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# Solved Example Problem: Thermodynamics(Chemistry)

Chemistry : Thermodynamics : Solved Example Problem

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. 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.

SOLUTION :

Given :

Ti = 298

Tf = 298.45 K

= 2.5 kJ K−1 m = 3.5g

Mm = 28

heat evolved = k ΔT

=k (Tf − Ti)

=2.5 kJ K−1 (298.45−298)K

=1.125kJ

ΔHc =(1.125/3.5)x28kJmol−1

ΔHc = 9 kJ 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. 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. 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. 58. Calculate the standard heat of formation of propane, if its heat of combustion is −2220.2 kJ mol−1. the heats of formation of CO2(g) and H2O(l) are −393.5 and −285.8 kJ mol−1 respectively.

SOLUTION :

Given

C3H8 + 5O2 3CO2 + 4H2O

DH0C = −2220.2 kJ mol−1 −−−−−(1)

C + O2 CO2

DH0f  = −393.5 kJ mol−1 −−−−−(2)

H2 + 1/2 O2 H2O

DH0f  = −285.8 kJ mol−1 −−−−− (3)

3C + 4H2 C3H8

DH0C  = ?

(2) × 3 3C + 3O2 3CO2

DH0f  = −1180.5 kJ−−−−−(4)

(3)× 4 4 H2 + 2O2

4H2O DH0f = −1143.2 kJ −−−−−(5)

(4) + (5) − (1) 3C + 3O2 + 4H2 +

2O2 + 3CO2 + 4H2O 3CO2 + 4H2 O + C3H8 + 5O2

DH0f  =−1180.5 − 1143.2− (−2220.2) kJ

3C + 4 H2 C3H8

DH0f = −103.5 kJ

Standard heat of formation of propane is DH0f (C3H 8) = −103.5 kJ

59. You are given normal boiling points and standard enthalpies of vapourisation. Calculate the en-tropy of vapourisation of liquids listed below.  ΔSV = + 91.76 J K−1 mol−1

60. For the reaction Ag2O(s) → 2Ag(s)+ ½ O2(g) : ΔH = 30.56 kJ mol−1 and ΔS=6.66JK−1mol−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.

Solution :

Given :

ΔH = 30.56 kJ mol−1

= 30560 J mol−1

ΔS = 6.66 × 10-3 kJK−1mol−1

T = ? at which ΔG=0

ΔG = ΔH − TΔS

0 = ΔH – TΔS

T= ΔH/ ΔS

T = 30560 J mol−1 / 6.66 × 10-3 kJK−1mol−1

T = 4589 K

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

ii. at temperature below 4598 K , ΔH > T ΔS

ΔG = ΔH − TΔS > 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 (g) 2NO(g) + Cl2(g),

given that ΔH0 = 77.2kJ mol−1 ;

and ΔS0=122JK−1mol−1.

Solution :

Given

= 400K ; ΔH0 = 77.2 kJ mol−1=77200

J mol−1 ; ΔS0 = 122 JK−1 mol−1

ΔG0 = −2.303 RT log Keq 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= ?

Solution :

Given

T = 298K ;

ΔU = − 742.4 kJmol−1

ΔH = ?

ΔH = ΔU + Δng RT

ΔH = ΔU + (np−nr)RT

ΔH = -742.4 + (2- 3/2)

= − 742.4 + (0.5 × 8.314 ×10-3 × 298)

=− 742.4 + 1.24

=− 741.16 kJ mol−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. 64. Calculate the lattice energy of CaCl2 from the given data

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

Atomisation : Ca(s) → Ca(g) ∆H10 = + 121 kJ mol−1

Ionisation : Ca(g) → Ca2+(g) + 2e− ∆H20 = + 2422 kJ mol−1

Dissociation : Cl2(g) → 2Cl(g) H30 = + 242.8 kJ mol−1

Electron affinity : Cl (g) + e− → Cl (g) H04 = −355 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 = −741 kJ

C + ½ O­2 → CO; ΔH = − 137.5 kJ

C + O­2 → CO2; ΔH = − 394.5 kJ

Solution :

Given :

ΔHf (Fe2O3) = −741 kJ mol−1

ΔHf (CO) = −137 kJ mol−1

Hf(CO2) = −394.5 kJ mol−1

Fe2O3 + 3CO 2Fe + 3CO2 ΔHr=?

ΔHr =  Σ(ΔHf)productsΣ (ΔHf)reactants

ΔHr =  [2 ΔHf (Fe) + 3ΔHf (CO2)] - [ΔHf (Fe2O3) + 3ΔHf (CO)]

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      ∆G10 = ?

B C      ∆G20 = ? 67. At 33K, N2O4 is fifty percent dissociated. Calculate the standard free energy change at this tem-perature and at one atmosphere. 68. The standard enthalpies of formation of SO2 and SO3 are −297 kJ mol−1 and −396 kJ mol−1 respec-tively. Calculate the standard enthalpy of reaction for the reaction: SO2 + ½ O2 → SO3 69. For the reaction at 298 K : 2A +B → C

ΔH=400 J mol−1 ; ΔS = 0.2 JK−1 mol−1 Determine the temperature at which the reaction would be spontaneous.

Solution

Given :

T = 298 K

ΔH = 400 J mol−1 = 400 J mol−1

ΔS = 0.2 J K−1 mol−1

ΔG = ΔH − TΔS

if T = 2000 K

ΔG = 400 − (0.2 × 2000) = 0

if T > 2000 K

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, 2 NH3(g) + CO2 (g) NH2CONH2 (aq) + H2O (l) Standard Gibbs energy change, G0r at the given temperature is –13.6 kJ mol−1. 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 :

Given :

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 ### 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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