3-chemical thermodynamics and energetics

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3. CHEMICAL THERMODYNAMICS

AND ENERGETICS1) Three mol es o f an i deal gas a re

expanded isothermally from a volumeof 300 cm3 to 2.5 L at 300 K against apressure of 1.9 atm. Calculate the workdone in L atm and joules.

Given :V

1= 300 cm3

= 300 ×××× 10–3 L= 0.3 L

V2

= 2.5 L

T = 300 KP = 1.9 atm

To Find :Work done (W)

Formula :W = –P∆∆∆∆V

Solution :∴∴∴∴ W = –P∆∆∆∆V

W = –1.9(2.5 – 0.3)= –1.9 ×××× 2.2= –4.18 L atm

Now, 1 L atm = 101.33 JSo –4.18 L atm= –423.56 J

∴∴∴∴ W = –423.56 J

2) One mole of an ideal gas is compressedfrom 500 cm3 against a constant pressureof 1.216 ×××× 105 Pa. The work involved inthe process is 36.50 J. Calculate the finalvolume.

Given :V

1= 500 cm3

= 0.5 LP = 1.216 ×××× 105 Pa

= 1.2 atmW = 36.50 J

= 0.36 L atmTo Find :

Find Volume (V2)


Solution :W = –P∆∆∆∆V

3) Calculate the maximum work when 24

g of oxygen are expanded isothermallyand reversibly from a pressure of1.6 ×××× 105 Pa to 100 kPa at 298 K.

Given :Mass = 24 gP

1= 1.6 ×××× 105 Pa

P2

= 100 K Pa= 100 ×××× 103 Pa= 105 Pa

n =24

32

= 0.75 molesR = 8.314 J K–1 mol–1

T = 298 KTo Find :

Maximum mark (Wmax

)Formula :

Wmax

= –2.303nRT log10

P

P1

2Solution :

Wmax

= –2.303nRT log10

P

P

1

2= –2.303 ×××× 0.75 ×××× 8.314 ×××× 298

×××× log10

××××1.6 10

10

5

5

= –2.303 ×××× 0.75 ×××× 8.314 ×××× 298×××× log

10 1.6

= –4273.39 ×××× log 1.6= –4279.39 ×××× 0.204

W max

= –873 J

∴∴∴∴ 0.36 = –1.2 ×××× (V2 – 0.5)

∴∴∴∴0.36

1.2= –(V

2 – 0.5)

∴∴∴∴ 0.3 = –V2 + 0.5

∴∴∴∴ V2

= 0.2 L

∴∴∴∴ V2

= 200 cm3

Chemical Thermodynamcs and Energetics



.. 2


4) Three mol es o f an i deal gas a re

compressed isothermally and reversiblyto a volume of 2L. The work done is2.983 kJ at 220C. Calculate the initialvolume of the gas.

Given :n = 3 molesV

2= 2 L

Wmax

= 2.983 kJ= 2.983 ×××× 103 J

T = 220C= 295 K

R = 8.314 JK–1 mol–1

To Find :Initial volume

Formula :

Wmax

= –2.303nRT log10

V

V2

1Solution :

Wmax

= –2.303nRT log10

V

V2

1∴∴∴∴ 2.983 ×××× 103 = –2.303 ×××× 3 ×××× 8.314

×××× 295 log10

V

V21

∴∴∴∴××××2.983 10

–16,945.22

3

= log10

2

V1

∴∴∴∴ –0.176 = log10

2

V1

Take antilog on both sides.

Antilog [–0.176] =2

V1

∴∴∴∴ 0.6668 =2

V1

∴∴∴∴ V1

=2

0.6668

∴∴∴∴ V1

= 2.999 L∴∴∴∴ V

1= 3 L

5) 2.8 ×××× 10–2 kg of nitrogen is expanded

isothermally and reversibly at 300 Kfrom 15.15 ×××× 105 Nm–2 when the workdone is found to be –17.33kJ. Find thefinal pressure.

Given :

n = × ×× ×× ×× ×2.8 10 10

28

–2 3

= 1 moleT = 300 kP

1= 15.15 ×××× 105 Nm–2

Wmax

= –17.33 kJ

= –17.33 ×××× 103 JTo Find :

Final pressureFormula :

Wmax

= –2.303nRT log10

P

P1

2Solution :

Wmax

= –2.303nRT logP

P1

2∴∴∴∴ –17.33 ×××× 103 = –2.303 ×××× 8.314

× × × × 300 log××××15.15 10

P

5

2

∴∴∴∴ 17.33 ×××× 103

= 5.744 ×××× 103 log××××15.15 10

P

5

2

∴∴∴∴17.33

5.744= log

××××15.15 10

P

5

2

∴∴∴∴ 3.017 = log

××××15.15 10

P

5

2

Taking antilog on both sides

Antilog [3.017] =××××15.15 10

P

5

2

∴∴∴∴ 1039.92 =××××15.15 10

P

5

2



.. 3


∴∴∴∴ P2 = ××××15.15 10

1039.92

5

P2

= 1456.8 Nm–2

6) A sample of gas absorbs 4000 kJ of heat.i) If volume remains constant, what is

∆∆∆∆U ?ii) Suppose that in addition to

absorption of heat by the sample, thesurroundings does 2000 kJ of workon the sample, what is ∆∆∆∆U ?

iii)Suppose that as the original sample

absorbs heat, it expands againstatmospheric pressure and does 600 kJof work on its surroundings. Whatis ∆∆∆∆U ?

Given :i) q = 4000 KJii) W = 2000 KJiii) w = 600 KJ

To Find :i) ∆U ii) ∆U iii)∆U

Solution :i) At constant volume, ∆∆∆∆H = ∆∆∆∆U

So, ∆∆∆∆U = 4000 kJ

ii) Surroundings does 2000 kJ of work onsampleSo, ∆∆∆∆U = 4000 + 2000

= 6000 kJiii) Original sample absorbs heat and expands

aginst atmospheric pressure. Work doneon surroundings.

∆∆∆∆U = 4000 – 600= 3400 kJ

7) Calculate the work done in each of thefollowing reactions. State whether workis done on the system or by the system..

i) The oxidation of one mole of SO2 at 500C

2SO2(g)

+ O2(g)

2SO3(g)

ii) Decomposition of 2 moles of NH4NO

3

at 1000CNH

4NO

3(s) N

2O

(g) + 2H

2O

(g)

To Find :

WFormula :

W = –∆∆∆∆nRTSolution :i) 2SO

2(g) + O

2(g) 2SO

3(g)

1 mole of SO2 will 0.5 mole of O

2 to give

1 mole of SO3.

W = –∆∆∆∆n.RT= –RT(n

2 – n

1)

W = –8.314 ×××× 323(1 – 1.5)= –8.314 ×××× 323 ×××× (–0.5)

W = 1342.7 J

As W is positive, work is done onsystem.

ii) NH4NO

3(s) N

2O

(g) + 2H

2O

(g)

2 moles of NH4NO

3(s) gives 2 moles of

N2O

(g) and 4 moles of H

2O

(g)

So, ∆∆∆∆n = (6 – 0)= 6

W = –∆∆∆∆n.RT= –6 ×××× 8.314 ×××× 373= –18606.75 J

W = –18.61 kJAs W is negative, work is done by the

system.

8) The enthalpy change for the reactionC

2H

4(g) + H

2(g) C

2H

6(g) is –620 J

when 100 mL of enthylene and 100 mLof H

2 react at 1 atm pressure. Calculate

the pressrure volume work and ∆∆∆∆U.Given :

∆∆∆∆H = –620 J,V

1= 200ml = 0.2 L

P = 1 atmV

2

= 100 ml = 0.1 L

To Find :Pressure – Volume work (W)∆∆∆∆U


Solution :W = –P∆∆∆∆V

= –1 ×××× (0.1 – 0.2)= –1 ×××× –0.1



.. 4


W = 0.1L. atm

= 0.1 ×××× 101.33 J= +10.13 J

∆∆∆∆H = ∆U + P∆VW = –P∆V10.13 = –P∆V

∴∴∴∴ P∆V = –10.13 J∆H = –620 J

∴∴∴∴ ∆U = ∆H – P∆V= –620 – (–10.13)= –620 + 10.13

∴∴∴∴ ∆∆∆∆U = –609.9 J

9) Calculate s tandard enthalpy of thereactionFe

2O

3(s) + 3CO

(g) 2Fe

(s) + 3CO

2(g)

From the following data :∆∆∆∆

fH0(Fe

2O

3)= –824.2 kJ mol–1 ,

∆∆∆∆fH0(CO) = –110.5 kJ mol–1 ,

∆∆∆∆fH0(CO

2) = –393.5 kJ mol–1

Given :∆∆∆∆

fH0(Fe

2O

3) = –824.2 kJ/mol

∆∆∆∆fH0(CO) = –110.5 kJ/mol

∆∆∆∆fH0(CO

2) = –393.5 kJ/mol

To Find :Standard enthalpy of reaction (∆∆∆∆H0)

Formula :∆∆∆∆H0 = ∑∑∑∑∆∆∆∆

fH0(products)

– ∑∑∑∑∆∆∆∆fH0(reactants)

Solution :Standard enthalpy of reaction (∆∆∆∆H0)

= ∆∆∆∆fH0(products)

– ∆∆∆∆fH0(reactants)

= [2 moles of Fe ×××× 0+ 3 moles of CO

2

×××× ∆∆∆∆

fH0

of CO2] –[1 mole of Fe2O

3 ×××× ∆∆∆∆

fH0 of

Fe2O

3 + 3 moles of CO×××× ∆∆∆∆

fH0 of CO]

= [0 + (–1180.5)]– [–824.2 + (3 ×××× –110.5)]

= –1180.5 + 824.2 + 331.5= –24.8 kJ

10) Calculate the standard enthalpy of

formation of C2H6 from the following data :2C

2H

6(g) + 7O

2(g) 4CO

2(g) + 6H

2O

(l) ,

∆∆∆∆H0 = –3119 kJ mol–1

∆∆∆∆fH0(CO

2) = –285.8 kJ mol–1.

Given :Standard enthalpy of reaction (∆∆∆∆H0)

= –3119 kJ∆∆∆∆

fH0(CO

2) = –393.5 kJ/mol

∆∆∆∆fH0(H

2O) = –285.8 kJ/mol

To Find :Standard enthalpy of formation of

C2H6(∆∆∆∆fH0)Formula :

∆∆∆∆H0 = ∑∑∑∑∆∆∆∆fH0(products)


Solution :Standard enthalpy is given by,∆∆∆∆H0 = ∑∑∑∑∆∆∆∆

fH0(products)


= [4 moles of CO2 ×××× ∆∆∆∆

fH0 of

CO2 + 6 moles of H

2O

×××× ∆∆∆∆fH0 of H

2O] –

[2 moles of C2H

6 ×××× ∆∆∆∆

fH0 of

C2H6 + 0]= [4 ×××× (–393.5) + 6 ×××× –285.8]

– [2 ×××× ∆∆∆∆fH0 of C

2H

6]

∴∴∴∴ –3119 = –1574 – 1714.8 – 2×××× ∆∆∆∆

fH0 of C

2H

6

∴∴∴∴ –3119 = –3288.8 – 2×××× ∆∆∆∆

fH0 of C

2H6

∴∴∴∴ 169.8 = –2 ×××× ∆∆∆∆fH0 of C2H6

∴∴∴∴ –84.9 = ∆∆∆∆fH0 of C

2H

6

∴∴∴∴ ∆∆∆∆fH0 of C

2H

6= –84.9 kJ/mol

11) How much heat is evolved when 12 g of CO reacts with NO2 according to the

following reaction,4CO

(g) + 2NO

2(g) 4CO

2(g) + N

2 ,

∆∆∆∆H0 = –1198 kJGiven :

∆∆∆∆H0 = –1198 kJTo Find :

Heat evolved



.. 5


Solution :

According to the reaction, +1198 kJ ofheat is evolved. When 4 moles of COreacts with NO

2. So heat evolved per

mole is +299.5 kJ.Number of moles of CO

=massof CO

MolarmassofCO

=12

28

= 0.43 moles

So, heat evolved when 0.43 moles of COreacts = 0.43 ×××× 299.5

= +128.8 kJ

12) 38.55 kJ of heat is absorbed when 6.0 g of O

2 react with ClF according to the

reation,2ClF

(g) + O

2(g) Cl

2O

(g) + OF

2(g)

What is the standard enthalpy of thereaction ?

Given :Heat absorbed = 38.55 kJ

Mass of O2 = 6.0 gTo Find :

Standard enthalpy of reactionSolution :

Number of moles of O2

=massofO

MolarmassofO2

2

=6

32

= 0.1875

38.55 kJ heat is absorbed when 0.1875moles react with ClF.

∴∴∴∴ Heat absorbed for 1 mole =38.55

0.1875

= 205.6 kJFrom reaction 2 moles of ClF reacts with1 mole of O

2 so standard enthalpy of

reaction is 205.6 kJ.

13) Calculate ∆∆∆∆H0 of the reaction

CH4(g) + O2(g) CH2O(g) + H2O(g)

From the following data :Bond C–H O=O C=O O–H∆∆∆∆H0 /kJ mol–1 414 499 745 464

Given :Bond C–H O=O C=O O–H∆H0/kJ mol–1 414 499 745 464

To Find :∆∆∆∆H0

Solution :∆∆∆∆H0 = ∑∑∑∑∆∆∆∆H0(reactant bonds)

– ∑∑∑∑∆∆∆∆H0(product bonds)= [4∆∆∆∆H0(C–H) + ∆∆∆∆H0(O–O)]

–[2∆∆∆∆H0(C–H) +∆∆∆∆H0 (C=O)]+ 2(O–H)

= [2 ∆∆∆∆H0 (C–H) + ∆∆∆∆H0 (O–O)]– ∆∆∆∆H0 (C=O) – 2 (O–H)

= 2 ×××× 414 + 499 – 745 – 2 ×××× 464= –346 kJ

14) Calculate C–CI bond enthalpy from thefollowing data :CH

3Cl

(g) + Cl

2(g) CH

2Cl

2(g) + HCl

(g)

∆∆∆∆H0 = –104 kJBond C–H Cl=Cl H=Cl

∆∆∆∆H0 /kJ mol–1 414 243 431

Given :C–H Cl=Cl H=Cl 414 243 431

To Find :∆∆∆∆H0C–Cl

Solution :∆∆∆∆H0 = ∑∑∑∑∆∆∆∆H0(reactant bonds)

–∑∑∑∑∆∆∆∆

H0

(product bonds)= [3∆∆∆∆H0(C–H) + ∆∆∆∆H0

(C–Cl) + ∆∆∆∆H0(Cl–Cl)]–[2∆∆∆∆H0(C–H) + 2∆∆∆∆H0

(C–Cl) +∆∆∆∆H0(H–Cl)]= [∆∆∆∆H0(C–H) – ∆∆∆∆H0(C–Cl)]

+ ∆∆∆∆H0(Cl–Cl)– ∆∆∆∆H0(H–Cl)∴∴∴∴ –104 = 414 – ∆∆∆∆H0(C–Cl) + 243

– 431 ∆∆∆∆H0(C–Cl) = 414 + 104 + 243 – 431

= 300 kJ/mol



.. 6


15) Calculate the standard enthalpy of the

reaction,2C

(graphite) + 3H

2(g) C

2H

6(g) from the

following ∆∆∆∆H0 values :

i) C2H

6(g)+

7

2O

2(g)

2CO2(g)

+ 3H2O

(l) , ∆∆∆∆H0 = –1560 kJ

ii) H2(g)

+1

2O

2(g)H

2O

(l) ,

∆∆∆∆H0 = –285.8 kJiii ) C(graphite)

+ O

2(g)CO

2(g) ,

∆∆∆∆

H0

= –393.5 kJ

Given :Equation are,

C2H

6(g)+

7

2O

2(g)

2CO2(g)

+ 3H2O

(l) , ∆∆∆∆H0 = –1560 kJ ...(i)

H2(g)

+1

2O

2(g)H

2O

(l) ,

∆∆∆∆H0 = –285.8 kJ...(ii)

C(graphite) + O2(g) CO2(g) ,∆∆∆∆H0 = –393.5 kJ

...(iii)To Find :

∆∆∆∆H0

Solution :Reverse equation (i)

2CO2(g)

+ 3H2O

(l)C

2H

6(g)+

7

2O

2(g)

∆∆∆∆H0 = 1560 kJ ...(iv)Multiply equation (ii) by 3 and (iii) by 2

then add (iv), (ii) and (iii)

2CO2(g)

+ 3H2O

(l)C

2H

6(g)+

7

2O

2(g)

∆∆∆∆H0 = +1560 kJ

3H2(g)

+3

2O

2(g)3H

2O

(l) ,

∆∆∆∆H0 = –857.4 kJ

2C + 2O

22CO

2

∆∆∆∆H0 = –787 kJ

2C(g)

+ 3H2(g)

C2H

6(g)

∆∆∆∆H0 = 1560 + (–857.4) + (–787)= –84.4 kJ

16) Given the following equation Calculatestandard enthalpy of the reaction,

2Fe(g)

+3

2O

2(g) Fe

2O

3(s) ∆∆∆∆H0 = ?

i) 2Al(s)

+ Fe2O

3(s)

2 Fe(s) + Al2O(3)(s) ∆∆∆∆H0 = –847.6 kJ

ii) 2Al(s)

+3

2O

2(g)Al

2O

(3)(s)

∆∆∆∆H0 = –1670 kJGiven :

Equation are,2Al

(s)+ Fe

2O

3(s)2Fe

(s) + Al

2O

(3)(s)

∆∆∆∆H0 = –847.6 kJ...(i)

2Al(s)

+ 3

2

O2(g)

Al2O

(3)(s)

∆∆∆∆H0 = –1670 kJ ...(ii)

To Find :∆∆∆∆H0

Solution :Reverse equation (i), so that it becomes2Fe

(s)+ Al

2O

3(s)2Al

(s)+ Fe

2O

3(s)

∆∆∆∆H0 = 847.6 kJ ...(iii)Now add equation (ii) to equation (iii)2Fe

(s)+ Al

2O

3(s)2Al

(s)+ Fe

2O

3(s)

∆∆∆∆H0 = 847.6 kJ

2Al(s)

+ 3

2O

2(g)Al

2O

(3)(s)

∆∆∆∆H0 = –1670 kJ

2Fe(s)

+ 3

2O

2(g)Fe

2O

3(s)

∆∆∆∆H0 = 847.6 + (–1670)= 847.6 – 1670= –822.4 kJ



.. 7


17) Given the following equation and ∆∆∆∆H0

values at 250

C,i) Si

(s)+ O

2(g)SiO

2(s)

∆∆∆∆H0 = –911 kJii) 2C

(graphite)+ O

2(g)2CO

(g)

∆∆∆∆H0 = –221 kJ

iii) Si(s)

+ C(graphite)

SiC(s)

∆ ∆ ∆ ∆H0 = –65.3 kJCalculate ∆∆∆∆H0 for the following reaction,SiO

2(s) + 3C

(graphite) SiC

(s)+ 2CO

(g),

Given :

Equation are,Si

(s)+ O

2(g)SiO

2(s)

∆∆∆∆H0 = –911 kJ ...(i)2C

(graphite) + O

2(g)2CO

(g)

∆∆∆∆H0 = –221 kJ ...(ii)Si

(s)+ C

(graphite)SiC

(s)

∆∆∆∆H0 = –65.3 kJ...(iii)To Find :

∆∆∆∆H0

Solution :Reverse equation (i), so that it becomes

SiO2(s) Si(s) + O2(g)

∆∆∆∆H0 = 911 kJ ...(iv)Add equations (iv), (ii), (iii),SiO

2(s)Si

(s)+ O

2(g)

∆∆∆∆H0 = 911 kJ2C

(graphite) + O

2(g)2CO

(g)

∆∆∆∆H0 = –221 kJ

Si(s)

+ C(graphite)

SiC(s)

∆ ∆ ∆ ∆H0 = –65.3 kJ

SiO2(s)

+ 3C(graphite)

SiC(s)

+ 2CO(g)

∆∆∆∆H0 = 911 + (–211) + (–65.3)= 624.7 kJ

18) Given the following equation and ∆∆∆∆H0

values at 250

C,i) 2H

3BO

3(aq)

B2O

3(s)+ 3H

2O

(l) , ∆∆∆∆H0 = +14.4 kJ

ii) H3BO

3(aq)

HBO2(aq)

+ H2O

(l) ∆∆∆∆H0 = –0.02 kJ

iii) H2B

4O

7(s)2B

2O

3(s) + H

2O

(l)

∆∆∆∆H0 = 17.3 kJCalculate ∆∆∆∆H0 for the following reaction,H

2B

4O

7(s) + H

2O

(l) 4HBO

2(aq)

Given :

Equation are,2H

3BO

3(aq)B

2O

3(s)+ 3H

2O

(l) ,

∆∆∆∆H0 = +14.4 kJ ...(i)H

3BO

3(aq)HBO

2(aq)+ H

2O

(l)

∆∆∆∆H0 = –0.02 kJ ...(ii)H

2B

4O

7(s)2B

2O

3(s) + H

2O

(l)

∆∆∆∆H0 = 17.3 kJ...(iii)To Find :

∆∆∆∆H0

Solution :Reverse equation (i) and multiply by (2)2B

2

O3(s)

+ 6H2

O(l)

4H3

BO3(aq)

∆∆∆∆H0 = 28.8 kJ ...(iv)Multiply equations (ii) by 44H

3BO

3(aq)

4HBO2(aq)

+ 4H2O

(l)

∆∆∆∆H0 = (–0.08) kJ ...(v)Now add equation (iv), (v) and (iii)2B

2O

3(s)+ 6H

2O

(l)4H

3BO

3

∆∆∆∆H0 = 28.8 kJ4H

3BO

3(aq)

4HBO2(aq)

+ 4H2O

(l)

∆∆∆∆H0 = (–0.08) kJ

H2B4O7(s) 2B2O3(s)

+ H2O(l)

∆∆∆∆H0 = 17.3 kJ

H2B

4O

7(s)+ H

2O

(l)4HBO

2(aq)

∆∆∆∆H0 = –28.8 + (–0.08) + 17.3= –11.58 kJ



.. 8


20) Kp for the reaction,MgCO

3(s)MgO

(s)+ CO

2(s)is

9 ×××× 10–10. Calculate ∆∆∆∆G0 for the reactionat 250C.

Given :K

p= 9 ×××× 10–10

T = 250C = 298 KTo Find :

∆∆∆∆G0

Formula :∆∆∆∆G0 = –2.303RT log

10 K

p

Solution :∆∆∆∆G0 = –2.303RT log

10 K

p

∴∴∴∴ ∆∆∆∆G0 = –2.303 ×××× 8.314 ×××× 298 ×××× log9×××× 10–10

= 51,613.96 J/mol= 51.61 kJ

19) Calculate Kp for the reaction,

C2H4(g) + H2(g) C2H6(g) ,∆∆∆∆G0 = –100 kJmol–1 at 250C

Given :∆∆∆∆G0 = –100 kJ/mol

= –100000J/molT = 250C = 298 K

To Find :K

p

Formula :∆∆∆∆G0 = –2.303RT log

10 K

p

Solution :∆∆∆∆G0 = –2.303RT log

10 K

p

∴∴∴∴ –100000 = –2.303 ×××× 3.314 ×××× 298×××× log

10 K

p

∴∴∴∴ 100000 = 5705.85 log10

Kp

∴∴∴∴100000

5705.85= log

10 K

p

Take Antilog on both sides∴∴∴∴ Anti [17.5258] = K

p

∴∴∴∴ 3.35 ×××× 1017 = Kp

∴∴∴∴ Kp

= 3.35 ×××× 1017

21) Calculate ∆∆∆∆G0 for the reaction at 250C

CO(g) + 2H2(g) CH3OH(g)

∆∆∆∆G0 = –24.8 kJmol–1

If PCO

= 4 atm, PH2

= 2 atm, PCH3OH

= 2 atmGiven :

∆∆∆∆G0 = –24.8kJ/molT = 250C = 298 KP

CO= 4 atm

PH2

= 2 atmP

CH3OH= 2 atm

To Find :∆∆∆∆G

Formula :∆∆∆∆G = ∆∆∆∆G0 + 2.303RT log

10 K

p

Solution :∆∆∆∆G = ∆∆∆∆G0 + 2.303RT log

10 K

p

∴∴∴∴ QP

=P

P P

CH OH32

CO H2××××

∴∴∴∴ QP

=( )

2

4 22

××××=

2

16= 0.125

∴∴∴∴ ∆∆∆∆G = –24.8 + [2.303 ×××× 8.314 ×××× 10–3

×××× 298

×××× log(0.125)]= –24.8 + [–5.153]

= –24.8 – 5.153= 29.95 kJ

22) Calculate ∆∆∆∆S(total)

and hence showwhether the following reaction isspontaneous at 250C.HgS

(s)+ O

2(g)Hg

(l)+ SO

2(g)

∆∆∆∆H0 = –238.6 kJ and ∆∆∆∆S0 = +36.7 JK–1

Given :T = 250C = 298 k

∆∆∆∆H0 = –238.6 kJ∆∆∆∆S0 = +36.7 JK–1

To Find :∆∆∆∆S

(total)

Formula :

∆∆∆∆Ssurr

=– H

T

0∆∆∆∆



.. 9


Solution :

∆∆∆∆Ssurr

=– H

T

0∆∆∆∆=

–238.6kJ

298k

= 0.8006 kJ/K= 800.6 J/K

∆∆∆∆Ssys

= ∆∆∆∆S0 = +36.7 J/K∴∴∴∴ ∆∆∆∆S

(total)= ∆∆∆∆S

sys+ ∆∆∆∆S

surr

= 36.7 + 800.6= 837.3 J/K

As ∆∆∆∆S(total)

is positive, the reaction isspontaneous at 298 K

23) Determine whether the reaction ∆∆∆∆H and∆∆∆∆S values are spontaneous ornonspontaneous. State whether they areexothermic or endothermic.

i) ∆∆∆∆H = –110 kJ and ∆∆∆∆S = + 40JK–1

at 400 Kii) ∆∆∆∆H = –50 kJ and ∆∆∆∆S = – 130 JK–1 at 400 K

Given :∆∆∆∆H = –110 kJ and ∆∆∆∆S = + 40JK–1 at 400 K∆∆∆∆H = –50 kJ and ∆∆∆∆S = – 130 JK–1 at 400 K

To Find :To state whether the reactions areexothermic or endothermic andspontaneous or nonspontaneous.

Formula :∆∆∆∆G = ∆∆∆∆H – T∆∆∆∆S

Solution :i) ∆∆∆∆G = ∆∆∆∆H – T∆∆∆∆S

∆∆∆∆S = +40 J/K = 0.04 kJ/K∆∆∆∆H = –110 kJ

∴∴∴∴ ∆∆∆∆G = –110 – (400 ×××× 0.04)= –110 –16

= –126 kJBecause ∆∆∆∆G is negative, thereactions is spontaneous and thenegative value ∆∆∆∆H indicates thatthe reaction is exothermic.

ii) ∆∆∆∆G = ∆∆∆∆H – T∆∆∆∆S∆∆∆∆H = +50 kJ∆∆∆∆S = –130 J/K = –0.130kJ/k

∴∴∴∴ ∆∆∆∆G = 50 – (250 ×××× –0.130)

= 50 + 32.5

= 82.5 kJAs ∆∆∆∆G is positive, the reactions isnonspontaneous and the positivevalue ∆∆∆∆H indicates that thereaction is endothermic.

24) For the certain reaction, ∆∆∆∆H0 = –224 kJand ∆∆∆∆S0 = –153 JK–1. At what temperaturewill it change from spontaneous tononspontaneous ?

Given :∆∆∆∆H0 = –224 kJ

∆∆∆∆S0 = –153 JK–1

= –0.153 kJ/KTo Find :

Temperature (T) = ?Formula :

T =∆

∆

H

S

0

0

Solution :

T =∆

∆

H

S

0

0

=–224

0.153

= 1464.05 KAs ∆∆∆∆H0 and ∆∆∆∆S0 both are negative,reaction is spontaneous at lowertemperature. So the reaction will bespontaneous below 1464.05 K andnonspontaneous above 1464.05 K. Thechange over between spontaneous andnonspontaneous occurs at 1464.05 K

25) Determine whether the following reaction is spontaneous ornonspontaneous under standardcondition ?Zn

(s) + Cu2+

(aq)Zn2+ + Cu

(s)

∆∆∆∆H0 = – 219 kJ, ∆∆∆∆S0 = – 21 JK–1

Given :∆∆∆∆H0 = – 219 kJ∆∆∆∆S0 = – 21 JK–1

T = 298 K



.. 10


26) Determine whether the following reaction is spontaneous under standardconditions.2H

2O

(l) + O

2(g)2H

2O

2(l)

∆∆∆∆H0 = + 196 kJ, ∆∆∆∆S0 = – 126 JK–1

Does it have a cross-over temperatuer ?Given :

∆∆∆∆H0 = 196 KJ, ∆∆∆∆S0 = –126 J/KSolution :

∆∆∆∆G0 = ∆∆∆∆H0 – T∆∆∆∆S0

= 196 – (298 ×××× –0.126)= 196 + (37.55)= 233.55 kJ

As ∆∆∆∆G0 is positive the reaction will benonspontaneous. As ∆∆∆∆H0 is positive andAs ∆∆∆∆S0 is negative. ∆∆∆∆G0 will be alwayspositive regardless of temperature. Thereaction is therefore nonspontaneous atall the temperature. So it will not havecross-over temperature.

To Find :

∆∆∆∆G0

Formula :∆∆∆∆G0 = ∆∆∆∆H0 – T∆∆∆∆S0

Solution :∴∴∴∴ ∆∆∆∆G0 = ∆∆∆∆H0 – T∆∆∆∆S0

= –219 – (298 ×××× –0.021)= –219 – (–6.258)= –219 + 6.258= –212.74 kJ

As ∆∆∆∆G0 is negative the reactionspontaneous.

27) Oxidation of propane is represented asC

3H

8(g) + 5O

2(g) 3CO

2(g) + 4H

2O

(g) ,

∆∆∆∆

H0

= – 2043 kJ.How much pressure volume work isdone and what is the value of ∆∆∆∆U atconstant pressure of 1 atm when volumechange i s+22.4 L.

Given :P = 1 atmV = +22.4 L

To Find :Work done (W)Value of ∆∆∆∆U

Formula :

W = – P∆∆∆∆VSolution :

W = – P∆∆∆∆V= –1 ×××× (22.4)= – 22.4 L atm= – 22.4 ×××× 101.33 J= –2269.8 J

Ans : W = –2.27 kJ∆∆∆∆H = ∆∆∆∆U + P∆∆∆∆V

∴∴∴∴ –P∆∆∆∆V = –2.27P∆∆∆∆V = 2.27

∴∴∴∴ ∆∆∆∆U = ∆∆∆∆H – P∆∆∆∆V

= –2043 – 2.27= –2045.27 kJ

∴∴∴∴ ∆∆∆∆U = –20445.27 kJ

28) What is the va lue of ∆∆∆∆Ssurr

for thefollowing reaction at 298 K ?

6CO2(g)

+ 6H2O

(l)C

6H

12O

6(s)

+ 6O2(g)

,

∆∆∆∆G0 = 2879 kJ mol–1

∆∆∆∆S0 = – 210 JK–1mol–1.Does it have a cross-over temperatuer ?

Given :∆∆∆∆G0 = 2879 kJ mol–1

∆∆∆∆S0 = – 210 JK–1mol–1

= –0.210 kJ/K/mol.T = 298 K

To Find :∆∆∆∆S

surr

Formula :∆∆∆∆G0 = ∆∆∆∆H0 – T∆∆∆∆S0

Solution :∆∆∆∆G0 = ∆∆∆∆H0 – T∆∆∆∆S0

2879 = ∆∆∆∆H0 – (298 ×××× –0.210)

2879 = ∆∆∆∆H0 – (–62.58)∆∆∆∆H0 = 2816.42 kJ

∆∆∆∆Ssurr

=2816.42

298

= 9.45 kJ/K



.. 11


29) Calculate the enthalpy change for the

reaction.H

2O

(g) 2H

(g)+ O

(g)

And hence, calculate the bond enthalpyof O – H bond H

2O from the following

data :

∆∆∆∆vap

H(H2O)= 44.0 kJ mol–1

∆∆∆∆fH(H

2O) = –285.8 kJ mol–1

∆∆∆∆aH(H

2) = 436.0 kJ mol–1

∆∆∆∆H(O2) = 498.0 kJ mol–1

where ∆∆∆∆aH is the enthalpy of

atomizationGiven :∆∆∆∆

vapH(H

2O) = 44.0 kJ mol–1

∆∆∆∆fH(H

2O) = –285.8 kJ mol–1

∆∆∆∆aH(H

2) = 436.0 kJ mol–1

∆∆∆∆H(O2) = 498.0 kJ mol–1

To Find :∆∆∆∆H(O–H)∆∆∆∆H

Solution :Reaction can be given as follows

H2O(l) H2O(g)

∆∆∆∆H = 44.0 kJ mol ...(i)

H2(g)

+1

2O

2(g)H

2O

∆∆∆∆H = –285.8 kJ mol ...(ii)H

2(g)2H

(g)

∆∆∆∆H = 436 kJ mol ...(iii)O

2(g)2(O)

(g)

∆∆∆∆H = 498 kJ mol ...(iv)Reverse equation (i) and (ii) and addthem to equation (iii) and (iv)

H2O(g) H2O(l)

∆∆∆∆H = –44.0 kJ/mol

H2O

(l)H

2(g)+

1

2O

2(g)

∆∆∆∆H = 285.8 kJ molH

2(g)2H

(g)

∆∆∆∆H = 436 kJ mol

1

2O

2(g)

2

22(O)

(g)

∆∆∆∆H =

498

2 kJ mol .

H2O

(g)2H

(g)+ (O)

(g)

∆∆∆∆H = –44 + 285.8 + 436 + 249= 926.8 kJ ...(i)

Bond enthalpy of O–H bond in H2O can

be calculated as follows :∆∆∆∆H = ΣΣΣΣ(∆∆∆∆H of reactants)

– ΣΣΣΣ(∆∆∆∆H of products)= [2∆∆∆∆H (O–H)] – [O]

∆∆∆∆H = 926.8 kJ ... from(i)

∴∴∴∴ 926.8 = 2 ∆∆∆∆H (O–H)

∴∴∴∴ ∆∆∆∆H (O–H) =926.8

2

= 463.4 kJ/mol

30) Calculate ∆∆∆∆Ssurr

when one mole ofmethanol (CH

3OH) is formed from its

elements under standard conditions if∆∆∆∆

fH0(CH

3OH) = –238.9 kJ mol–1.

Given :

( )

∆H0

CH OH3 = –238.9 KJ/molTo Find :

∆∆∆∆Ssum

Solution :

∆∆∆∆Ssurr

=H

T

0∆∆∆∆

∆∆∆∆fH0 = –238.9 kJ mol

= –238900 J/mol

∴∴∴∴ ∆∆∆∆Ssurr

=238900

– –298

= 801.7 J/K

31) Calculate the total heat to melt 180g ofice at 00C, heat it to 1000C and thenvaporize it at that temperature. ∆∆∆∆

fusH(ice)

= 6.01 kJ mol–1 at 00C, ∆∆∆∆vap

H(H2O) = 40.7

kJ mol–1 at 1000C. Specific heat of water= 4.28 J g –1 K–1.



.. 12


Given :

Wice = 180 g∆∆∆∆

fusH(ice) = 6.01 KJ/ mol

( )∆∆∆∆ Hvap H O2 = 40.7 KJ/mol

sp. heat of H2O = 4.28 Jg–1k–1

To Find :Heat required to melt ice.

Solution :

–H2O

(g)Latent heat

of fusion H

2O

(l)Heating

00C 00C

H2

O(l)

Latent

heat of vaporization H

2

O(g)

1000C 1000CPart I :

H2O

(g)H

2O

(l)

Heat required = Latent heat for 180 g.1 mol of H

2O = 6.01 kJ

1 mol of H2O = 18 g

∴∴∴∴ 180 g = ? moles∴∴∴∴ 180 g of H

2O = 10 moles of H

2O

∴∴∴∴ 10 mol of H2O = 60.1 kJ

∴∴∴∴ Heat required = 60.1 kJ ...(i)Part II :

H2O(l) H2O(l)

00C 00CHeat required = mass ×××× Specific heat

×××× ∆∆∆∆T= 180 ×××× 4.18 ×××× 100= 75240 J= 75.240 kJ ...(ii)

Part III :H

2O

(l)H

2O

(g)

1000C 1000CHeat required = Latent heat of

vaporization1 mol of H

2O = 40.7 kJ

∴∴∴∴ 1 mol of H2O = 18 g

∴∴∴∴ 180 g of H2O = 10 moles of H

2O

∴∴∴∴ Heat required = 407 kJ ...(iii)From (i), (ii) and (iii)Heat required to melt ice

= 60.1 + 75.240 + 407= 542.34 kJ

Heat required to melt ice = 542.34 kJ

32) 6.24 g of ethanol are vaporized by

supplying 5.89 kJ of heat energy. Whatis enthalpy of vaporization of ethanol ?

Given :W

ethanol= 6.249

q = 5.89 KJTo Find :

∆∆∆∆vap

HSolution :

Number of moles of ethanol

=Massof ethanol

Molarmassofethanol

=6.24

46

= 0.1356 moles5.89 kJ heat is requiered to heat 0.1356moles.Then heat required to vaporize 1 molewill be,

=5.89

0.1356

= 43.44 kJ

∴∴∴∴ Enthalpy of vaporization of ethanol= 43.44 kJ

33) Enthalpy of fusion of ice is 6.01 kJ mol–1.The enthalpy of vaporization of wateris 45.07 kJ–1. What is enthalpy ofsublimation of ice ?

Given :∆∆∆∆

fusH = 6.01 kJ/mol

∆∆∆∆vap

H = 45.07 kJ/mol

To Find :∆∆∆∆

subH

Formula :∆∆∆∆

subH= ∆∆∆∆

fusH + ∆∆∆∆

vapH

Solution :∆∆∆∆

subH= ∆∆∆∆

fusH + ∆∆∆∆

vapH

∴∴∴∴ ∆∆∆∆sub

H= 6.01 + 45.07= 51.08 kJ/mol51.08 kJ

3-chemical thermodynamics and energetics

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