/K mol) = 30.093 – 4.944 x 10-3

T. If one mole of

mercury is raised from 0°C to 100 °C at constant pressure, calculate ∆H and ∆S.

ANSWER:

Karena reaksi dilaksanakan pada tekanan tetap, dan Cp sebagai fungsi temperatur sehingga rumus

yang dipakai adalah:

∆𝐻 = 𝐶𝑝𝑑𝑇𝑇2

𝑇1

∆𝐻 = (30.093 − 0.004944𝑇)𝑑𝑇𝑇2

𝑇1

= 30.093(T2–T1) – ½(0.004944)(T22 – T1

2)

Karena T1 = (0 + 273.15)= 273.15°K dan T2 = (100 + 273.15) = 373.15°K

= 30.093(100) – ½ (0.004944)(373.152 – 273.15

2)

= 3009.3 – ½ (0.004944)(6429.5)

= 3009.3 – 159.635

= 2849.665 joule/mol

∆𝑆 = 𝐶𝑝

𝑇

𝑇2

𝑇1

𝑑𝑇

∆𝑆 = (30.093 − 0.004944𝑇)

𝑇

373.15

273.15

𝑑𝑇

= 30.093 𝑑𝑇

𝑇

373.15

273.15 + 0.004944 𝑑𝑇

373.15

273.15

= 30.093 𝑙𝑛373.15

273.15 + 0.004944 (373.15 – 273.15)

= (30.093)(0.31196) + 0.4944

= 9.882 joule/K mol

9.16 One mole of an ideal gas, initially at 25 °C and 1 atm is transformed to 40°C and 0.5 atm. In the

transformation 300 J of work are produced in the surroundings. If Cv = 3/2R, calculate Q, ∆U, ∆H,

and ∆S.

ANSWER:

∆U = Cv ∆T

= 3/2R (313 – 298)

= 22.5R

= (22.5)(8.314)

= 187.065 joule/mol

Menurut Hukum I termodinamika ∆U = Q – W, sehingga

∆U = Q – W

187.065 = Q – 300

Q = 187.065 + 300

= 487.065 joule/mol

∆H = Cp ∆T padahal Cp – Cv = R Cp = R + Cv

= (R + Cv) ∆T

= (R + 3/2R) ∆T

= (5/2R)(15)

= 37.5R

= (37.5)(8.314)

= 311.775 joule/mol

∆S = Cp ln T2

T1 −𝑛𝑅 ln

𝑃2

𝑃1

= (R+Cv) ln T2

T1 −𝑛𝑅 ln

𝑃2

𝑃1

= (R +3/2R) ln

313.15

298.15 − 1 (𝑅) ln

0.5

1

= (5/2R) ln(1.050) – (R) ln(0.5)

= (5/2R)(0.04879) + 0.6932R

= 0.815175R

= (0.815175)(8.314)

= 6.777 joule/K mol

9.18 Consider one mole of an ideal gas, Cv = 3/2R, in the initial state : 300 K, 1 atm. For each

transformation, (a) through (g), calculate Q, W, ∆U, ∆H, and ∆S ; compare ∆S to Q/T.

a) At constant volume, the gas is heated to 400 K.

b) At constant pressure, 1 atm, the gas is heated to 400 K.

c) The gas is expanded isothermally and reversibly until the pressure drops to ½ atm.

d) The gas is expanded isothermally against a constant external pressure equal to ½ atm until the

gas pressure reaches ½ atm.

e) The gas is expanded isothermally against zero opposing pressure (Joule expansion) until the

pressure of the gas is ½ atm.

f) The gas is expanded adiabatically against a constant pressure of ½ atm until the final pressure is

½ atm.

g) The gas is expanded adiabatically and reversibly until the final pressure is ½ atm.

ANSWER:

a. Pada keadaan Volume konstan dan suhu 400 K, rumus yang digunakan adalah

∆U = Cv ∆T

= ³/2R (400–300)

= 150R

= (150)(8.314)

= 1247.1 joule/mol

∆U = Qv Qv = 1247.1 joule/mol

Menurut Hukum I Termodinamika ∆U = Q – W, padahal pada kondisi volume tetap, harga

∆U = Q sehingga 1247.1 = 1247.1 – W

W = 0 joule/mol

∆H = Cp ∆T padahal Cp – Cv = R Cp = R + Cv

= (R + Cv) ∆T

= (R + 3/2R) (400–300)

= (5/2R)(100)

= 250R

= (250)(8.314)

= 2078.5 joule/mol

∆𝑆 = 𝐶𝑣𝑇

𝑇2

𝑇1

𝑑𝑇

∆𝑆 =

32𝑅

𝑇

𝑇2

𝑇1

𝑑𝑇

= 3/2R ln

400

300

= (1.5R)(0.287)

= 0.4305R

= (0.4305)(8.314)

= 3.579 joule/K mol

b. Pada keadaan Tekanan konstan dan suhu 400°K. sehingga rumus yang digunakan adalah

∆H = Cp ∆T sedangkan Cp – Cv = R Cp = R + Cv

= (R + 3/2R)(400 – 300)

= (5/2R)(100)

= 250R

= (250)(8.314)

= 2078.5 joule/mol

∆H = Qp Qp = 2078.5 joule/mol

Untuk mencari W pada tekanan tetap, kita bantu dengan harga ∆U pada volume tetap sbb:

∆U = 1247.1 joule/mol

∆U = Qp – W

1247.1 joule/mol = 2078.5 joule/mol – W

W = 2078.5 – 1247.1

= 831.4 joule/mol

∆𝑆 = 𝐶𝑝

𝑇

𝑇2

𝑇1

𝑑𝑇

sedangkan Cp – Cv = R Cp = R + Cv Cp = R + 3/2R =

5/2R

∆𝑆 =

52𝑅

𝑇

400

300

𝑑𝑇

= 5/2R ln

400

300

= (5/2R)(0.2877)

= 0.719R

= (0.719)(8.314)

= 5.979 joule/K mol

c. Karena gas diekspansi pada suhu tetap (isotermal) dan reversibel sehingga tekanannya turun

menjadi ½ atm, maka

∆S = 𝐶𝑝 ln T2

T1 −𝑛𝑅 ln

𝑃2

𝑃1

∆S = − nR ln P2

P1

= − (1)(8.314) ln 0.5

1

= –{(8.314)(–0.693)}

= 5.763 joule/K mol

∆S = 𝑄𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑒𝑙

𝑇

5.763 = 𝑄𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑒𝑙

300

Qreversibel = (5.763)(300)

= 1728.9 joule/mol

∆U = Cv ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆U = 0

∆H = Cp ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆H = 0

Menurut Hukum I Termodinamika :

∆U = Q – W, padahal ∆U = 0, sehingga 0 = Q – W atau Q = W W = 1728.9 J/mol

Perbandingan antara = 𝑄

𝑇=

1728 .9

300 = 5.763 joule/K mol

d. The gas is expanded isothermally against a constant external pressure equal to ½ atm until the

gas pressure reaches ½ atm

∆S = 𝐶𝑝 ln T2

T1 −𝑛𝑅 ln

𝑃2

𝑃1

∆S = − nR ln P2

P1

= − (1)(8.314) ln 0.5

1

= –{(8.314)(–0.693)}

= 5.763 joule/K mol

∆U = Cv ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆U = 0

∆H = Cp ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆H = 0

Menurut persamaan ∆U = T∆S – P∆V atau ∆U = T∆S – W

0 = (300)(5.763) – W

0 = 1728.9 – W

W = 1728.9

Menurut Hukum I Termodinamika : ∆U = Q – W , sehingga Q = 1728.9

𝑄

𝑇=

1728.9

300= 5.763

e. The gas is expanded isothermally against zero opposing pressure (Joule expansion) until the

pressure of the gas is ½ atm

∆S = 𝐶𝑝 ln T2

T1 −𝑛𝑅 ln

𝑃2

𝑃1

∆S = − nR ln P2

P1

= − (1)(8.314) ln 0.5

1

= –{(8.314)(–0.693)}

= 5.763 joule/K mol

∆U = Cv ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆U = 0

∆H = Cp ∆T, tetapi karena T2 dan T1 sama, maka ∆T = 0, sehingga ∆H = 0

W = 0 maka Q = 0

𝑄

𝑇= 0

f. The gas is expanded adiabatically against a constant pressure of ½ atm until the final pressure is

½ atm

Pada proses adiabatik, maka harga Q = 0

𝑄

𝑇= 0

g. The gas is expanded adiabatically and reversibly until the final pressure is ½ atm.

Pada proses adiabatik, maka harga Q = 0

∆𝑆 =𝑄

𝑇= 0

𝑄

𝑇= 0

Soal Q W ∆U ∆H ∆S Q/T

a. 1247.1 0 1247.1 2078.5 3.579 -

b. 2078.5 831.4 1247.1 2078.5 5.979 -

c. 1728.9 1728.9 0 0 5.763 5.763

d. 1250 1250 0 0 5.763 5.763

e. 0 0 0 0 5.763 0

f. 0 ? ? ? ? 0

g. 0 ? ? ? 0 0