Solved Numerical Problem: Thermodynamics(Chemistry)

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

V_i = 500ml = 0.5lit

V_f = 2lit

T = 25°C = 298K

w = −2.303 nRT log (V_f / V_i)

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

T_i = 298K

T_f = 298.45K

k = 2.5 kJK^−1

m = 3.5g

M_m = 28

Heat evolved = kΔT

= k(T_f − T_i)

= 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

ΔH_C = 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:

T_sys = 77^oC = (77 + 273 ) = 350K

T_surr = 33^oC = (33 + 273) = 306K

q = 245J

Solution:

ΔS_sys = q/T_sys = −245 / 350 = − 0.7 JK^−1

ΔS_surr = q/T_sys = +245 / 350 = +0.8 JK^−1

ΔS = ΔS_sys + ΔS_surr

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

ΔS_expansion = q_expansion / T = 3710J / 100k = 37.1 JK^−1

ΔS_expansion = 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 :

ΔH_fusion = 30.4 kJ mol ^−1

ΔS_fusion = 284.4 JK^−1 mol^−1

Solution:

ΔS_fusion = ΔH_fusion / T_f

Melting point of NaCl, T_f = ΔH_fusion / ΔS_fusion

= 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, T_f = 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 :

ΔH⁰_{f CO2} = −393.5 kJ mol^−1

ΔH⁰_{f H2O} = −285.8 kJ mol^−1

ΔH⁰_{f C3H8} = −2220.2 kJ mol^−1

Solution:

Combustion of propane

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

ΔH⁰_c = ∑ ΔH⁰_{f p} − ∑ ΔH⁰_{f r}

 âˆ’2220.2 = −1180.5 − 1143.2 − ΔH⁰_{f C3H8}

ΔH⁰_{f C3H8} = −103.5 kJ mol^−1

ΔH⁰_{f 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

T_b = 78.4^oC = (78.4 + 273) = 351.4K

ΔH_V (Ethanol) = + 42.4kJ mol^−1

ΔS_V = ΔH_V/T_b

ΔS_V = + 42.4kJ mol^−1 / 351.4K

ΔS_V = +42400 J mol^−1 / 351.4K

ΔS_V = +120.66 JK^−1mol^−1

For Toluene:

Given

T_b =110.6°C = (110.6 +273) = 383.6K

ΔH_V (Toluene) = + 35.2 kJ mol^−1

ΔS_V = ΔH_V / T_b

ΔS_V = 35.2 kJ mol^−1 / 383.6 K

ΔS_V = +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) + Cl_2(g)

given that ΔH^o = 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 K_eq

log K_eq = − ΔG° / 2.303 RT

log K_eq = − (ΔH° − T ΔS°) / 2.303 RT

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

log K_eq = − (28400 / 7659)

log K_eq = − 3.7080

K_eq = antilog (−3.7080)

K_eq = 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 + Δn_gRT

ΔH = ΔU+ (n_p − n_r) 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:

E_C-H = 414kJ mol^−1

E_c-c = 347 kJ mol^−1

E_c-c = 618kJ mol^−1

E_H-H = 435 kJ mol^−1

Solution:

ΔH_r = ∑(Bond energy)_r − ∑_p (Bond energy)_p

ΔH_r = (E_c-c + 4E_C-H + E_H-H) − (E_c-c + 6E_C-H)

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

ΔH_r = 2709 − 2831

ΔH_r = −122 kJ mol^−1

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

Ca (s)+Cl₂(g) â†’ CaCl₂(s) âˆ†H⁰_f  = − 795 kJ mol^−1

Ca_(s) + Cl_2(g) → CaCl_2(s) ΔH°_f = −795 kJ mol^−1

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

Ionisation: Ca_(g) → Ca²⁺_(g) + 2e^−  Î”H°₂ = +2422 kJ mol^−1

Dissociation : Cl_2(g) → 2Cl_(g) ΔH°₃ = +242.8 kJ mol^−1

Electron affinity : Cl_(g) + e^− → Cl^−_(g) ΔH°₄ = −355 kJ mol^−1

Solution:

ΔH_f = 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;       Î”H = -741kJ

C + 1/2 O₂ → CO;            Î”H = −137kJ

C+ O₂ → CO₂; ΔH = − 394.5kJ

­­­­­­­­­­­­­­­­­­Given :

ΔH_f (Fe₂O₃) = − 741kJmol^−l

ΔH_f (CO) = − 137kJmol^−1

ΔH_r = ?

Solution:

Fe₂O₃ +3CO → 2Fe + 3CO₂

ΔH_r = ∑( ΔH_f) _products − ∑ (ΔH_f )_reactants

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

ΔH_r = [−1183.5] − [−1152]

ΔH_r = −1183.5 + −1152

ΔH_r = −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            Î”G⁰₁ = ?

B ⇌ C            Î”G⁰₂ = ?

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

P_N2O4 = (0.5/1.5) × 1 = 0.5 / 1.5

P_NO2 = (1/1.5) × 1 = 1/ 1.5

K_P = P_NO2 / P_N2O4 = (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

[Alternative Answer]

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:

ΔH^o_f (SO₂) = −297 kJ mol^−1

ΔH^o_f (SO₂) = −396 kJ mol^−1

SO₂ + ½ O₂ → SO₃

ΔH^o₁ = ?

Solution:

ΔH^o_r = (ΔH^o_f)_compound

 âˆ’ ∑ (ΔH_f)_elements

ΔH^o_r = ΔH^o_f (SO₃) − ( ΔH^o_f (SO₂) + 1/2 ΔH^o_f (O₂) )

ΔH^o_r = −396 kJ mol^−1

 = − (−297kJ mol^−1 + 0)

ΔH^o_r = −396 kJ mol^−1 + 297

ΔH^o_r = − 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, ΔG_r⁰ at the given temperature is −13.6 kJ mol^−1.

Solution :

Given:

T = 298K

ΔG_r⁰ = −13.6 kJ mol^−1

ΔG⁰ = −2.303 RT log K_eq

log K_eq = −ΔG⁰ / 2.303RT

log K_eq = [−(−13.6)] / [2.303 × 8.314×10^–3 × 298]

log K_eq = 2.38

K_eq = antilog (2.38)

K_eq = 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. (ΔH_c(CH₄) = - 890 kJ mol^−1 and (ΔH_c(C₂H₄) = −1423 kJ mol^−1

Solution:

ΔH_C (CH₄) = − 890 kJ mol^−1

ΔH_C (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

ΔH_c = [(ΔH_c (CH₄) / 22.4) × (x)] + [ (ΔH_c (C₂H₄) / 22.4) × (3.67 - x) ]

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

ΔH = [ −48.87kJmol^−1] + [−155kJmol^−1]

ΔH_c = −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) â†’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) â†’ 2CO2(g) + 3H2O(l)