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# Solved Numerical Problem: Thermodynamics(Chemistry)

Chemistry : Thermodynamics : Solved Example Problem, Numerical Problems Questions with Answers, Solution

Thermodynamics(Chemistry)

Numerical Problems Questions with Answers, Solution

53. Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25┬░C and normal pressure.

n = 2 moles

Vi = 500ml = 0.5lit

Vf = 2lit

T = 25┬░C = 298K

w = ŌłÆ2.303 nRT log (Vf / Vi)

w = ŌłÆ2.303 ├Ś 2 ├Ś 8.314 ├Ś 298 ├Ś log(2/0.5)

w = ŌłÆ2.303 ├Ś 2 ├Ś 8.314 ├Ś 298 ├Ś log(4)

w = ŌłÆ 2.303 ├Ś 2 ├Ś 8.314 ├Ś 298 ├Ś 0.6021

w = ŌłÆ 6871J

w = ŌłÆ 6.871kJ

54. In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5 kJ KŌłÆ1. Calculate the enthalpy of combustion of the gas in kJ molŌłÆ1.

Given

Ti = 298K

Tf = 298.45K

k = 2.5 kJKŌłÆ1

m = 3.5g

Mm = 28

Heat evolved = k╬öT

= k(Tf ŌłÆ Ti)

= 2.45 kJKŌłÆ1 ├Ś (298.45 ŌłÆ 298)K

= 1.12 kJ

Heat evolved for 3.5 g of a gas = 1.12 kJ

Ōł┤ Heat evolved for 28g of gas ( 1 mole)

= (1.125├Ś28) / 3.5

The enthalpy of combustion of the gas

╬öHC = 9JK molŌłÆ1

55. Calculate the entropy change in the system, and surroundings, and the total entropy change in the universe during a process in which 245 J of heat flow out of the system at 77┬░C to the surrounding at 33┬░C.

Given:

Tsys = 77oC = (77 + 273 ) = 350K

Tsurr = 33oC = (33 + 273) = 306K

q = 245J

Solution:

╬öSsys = q/Tsys = ŌłÆ245 / 350 = ŌłÆ 0.7 JKŌłÆ1

╬öSsurr = q/Tsys = +245 / 350 = +0.8 JKŌłÆ1

╬öS = ╬öSsys + ╬öSsurr

╬öS univ = ŌłÆ0.7 JKŌłÆ1 + 0.8 JKŌłÆ1

╬öS univ = ŌłÆ0.1 JKŌłÆ1

56. 1 mole of an ideal gas, maintained at 4.1 atm and at a certain temperature, absorbs heat 3710J and expands to 2 litres. Calculate the entropy change in expansion process.

Given:

p = 4.1 atm,

v = 2 lit,

q = +3710 J,

n = 1 mol

Solution:

The ideal gas equation is = ( 4.1 atm ├Ś 2 lit ) / ( 0.0821 lit atm molŌłÆ1 KŌłÆ1 ├Ś 1 mol )

T = 100 K

╬öSexpansion = qexpansion / T = 3710J / 100k = 37.1 JKŌłÆ1

╬öSexpansion = 37.1 JKŌłÆ1

57. 30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JKŌłÆ1 molŌłÆ1 . Calculate the melting point of sodium chloride.

Given :

╬öHfusion = 30.4 kJ mol ŌłÆ1

╬öSfusion = 284.4 JKŌłÆ1 molŌłÆ1

Solution:

╬öSfusion = ╬öHfusion / Tf

Melting point of NaCl, Tf = ╬öHfusion / ╬öSfusion

= 30.4 kJ molŌłÆ1 / 28.4 JKŌłÆ1 molŌłÆ1 = ( 30.4 ├Ś 1000 J molŌłÆ1 ) / ( 28.4 JKŌłÆ1 molŌłÆ1)

Melting point of NaCl, Tf = 1070.4 K

58. Calculate the standard heat of formation of propane, if its heat of combustion is ŌłÆ2220.2 KJmolŌłÆ1 the heats of formation of CO2 (g) and H2O(1) are ŌłÆ393.5 and ŌłÆ285.8 kJ molŌłÆ1 respectively.

Given :

╬öH0f CO2 = ŌłÆ393.5 kJ molŌłÆ1

╬öH0f H2O = ŌłÆ285.8 kJ molŌłÆ1

╬öH0f C3H8 = ŌłÆ2220.2 kJ molŌłÆ1

Solution:

Combustion of propane

C3H8 + 5O2 ŌåÆ 3CO2 + 4H2O

╬öH0c = Ōłæ ╬öH0f p ŌłÆ Ōłæ ╬öH0f r ŌłÆ2220.2 = ŌłÆ1180.5 ŌłÆ 1143.2 ŌłÆ ╬öH0f C3H8

╬öH0f C3H8 = ŌłÆ103.5 kJ molŌłÆ1

╬öH0f C3H8 = ŌłÆ103.5 KJ

59. You are given normal boiling points and standard enthalpies of vapourisation Calculate the entropy of vapourisation of liquids listed below. Solution

For Ethanol:

Given

Tb = 78.4oC = (78.4 + 273) = 351.4K

╬öHV (Ethanol) = + 42.4kJ molŌłÆ1

╬öSV = ╬öHV/Tb

╬öSV = + 42.4kJ molŌłÆ1 / 351.4K

╬öSV = +42400 J molŌłÆ1 / 351.4K

╬öSV = +120.66 JKŌłÆ1molŌłÆ1

For Toluene:

Given

Tb =110.6┬░C = (110.6 +273) = 383.6K

╬öHV (Toluene) = + 35.2 kJ molŌłÆ1

╬öSV = ╬öHV / Tb

╬öSV = 35.2 kJ molŌłÆ1 / 383.6 K

╬öSV = +91.76 JKŌłÆ1 molŌłÆ1

60. For the reaction Ag2O(s) ŌĆö> 2Ag(s) + ┬Į O2(g) : ╬öH = 30.56 kJ molŌłÆ1 and ╬öS=6.66JKŌłÆ1 molŌłÆ1 (at 1 atm). Calculate the temperature at which ╬öG is equal to zero. Also predict the direction of the reaction (i) at this temperature and (ii) below this temperature.

Given:

╬öH = 30.56 kJmolŌłÆ1

= 30560JmolŌłÆ1

╬öS = 6.66 ├Ś10ŌłÆ3kJKŌłÆ1mo1ŌłÆ1

T = ? at which ╬öG = 0

╬öG = ╬öH ŌłÆ T╬öS

0 = ╬öH ŌłÆ T╬öS

T = ╬öH / ╬öS

T = (30.56 kJmolŌłÆ1) / (6.66├Ś10ŌłÆ3 kJKŌłÆ1molŌłÆ1)

T = 4589K

(i) At 4589K ; ╬öG = 0 the reaction is in equilibrium.

(ii) At temperature below 4589k, ╬öH > T ╬öS ╬öG = ╬öH ŌłÆT ╬öS > 0, the reaction in the forward direction, is non spontaneous. In other words the reaction occurs in the backward direction.

61. What is the equilibrium constant Keq for the following reaction at 400K.

2NOCl Ōćī 2NO(g) + Cl2(g)

given that ╬öHo = 77.2 kJ molŌłÆ1 ; and ╬öS┬░ = 122 JKŌłÆ1 molŌłÆ1

Given:

T = 400 K

╬öH┬░c = 77.2 KJmolŌłÆ1

= 77200 JmolŌłÆ1

╬öG┬░ = ŌłÆ 2.303RT log Keq

log Keq = ŌłÆ ╬öG┬░ / 2.303 RT

log Keq = ŌłÆ (╬öH┬░ ŌłÆ T ╬öS┬░) / 2.303 RT

log Keq = ŌłÆ ( [77200 ŌĆō 400 ├Ś 122] / [2.303 ├Ś 8.314 ├Ś 400] )

log Keq = ŌłÆ (28400 / 7659)

log Keq = ŌłÆ 3.7080

Keq = antilog (ŌłÆ3.7080)

Keq = 1.95 ├Ś10-4

62. Cyanamide (NH2CN) is completely burnt in excess oxygen in a bomb calorimeter, ╬öU was found to be -742.4 kJ molŌłÆ1, calculate the enthalpy change of the reaction at 298K. NH2CN(s) + 3/2 O2(g) ŌåÆ N2 (g) + CO2 (g) + H2O(l) ╬öH= ?

Given

T = 298K; ╬ö U= ŌłÆ742.4kJmolŌłÆ1

╬öH = ?

Solution:

╬öH = ╬öU + ╬öngRT

╬öH = ╬öU+ (np ŌłÆ nr) RT

╬öH = -742.4 + (2- 3/2) ├Ś 8.314 ├Ś l0ŌłÆ3 ├Ś 298

= ŌłÆ742.4 + (0.5 ├Ś 8.314 ├Ś 10ŌłÆ3 ├Ś 298)

= ŌłÆ742.4 + 1.24

= ŌłÆ741.16 kJmolŌłÆ1

63. Calculate the enthalpy of hydrogenation of ethylene from the following data. Bond energies of C-H, C-C, C = C and H - H are 414, 347, 618 and 435 kJ molŌłÆ1

Given:

EC-H = 414kJ molŌłÆ1

Ec-c = 347 kJ molŌłÆ1

Ec-c = 618kJ molŌłÆ1

EH-H = 435 kJ molŌłÆ1

Solution: ╬öHr = Ōłæ(Bond energy)r ŌłÆ Ōłæp (Bond energy)p

╬öHr = (Ec-c + 4EC-H + EH-H) ŌłÆ (Ec-c + 6EC-H)

╬öHr = (618 + (4 ├Ś 414) = 435) ŌłÆ (347 + (6 ├Ś 414))

╬öHr = 2709 ŌłÆ 2831

╬öHr = ŌłÆ122 kJ molŌłÆ1

64. Calculate the lattice energy of CaCl2 from the given data

Ca (s)+Cl2(g) ŌåÆ CaCl2(s) ŌłåH0f  = ŌłÆ 795 kJ molŌłÆ1

Ca(s) + Cl2(g) ŌåÆ CaCl2(s) ╬öH┬░f = ŌłÆ795 kJ molŌłÆ1

Atomisation : Ca(s) ŌåÆ Ca(g) ╬öH┬░1 = +121 kJ molŌłÆ1

Ionisation: Ca(g) ŌåÆ Ca2+(g) + 2eŌłÆ  ╬öH┬░2 = +2422 kJ molŌłÆ1

Dissociation : Cl2(g) ŌåÆ 2Cl(g) ╬öH┬░3 = +242.8 kJ molŌłÆ1

Electron affinity : Cl(g) + eŌłÆ ŌåÆ ClŌłÆ(g) ╬öH┬░4 = ŌłÆ355 kJ molŌłÆ1

Solution: ╬öHf = ╬öH1 + ╬öH2 + ╬öH3 + 2 ╬öH4 + u

ŌłÆ795 = 121 + 2422 + 242.8 + (2 ├Ś ŌłÆ355) + u

ŌłÆ795 = 2785.8 ŌłÆ 710 + u

ŌłÆ795 = 2075.8 + u

u = ŌłÆ 795 ŌłÆ 2075.8

u = ŌłÆ2870.8 kJ molŌłÆ1

65. Calculate the enthalpy change for the reaction Fe2O3 + 3CO ŌåÆ 2Fe + 3CO2 from the following data.

2Fe + 3/2 O2 ŌåÆ Fe2O3;       ╬öH = -741kJ

C + 1/2 O2 ŌåÆ CO;            ╬öH = ŌłÆ137kJ

C+ O2 ŌåÆ CO2; ╬öH = ŌłÆ 394.5kJ

┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬Ł┬ŁGiven :

╬öHf (Fe2O3) = ŌłÆ 741kJmolŌłÆl

╬öHf (CO) = ŌłÆ 137kJmolŌłÆ1

╬öHr = ?

Solution:

Fe2O3 +3CO ŌåÆ 2Fe + 3CO2

╬öHr = Ōłæ( ╬öHf) products ŌłÆ Ōłæ (╬öHf )reactants

╬öHr = [0+ 3(ŌłÆ 394.5)] ŌłÆ [ŌłÆ741 + 3(ŌłÆ 137)]

╬öHr = [ŌłÆ1183.5] ŌłÆ [ŌłÆ1152]

╬öHr = ŌłÆ1183.5 + ŌłÆ1152

╬öHr = ŌłÆ31.5 kJ molŌłÆ1

66. When 1-pentyne (A) is treated with 4N alcoholic KOH at 175┬░C, it is converted slowly into an equilibrium mixture of 1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of 1,2 pentadiene (C) the equilibrium was maintained at 175┬░C, calculate ╬öG0 for the following equilibria.

B Ōćī A            ╬öG01 = ?

B Ōćī C            ╬öG02 = ?

Given :

T= 175┬░ C = 175 + 273 = 448K

Concentration of 1-pentyne [A] = 1.3%

Concentration of 2-pentyne [B] = 95.2%

Concentration of 1, 2-pentadiene [C] = 3.5%

B [95.2%] Ōćī A [3.5%]

K1 = 1.3 / 95.2 = 0.0136

B [95.2%] Ōćī C [3.5%]

K2 = 3.5 / 95.2 = 0.0367

ŌćÆ ╬öG01 = ŌłÆ2.303 RT logK1

╬öG01 = ŌłÆ2.303 ├Ś 8.314 ├Ś 448 ├Ś log0.0136

╬öG01 = ŌłÆ2.303 ├Ś 8.314 ├Ś 448 ├Ś ŌłÆ1.8664

╬öG01 = +16010J

╬öG01 = +16 kJ

╬öG02 = ŌłÆ2.303 RT logK2

╬öG02 = ŌłÆ2.303 ├Ś 8.314 ├Ś 448 ├Ś log 0.0367

╬öG02 = ŌłÆ2.303 ├Ś 8.314 ├Ś 448 ├Ś ŌłÆ1.4353

╬öG02 = +12312 J

╬öG02 = +12.312 kJ

67. At 33K, N2O4 is fifty percent dissociated, calculate the standard free energy change at this temperature and at one atmosphere

Given:

N2O4 Ōćī 2NO2

Initial mole 1

At Equilibrium remaining mole Total moles at equilibrium 0.5 + 1 = 1.5

Total pressure = 1atm

Partial pressure = mole fraction ├Ś total pressure

PN2O4 = (0.5/1.5) ├Ś 1 = 0.5 / 1.5

PNO2 = (1/1.5) ├Ś 1 = 1/ 1.5

KP = PNO2 / PN2O4 = (1/1.5)2 / (0.5/1.5) = 1.33

We know that,

╬öG┬░ = ŌłÆ2.303RT log Keq

= ŌłÆ2.303 ├Ś 8.314 ├Ś 331og 1.33

= ŌłÆ2.303 ├Ś 8.314 ├Ś 33 ├Ś 0.1239

╬öG┬░ = ŌłÆ78.29JmolŌłÆ1 68. The standard enthalpies of formation of SO2and SO3are ŌłÆ297 kJ molŌłÆ1 and ŌłÆ396 kJ molŌłÆ1 respectively. Calculate the standard enthalpy of reaction for the reaction: SO2 + ┬Į O2 ŌåÆ SO3

Given:

╬öHof (SO2) = ŌłÆ297 kJ molŌłÆ1

╬öHof (SO2) = ŌłÆ396 kJ molŌłÆ1

SO2 + ┬Į O2 ŌåÆ SO3

╬öHo1 = ?

Solution:

╬öHor = (╬öHof)compound

ŌłÆ Ōłæ (╬öHf)elements

╬öHor = ╬öHof (SO3) ŌłÆ ( ╬öHof (SO2) + 1/2 ╬öHof (O2) )

╬öHor = ŌłÆ396 kJ molŌłÆ1

= ŌłÆ (ŌłÆ297kJ molŌłÆ1 + 0)

╬öHor = ŌłÆ396 kJ molŌłÆ1 + 297

╬öHor = ŌłÆ 99kJ molŌłÆ1

69. For the reaction at 298 K: 2A +B ŌĆö> C

╬öH = 400 J molŌłÆ1; ╬öS = 0.2 JKŌłÆ1molŌłÆ1

Determine the temperature at which the reaction would be spontaneous.

Given :

╬öH = 400 J molŌłÆ1

╬öS = 0.2 J KŌłÆ1 molŌłÆ1

T = 298 K

Solution:

We know that ╬öG = ╬öH ŌłÆ T╬öS

At equilibrium, ╬öG = 0

Ōł┤ T╬öS = ╬öH

T = ╬öH / ╬öS = ( 400 J molŌłÆ1 ) / ( 0.2 JkŌłÆ1 molŌłÆ1)

T = 2000K

╬öG = 400 ŌłÆ (2000 ├Ś 0.2)

= 0

if T > 2000K ╬öG will be negative

The reaction would be spontaneous only beyond 2000K

70. Find out the value of equilibrium constant for the following reaction at 298K, 2NH3(g) + CO2(g) Ōćī NH2CONH2(aq) + H2O (1) Standard Gibbs energy change, ╬öGr0 at the given temperature is ŌłÆ13.6 kJ molŌłÆ1.

Solution :

Given:

T = 298K

╬öGr0 = ŌłÆ13.6 kJ molŌłÆ1

╬öG0 = ŌłÆ2.303 RT log Keq

log Keq = ŌłÆ╬öG0 / 2.303RT

log Keq = [ŌłÆ(ŌłÆ13.6)] / [2.303 ├Ś 8.314├Ś10ŌĆō3 ├Ś 298]

log Keq = 2.38

Keq = antilog (2.38)

Keq = 239.88

71. A gas mixture of 3.67 lit of ethylene and methane on complete combustion at 25┬░C and at 1 atm pressure produce 6.11 lit of carbondioxide. Find out the amount of heat evolved in kJ, during this combustion. (╬öHc(CH4) = - 890 kJ molŌłÆ1 and (╬öHc(C2H4) = ŌłÆ1423 kJ molŌłÆ1

Solution:

╬öHC (CH4) = ŌłÆ 890 kJ molŌłÆ1

╬öHC (C2H4) = ŌłÆ 1423 kJ molŌłÆ1

Let the mixture contain x lit of CH4 and (3.67 ŌłÆ x) lit of ethylene.

CH4 + 2O2 ŌåÆ CO2 + 2H2O

x lit                 x lit

C2H4 + 3O2 ŌåÆ 2CO2 + 2H2O

(3.67-x) lit             2 (3.67 - x) lit

Volume of Carbondioxide formed

= x + 2 (3.67ŌłÆ x) = 6.11 lit

x + 7.34 ŌłÆ 2x = 6.11

7.34 - x = 6.11

x = 1.23 lit

Given mixture contains 1.23 lit of methane and 2.44 lit of ethylene, hence ╬öHc = [(╬öHc (CH4) / 22.4) ├Ś (x)] + [ (╬öHc (C2H4) / 22.4) ├Ś (3.67 - x) ]

╬öHc = [ (ŌłÆ890kJmolŌłÆ1 / 22.4) ├Ś 1.23] + [ (ŌłÆ1423/22.4) ├Ś (3.67 ŌłÆ 1.23) ]

╬öH = [ ŌłÆ48.87kJmolŌłÆ1] + [ŌłÆ155kJmolŌłÆ1]

╬öHc = ŌłÆ203.87 kJmolŌłÆ1

### Problem

The standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l) are - 277, -393.5 and -285.5 kJ mol-1 respectively.

Calculate the standard enthalpy change for the reaction

C2H5OH(l)+3O2(g) ŌåÆ2CO2(g)+ 3 H2O(l)

The enthalpy of formation of O2(g) in the standard state is Zero, by definition

### Solution:

For example, the standard enthalpy change for the combustion of ethanol can be calculated from the standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l). The enthalpies of formation are ŌĆō277, ŌĆō 393.5 and ŌĆō285.5 kJ molŌĆō1 respectively.

C3H5OH(l) + 3O2(g) ŌåÆ 2CO2(g) + 3H2O(l) =[ ŌłÆ787 ŌłÆ 856 . 5] ŌłÆ [ ŌłÆ277]

= - 1643 .5 + 277

ŌłåH0r = ŌłÆ1366 . 5 KJ

### Problem

Calculate the value of ŌłåU and ŌłåH on heating 128.0 g of oxygen from 0o C to 1000 C. CV and CP on an average are 21 and 29 J mol-1 K-1. (The difference is 8Jmol-1 K-1 which is approximately equal to R)

### Solution.

We know

ŌłåU   = n Cv (T2-T1)

ŌłåH   = n CP (T2- T1)

Here

n= 128/32 4 moles ;

T2 = 1000

C =373K;

T1 = 00

C = 273K

ŌłåU   = n Cv (T2-T1)

ŌłåU =    4 x 21 x (373 - 273)

ŌłåU =    8400 J

ŌłåU =     8.4 kJ

ŌłåH   = n Cp (T2- T1)

ŌłåH =    4 ├Ś 29 ├Ś (373- 273)

ŌłåH =    11600 J

ŌłåH =    11.6 kJ

### Problem:

If an automobile engine burns petrol at a temperature of 816o C and if the surrounding temperature is 21o C, calculate its maximum possible efficiency.

Solution: Here

Th = 816+273= 1089 K;

Tc= 21+273= 294K

%Efficiency=( 1089-294 / 1089) x100

%Efficiency=73%

### Problem:

Calculate the standard entropy change for the following reaction( ŌłåS0f ), given the standard entropies of CO2(g), C(s),O2(g) as 213.6 , 5.740 and 205 JKŌłÆ1 respectively.

C(g) + O2(g) ŌåÆCO2(g)

S0r = Ōłæ S0products ŌłÆ Ōłæ Sreac0 tan ts

S0r = {S0CO 2 } ŌłÆ {SC0 + S0O2 }

S0r = 213.6 ŌłÆ [5.74 + 205]

S0r = 213.6 ŌłÆ[210.74]

S0r = 2.86 JKŌłÆ1

### Problem:

Calculate the entropy change during the melting of one mole of ice into water at 00 C and 1 atm pressure. Enthalpy of fusion of ice is 6008 J mol-1

### Given:

ŌłåHfusion = 6008 JmolŌłÆ1

Tf = 0 0

C = 273 K

2O(S) --273 KŌåÆ H 2O ( l) fusion = 22 .007 J K ŌłÆ1 moleŌłÆ1

### Problem:

Calculate ╬öG0  for conversion of oxygen to ozone 3/2 OŌåö O3(g) at 298 K, if Kp for this conversion is 2.47 u 10ŌłÆ29 in standard pressure units.

### Solution:

╬öG0 = ŌłÆ 2.303 RT log Kp

Where

R = 8.314 JKŌłÆ1molŌłÆ1

Kp = 2.47 x10ŌłÆ29

T = 298K

╬öG0=ŌłÆ2.303(8.314)(298)log(2.47u10ŌłÆ29)

╬öG0 = 16300 JmolŌłÆ1

╬öG0 = 16.3 KJ molŌłÆ1

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11th Chemistry : UNIT 7 : Thermodynamics : Solved Numerical Problem: Thermodynamics(Chemistry) |