Tugas Kimia Fisika-Bp-KASMADI

TUGAS KIMIA FISIKA LANJUT Dosen Pengampu 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

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7.37 At 25 °C and and 1 atm pressure, pressure, the data are : Substance H2 (g) C (graphite) o ∆H combustion (kJ /mol) – 285.83 285.83 – 393.51 393.51 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 O 2 (g)



6CO2 (g) + 3H2O (g)



3H2(g) + 6C(s)



∆Hf ° C6H6

C2H2(g) – 1299.58 1299.58

H2O (g)

∆H°= – 285.83 285.83 kJ/mol } x 3

CO2 (g)

∆H°= – 393.51 – 393.51 kJ/mol } x 6

C6H6(l) + 15/ 2O2 (g)

∆H°= +3267.62 +3267.62 kJ/mol



{C (s) + O2 (g)

C6H6 (l) – 3267.62 3267.62

C6H6 (l)

= (3)(– (3)(– 285.83) 285.83) +(6)(– +(6)(– 393.51) 393.51) + 3267.62 – 2361.06 + 3267.62 = – 857.49 857.49 – 2361.06 = 49.07 kJ/mol

b.

{C2H2(g) + 5/ 2O2



2CO2(g) + H2O

∆H°= – 1299.58 1299.58 kJ/mol } x 3

6CO2 (g) + 3H2O



C6H6 (l) + 15/ 2O2

∆H°= +3267.62 +3267.62 kJ/mol

3C2H2 (g)



C6H6(l)

∆Hreaksi

= (3)(– (3)(– 1299.58) 1299.58) + 3267.62 = – 3898.74 3898.74 + 3267.62 = – 631.12 631.12 kJ/mol

7.38 For the following reactions at 25 °C ∆H° (kJ/mol) (kJ/mol) – 127.9 CaC2(s) + 2H2O (l)  Ca(OH)2(s) + C2H2(g) 127.9 – 635.1 Ca(s) + ½ O2(g)  CaO 635.1 CaO(s) + H2O(l)  Ca(OH)2(s) – 65.2 65.2 The heat of combustion of graphite is – 393.51 393.51 kJ/mol, and that of C 2H2(g) is – 1299.58 1299.58 kJ/mol. Calculate the heat of formation of CaC2(s) at 25°C. ANSWER:

Ca(OH)2(s) + C2H2(g)



CaC2(s) + 2H2O (l)

∆H° (kJ/mol) +127.9

Ca(s)

+ ½ O2(g)



CaO

– 635.1 635.1

CaO(s)

+ H2O(l)



Ca(OH)2(s)

– 65.2 65.2

{C(s)

+ O2



CO2(g)

– 393.51} 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) 635.1) + (– (– 65.2) 65.2) + (2)(– 393.51) + (1299.58) = – 59.84 59.84 kJ/mol





7.39 A sample of sucrose, C 12H22O11, 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 the heat of combustion and appropriate appropriate data in Table A-V calculate the heat of formation of sucrose. c) If the temperature increment increment in the experiment is 1.743°C, what is the heat capacity of the calorimeter and contents ? ANSWER:

a.

Mol sukrosa =

0.1265 342

 = 3.699 x 10

ΔHcombustion C12H22O11 = –

-4

mol

2082.3

3.669

 10−

4

– 5676.4 – 5676.4 kJ/mol

=

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



12CO2(g) + 11H2O(g)

ΔHcombustion C12H22O11 = (12ΔHf  CO2 + 11ΔHf  H2O) – (ΔH – (ΔHf  C12H22O11 + 12ΔHf  O2) – 5676.4 5676.4

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

– 5676.4 5676.4

= (– 4722.12 4722.12 – 3144.13) – 3144.13) – ΔH – ΔHf  C12H22O11

– 5676.4 5676.4

– ΔHf  C12H22O11 = – 7866.25 7866.25 – ΔH

ΔHf  C12H22O11 = 5676.4 – 7866.25 7866.25 – 2188.85 kJ/mol = – 2188.85 c.

∆ =    2

1

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. combustion for ethyl alcohol at 25°C. a) Calculate the molar ∆H° of combustion b) If the ∆H f ° of CO 2(g) and H2O(l) are – 393.51 393.51 kJ/mol and –285.83 kJ/mol, calculate the ∆H f ° of ethyl alcohol. ANSWER:

– 90.447 kJoule Q = – 90.447 a.



Mol C2H5OH =

Harga ΔHcombustion C2H5OH =

b. C2H5OH(l) + 3O2(g)



−90.447 = 0.0664

3.0539 46

 = 0.0664 

– 1362.2 1362.2 kJ/mol

2CO2(g) + 3H2O(g)

ΔHcombustion

– ΔHf  C2H5OH = {2ΔHf  CO2 + 3ΔHf  H2O} – ΔH

– 1362.2 1362.2

= 2(– 2(– 393.51) 393.51) + 3(– 3(– 285.83) 285.83) – ΔHf  C2H5OH – 857.49) – ΔH – ΔHf  C2H5OH = (– 787.02 787.02 – 857.49)

ΔHf  C2H5OH = (– (– 787.02 787.02 – 857.49) – 857.49) + 1362.2 = – 1644.51 – 1644.51 + 1362.2





7.41 From the data at 25 °C °C : Fe2O3(s) + 3C(graphite)  2Fe(s) + 3CO(g), FeO(s) + C(graphite)  Fe(s) + CO(g), C(graphite) + O2(g)  CO2(g), CO(g) + ½ O2(g)  CO2(g), Compute the standard heat of formation of FeO(s) and of Fe2O3(s). ANSWER: Fe(s) + CO(g)



FeO(s)

C(graphite) + O2(g)



CO2(g)

CO2(g)



CO(g)



FeO(s)

Fe(s)

+ ½ O2(g)

+ C(graphite)

∆H° = 492.6 kJ/mol ; ∆H° = 155.8 kJ/mol ; ∆H° = – 393.51 393.51 kJ/mol ; ∆H° = – 282.98 282.98 kJ/mol.

∆H° = – 155.8 155.8 kJ/mol ∆H° = – 393.51 393.51 kJ/mol

+ ½ O2(g)

∆H° = +282.98 +282.98 kJ/mol

∆Hf ° FeO(s) = (– 155.8) 155.8) + (– (– 393.51) 393.51) + (282.98) = – 266.33 266.33 kJ/mol



Fe2O3(s) + 3 C(graphite)

∆H° = – 492.6 – 492.6 kJ/mol

{C(graphite) + O2(g)



CO2(g)

∆H° = – 393.51 393.51 kJ/mol } x 3

{CO2(g)



CO(g)



Fe2O3(s)

2Fe(s)

+ 3CO(g)

+ 3/ 2 O2(g)

2Fe(s)

+ ½ O2(g)

∆H° = +282.98 kJ/mol } x 3

∆Hf ° Fe2O3(s) = (– (– 492.6) 492.6) + (3)(– (3)(– 393.51) 393.51) + (3)(282.98) = – 824.19 824.19 kJ/mol – 4.944 x 10 -3T. If one mole of 9.12 Between O°C and 100°C liquid mercury has has Cp/(J/K mol) = 30.093 30.093 – 4.944 mercury is raised from 0°C to 100 °C at constant pressure, calculate ∆H and ∆S. ANSWER: Karena reaksi dilaksanakan dilaksanakan pada tekanan tetap, dan Cp sebagai fungsi fungsi temperatur sehingga sehingga rumus yang dipakai adalah:

 ∆ =     2

1

∆ = 

2

(30.093

1

− 0.004944)

= 30.093(T2– T1) – ½(0.004944)(T – ½(0.004944)(T22 – T – T12) Karena T1 = (0 + 273.15)= 273.15°K

dan T 2 = (100 + 273.15) = 373.15°K

= 30.093(100) – ½ – ½ (0.004944)(373.15 (0.004944)(373.152 – 273.15 – 273.15 2) – ½ (0.004944)(6429.5) = 3009.3 – ½ = 3009.3 – 159.635 – 159.635 = 2849.665 joule/mol

  ∆ =    2

1

373.15

∆ = 

273.15

= 30.093

− 0.004944)    + 0.004944   

(30.093



373.15 273.15

373.15 273.15





9.16 One mole of an ideal gas, initially at 25 °C and 1 atm atm is transformed transformed to 40°C and 0.5 atm. In the transformation 300 J of work are produced in the surroundings. If C v = 3/ 2R, calculate Q, ∆U, ∆H, and ∆S. ANSWER:

∆U = Cv ∆T =

3

– 298) / 2R (313 – 298)

= 22.5R = (22.5)(8.314) = 187.065 joule/mol Menurut Hukum I termodinamika ∆U = Q – W, – W, sehingga ∆U

– W = Q – W

– 300 187.065 = Q – 300 Q = 187.065 + 300 = 487.065 joule/mol

∆H = Cp

∆T

padahal C p – C – Cv = R

= (R + Cv) ∆T 3

= (R + / 2R) ∆T 5

= ( / 2R)(15) = 37.5R = (37.5)(8.314) = 311.775 joule/mol

 − ln    (R+C ) ln  − ln    (R + / R) ln   − 1() ln  

∆S = Cp ln = =

T2

2

T1

1

v

3

2

T2

2

T1

1

313.15

0.5

298.15

1

– (R) ln(0.5) = (5/ 2R) ln(1.050) – (R) = (5/ 2R)(0.04879) + 0.6932R = 0.815175R = (0.815175)(8.314) = 6.777 joule/K mol



Cp = R + C v





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), (g), calculate Q, W, ∆U, ∆H, ∆ H, and ∆S ; compare ∆ S to Q/T. a) At constant volume, volume, the gas is heated to 400 K. b) At constant pressure, pressure, 1 atm, the gas is heated to 400 K. c) The gas is expanded isothermally isothermally and reversibly until until the pressure drops to ½ atm. d) The gas is expanded isothermally isothermally against a constant external external pressure equal to ½ atm until the gas pressure reaches ½ atm. e) The gas is expanded isothermally isothermally against zero opposing pressure pressure (Joule expansion) expansion) until the pressure of the gas is ½ atm. f) The gas is expanded adiabatically adiabatically against a constant pressure of ½ atm until until the final pressure is ½ atm. g) The gas is expanded adiabatically 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– (400– 300) 300) = 150R = (150)(8.314) = 1247.1 joule/mol ∆U = Qv



Qv = 1247.1 joule/mol

Menurut Hukum I Termodinamika ∆U = Q – W, – W, padahal pada kondisi volume tetap, harga ∆U = Q sehingga sehingga 1247.1 = 1247.1 1247.1 – W – W W = 0 joule/mol ∆H = Cp

∆T

padahal C p – C – Cv = R



Cp = R + C v

= (R + Cv) ∆T 3

= (R + / 2R) (400– (400– 300) 300) 5

= ( / 2R)(100) = 250R = (250)(8.314) = 2078.5 joule/mol

  ∆ =     32  ∆ =    2

1

2

1

= 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 – C – Cv = R  Cp = R + Cv 3

= (R + / 2R)(400 – 300) – 300)







∆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

=

1247.1 joule/mol = W

Qp – W 2078.5 joule/mol – W

=

2078.5 – 1247.1 – 1247.1

=

831.4 joule/mol

  ∆ =    2

1

– Cv = R sedangkan Cp – C



3

5

Cp = R + Cv  Cp = R + / 2R = / 2R

∆ =    400

5 2

300

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 t urun menjadi ½ atm, maka ∆S = ∆S = =

 ln  − ln   − nRln   − (1)(8.31 (1)(8.314) 4) ln   T2

2

T1

1

P2 P1

0.5 1

= – {(8.314)(– {(8.314)(– 0.693)} 0.693)} = 5.763 joule/K mol ∆S 5.763

   =  =

300

Qreversibel = (5.763)(300) = 1728.9 joule/mol ∆U = C ∆T,

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





=

− (1)(8.31 (1)(8.314) 4) ln   0.5 1

= – {(8.314)(– {(8.314)(– 0.693)} 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, sama, maka ∆T = 0, sehingga ∆H = 0

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

∆U = T∆S – W – W

0 = (300)(5.763) – W – W – W 0 = 1728.9 – W W = 1728.9 Menurut Hukum Hukum I Termodinamika Termodinamika : ∆U = Q – W – W , sehingga

 = 1728.9 = 5.763  300

Q = 1728.9

e. The gas is expanded isothermally against zero o pposing pressure pressure (Joule expansion) until the pressure of the gas is ½ atm ∆S = ∆S = =

 ln  − ln   − nRln   − (1)(8.31 (1)(8.314) 4) ln   T2

2

T1

1

P2 P1

0.5 1

= – {(8.314)(– {(8.314)(– 0.693)} 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

 =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

 =0 

g. The gas is expanded adiabatically and reversibly until the final pressure is ½ atm. Pada proses adiabatik, maka harga Q = 0

∆ =  = 0

Tugas Kimia Fisika-Bp-KASMADI

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