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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 = −2.303 nRT log (Vf / Vi)

w = −2.303 × 2 × 8.314 × 298 × log(2/0.5)

w = −2.303 × 2 × 8.314 × 298 × log(4)

w = − 2.303 × 2 × 8.314 × 298 × 0.6021

w = − 6871J

w = − 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 K−1. Calculate the enthalpy of combustion of the gas in kJ mol−1.

Given

Ti = 298K

Tf = 298.45K

k = 2.5 kJK−1

m = 3.5g

Mm = 28

Heat evolved = kΔT

= k(Tf − Ti)

= 2.45 kJK−1 × (298.45 − 298)K

= 1.12 kJ

Heat evolved for 3.5 g of a gas = 1.12 kJ

Heat evolved for 28g of gas ( 1 mole)

= (1.125×28) / 3.5

The enthalpy of combustion of the gas

ΔHC = 9JK mol−1

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:

ΔSsys = q/Tsys = −245 / 350 = − 0.7 JK−1

ΔSsurr = q/Tsys = +245 / 350 = +0.8 JK−1

ΔS = ΔSsys + ΔSsurr

ΔS univ = −0.7 JK−1 + 0.8 JK−1

ΔS univ = −0.1 JK−1

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 × 2 lit ) / ( 0.0821 lit atm mol−1 K−1 × 1 mol )

T = 100 K

ΔSexpansion = qexpansion / T = 3710J / 100k = 37.1 JK−1

ΔSexpansion = 37.1 JK−1

57. 30.4 kJ is required to melt one mole of sodium chloride. The entropy change during melting is 28.4 JK−1 mol−1 . Calculate the melting point of sodium chloride.

Given :

ΔHfusion = 30.4 kJ mol −1

ΔSfusion = 284.4 JK−1 mol−1

Solution:

ΔSfusion = ΔHfusion / Tf

Melting point of NaCl, Tf = ΔHfusion / ΔSfusion

= 30.4 kJ mol−1 / 28.4 JK−1 mol−1 = ( 30.4 × 1000 J mol−1 ) / ( 28.4 JK−1 mol−1)

Melting point of NaCl, Tf = 1070.4 K

58. Calculate the standard heat of formation of propane, if its heat of combustion is −2220.2 KJmol−1 the heats of formation of CO2 (g) and H2O(1) are −393.5 and −285.8 kJ mol−1 respectively.

Given :

ΔH0f CO2 = −393.5 kJ mol−1

ΔH0f H2O = −285.8 kJ mol−1

ΔH0f C3H8 = −2220.2 kJ mol−1

Solution:

Combustion of propane

C3H8 + 5O2 → 3CO2 + 4H2O

ΔH0c = ∑ ΔH0f p ∑ ΔH0f r −2220.2 = −1180.5 − 1143.2 − ΔH0f C3H8

ΔH0f C3H8 = −103.5 kJ mol−1

ΔH0f C3H8 = −103.5 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

ΔHV (Ethanol) = + 42.4kJ mol−1

ΔSV = ΔHV/Tb

ΔSV = + 42.4kJ mol−1 / 351.4K

ΔSV = +42400 J mol−1 / 351.4K

ΔSV = +120.66 JK−1mol−1

For Toluene:

Given

Tb =110.6°C = (110.6 +273) = 383.6K

ΔHV (Toluene) = + 35.2 kJ mol−1

ΔSV = ΔHV / Tb

ΔSV = 35.2 kJ mol−1 / 383.6 K

ΔSV = +91.76 JK−1 mol−1

60. For the reaction Ag2O(s) —> 2Ag(s) + ½ O2(g) : ΔH = 30.56 kJ mol−1 and ΔS=6.66JK−1 mol−1 (at 1 atm). Calculate the temperature at which ΔG is equal to zero. Also predict the direction of the reaction (i) at this temperature and (ii) below this temperature.

Given:

ΔH = 30.56 kJmol−1

= 30560Jmol−1

ΔS = 6.66 ×10−3kJK−1mo1−1

T = ? at which ΔG = 0

ΔG = ΔH − TΔS

0 = ΔH − TΔS

T = ΔH / ΔS

T = (30.56 kJmol−1) / (6.66×10−3 kJK−1mol−1)

T = 4589K

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

(ii) At temperature below 4589k, ΔH > T ΔS ΔG = ΔH −T ΔS > 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 2NO(g) + Cl2(g)

given that ΔHo = 77.2 kJ mol−1 ; and ΔS° = 122 JK−1 mol−1

Given:

T = 400 K

ΔH°c = 77.2 KJmol−1

= 77200 Jmol−1

ΔG° = − 2.303RT log Keq

log Keq = − ΔG° / 2.303 RT

log Keq = − (ΔH° − T ΔS°) / 2.303 RT

log Keq = − ( [77200 – 400 × 122] / [2.303 × 8.314 × 400] )

log Keq = − (28400 / 7659)

log Keq = − 3.7080

Keq = antilog (−3.7080)

Keq = 1.95 ×10-4

62. Cyanamide (NH2CN) is completely burnt in excess oxygen in a bomb calorimeter, ΔU was found to be -742.4 kJ mol−1, calculate the enthalpy change of the reaction at 298K. NH2CN(s) + 3/2 O2(g) → N2 (g) + CO2 (g) + H2O(l) ΔH= ?

Given

T = 298K; Δ U= −742.4kJmol−1

ΔH = ?

Solution:

ΔH = ΔU + ΔngRT

ΔH = ΔU+ (np − nr) RT

ΔH = -742.4 + (2- 3/2) × 8.314 × l0−3 × 298

= −742.4 + (0.5 × 8.314 × 10−3 × 298)

= −742.4 + 1.24

= −741.16 kJmol−1

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 mol−1

Given:

EC-H = 414kJ mol−1

Ec-c = 347 kJ mol−1

Ec-c = 618kJ mol−1

EH-H = 435 kJ mol−1

Solution: ΔHr = ∑(Bond energy)r − ∑p (Bond energy)p

ΔHr = (Ec-c + 4EC-H + EH-H) − (Ec-c + 6EC-H)

ΔHr = (618 + (4 × 414) = 435) − (347 + (6 × 414))

ΔHr = 2709 − 2831

ΔHr = −122 kJ mol−1

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

Ca (s)+Cl2(g) → CaCl2(s) ∆H0f  = − 795 kJ mol−1

Ca(s) + Cl2(g) → CaCl2(s) ΔH°f = −795 kJ mol−1

Atomisation : Ca(s) → Ca(g) ΔH°1 = +121 kJ mol−1

Ionisation: Ca(g) Ca2+(g) + 2e  ΔH°2 = +2422 kJ mol−1

Dissociation : Cl2(g) → 2Cl(g) ΔH°3 = +242.8 kJ mol−1

Electron affinity : Cl(g) + eCl(g) ΔH°4 = −355 kJ mol−1

Solution: ΔHf = ΔH1 + ΔH2 + ΔH3 + 2 ΔH4 + u

−795 = 121 + 2422 + 242.8 + (2 × −355) + u

−795 = 2785.8 − 710 + u

−795 = 2075.8 + u

u = − 795 − 2075.8

u = −2870.8 kJ mol−1

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

2Fe + 3/2 O2 → Fe2O3;       ΔH = -741kJ

C + 1/2 O2 → CO;            ΔH = −137kJ

C+ O2 → CO2; ΔH = − 394.5kJ

­­­­­­­­­­­­­­­­­­Given :

ΔHf (Fe2O3) = − 741kJmol−l

ΔHf (CO) = − 137kJmol−1

ΔHr = ?

Solution:

Fe2O3 +3CO → 2Fe + 3CO2

ΔHr = ∑( ΔHf) products − ∑ (ΔHf )reactants

ΔHr = [0+ 3(− 394.5)] − [−741 + 3(− 137)]

ΔHr = [−1183.5] − [−1152]

ΔHr = −1183.5 + −1152

ΔHr = −31.5 kJ mol−1

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 ΔG0 for the following equilibria.

B A            ΔG01 = ?

B C            ΔG02 = ?

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%] A [3.5%]

K1 = 1.3 / 95.2 = 0.0136

B [95.2%] C [3.5%]

K2 = 3.5 / 95.2 = 0.0367

ΔG01 = −2.303 RT logK1

ΔG01 = −2.303 × 8.314 × 448 × log0.0136

ΔG01 = −2.303 × 8.314 × 448 × −1.8664

ΔG01 = +16010J

ΔG01 = +16 kJ

ΔG02 = −2.303 RT logK2

ΔG02 = −2.303 × 8.314 × 448 × log 0.0367

ΔG02 = −2.303 × 8.314 × 448 × −1.4353

ΔG02 = +12312 J

ΔG02 = +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 2NO2

Initial mole 1

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

Total pressure = 1atm

Partial pressure = mole fraction × total pressure

PN2O4 = (0.5/1.5) × 1 = 0.5 / 1.5

PNO2 = (1/1.5) × 1 = 1/ 1.5

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

We know that,

ΔG° = −2.303RT log Keq

= −2.303 × 8.314 × 331og 1.33

= −2.303 × 8.314 × 33 × 0.1239

ΔG° = −78.29Jmol−1 68. The standard enthalpies of formation of SO2and SO3are −297 kJ mol−1 and −396 kJ mol−1 respectively. Calculate the standard enthalpy of reaction for the reaction: SO2 + ½ O2 → SO3

Given:

ΔHof (SO2) = −297 kJ mol−1

ΔHof (SO2) = −396 kJ mol−1

SO2 + ½ O2 → SO3

ΔHo1 = ?

Solution:

ΔHor = (ΔHof)compound

− ∑ (ΔHf)elements

ΔHor = ΔHof (SO3) − ( ΔHof (SO2) + 1/2 ΔHof (O2) )

ΔHor = −396 kJ mol−1

= − (−297kJ mol−1 + 0)

ΔHor = −396 kJ mol−1 + 297

ΔHor = − 99kJ mol−1

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

ΔH = 400 J mol−1; ΔS = 0.2 JK−1mol−1

Determine the temperature at which the reaction would be spontaneous.

Given :

ΔH = 400 J mol−1

ΔS = 0.2 J K−1 mol−1

T = 298 K

Solution:

We know that ΔG = ΔH − TΔS

At equilibrium, ΔG = 0

TΔS = ΔH

T = ΔH / ΔS = ( 400 J mol−1 ) / ( 0.2 Jk−1 mol−1)

T = 2000K

ΔG = 400 − (2000 × 0.2)

= 0

if T > 2000K ΔG 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) NH2CONH2(aq) + H2O (1) Standard Gibbs energy change, ΔGr0 at the given temperature is −13.6 kJ mol−1.

Solution :

Given:

T = 298K

ΔGr0 = −13.6 kJ mol−1

ΔG0 = −2.303 RT log Keq

log Keq = ΔG0 / 2.303RT

log Keq = [(−13.6)] / [2.303 × 8.314×10–3 × 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. (ΔHc(CH4) = - 890 kJ mol−1 and (ΔHc(C2H4) = −1423 kJ mol−1

Solution:

ΔHC (CH4) = − 890 kJ mol−1

ΔHC (C2H4) = − 1423 kJ mol−1

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

CH4 + 2O2 → CO2 + 2H2O

x lit                 x lit

C2H4 + 3O2 → 2CO2 + 2H2O

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

Volume of Carbondioxide formed

= x + 2 (3.67− x) = 6.11 lit

x + 7.34 − 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 ΔHc = [(ΔHc (CH4) / 22.4) × (x)] + [ (ΔHc (C2H4) / 22.4) × (3.67 - x) ]

ΔHc = [ (−890kJmol−1 / 22.4) × 1.23] + [ (−1423/22.4) × (3.67 − 1.23) ]

ΔH = [ −48.87kJmol−1] + [−155kJmol−1]

ΔHc = −203.87 kJmol−1

### 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) →2CO2(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 –277, – 393.5 and –285.5 kJ mol–1 respectively.

C3H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(l) =[ −787 − 856 . 5] − [ −277]

= - 1643 .5 + 277

∆H0r = −1366 . 5 KJ

### Problem

Calculate the value of ∆U and ∆H 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

∆U   = n Cv (T2-T1)

∆H   = n CP (T2- T1)

Here

n= 128/32 4 moles ;

T2 = 1000

C =373K;

T1 = 00

C = 273K

∆U   = n Cv (T2-T1)

∆U =    4 x 21 x (373 - 273)

∆U =    8400 J

∆U =     8.4 kJ

∆H   = n Cp (T2- T1)

∆H =    4 × 29 × (373- 273)

∆H =    11600 J

∆H =    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( ∆S0f ), given the standard entropies of CO2(g), C(s),O2(g) as 213.6 , 5.740 and 205 JK−1 respectively.

C(g) + O2(g) →CO2(g)

S0r = ∑ S0products − ∑ Sreac0 tan ts

S0r = {S0CO 2 } − {SC0 + S0O2 }

S0r = 213.6 − [5.74 + 205]

S0r = 213.6 −[210.74]

S0r = 2.86 JK−1

### 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:

∆Hfusion = 6008 Jmol−1

Tf = 0 0

C = 273 K

2O(S) --273 K→ H 2O ( l) fusion = 22 .007 J K −1 mole−1

### Problem:

Calculate ΔG0  for conversion of oxygen to ozone 3/2 O↔ O3(g) at 298 K, if Kp for this conversion is 2.47 u 10−29 in standard pressure units.

### Solution:

ΔG0 = − 2.303 RT log Kp

Where

R = 8.314 JK−1mol−1

Kp = 2.47 x10−29

T = 298K

ΔG0=−2.303(8.314)(298)log(2.47u10−29)

ΔG0 = 16300 Jmol−1

ΔG0 = 16.3 KJ mol−1

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

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