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NCERT Class 11 Chemistry Chapter 6 Thermodynamics
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Thermodynamics
Chapter: 6
| Part – I |
Choose the correct answer.
1. A thermodynamic state function is a quantity.
(i) used to determine heat changes.
(ii) whose value is independent of path.
(iii) used to determine pressure volume work.
(iv) whose value depends on temperature only.
Ans: (ii) whose value is independent of path.
2. For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0
Ans: (iii) q = 0
3. The enthalpies of all elements in their standard states are:
(i) unity.
(ii) zero.
(iii) < 0
(iv) different for each element.
Ans: (ii) zero.
4. ∆U of combustion of methane is – X kJ mol–1. The value of ∆H is:
(i) = ∆U°
(ii) > ∆U°
(iii) < ∆U°
(iv) = 0
Ans: (iii) < ∆U°
5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are – 890.3 kJ mol–1, – 393.5 kJ mol–1, and – 285.8 kJ mol–1 respectively. Enthalpy of formation of CH4(g) will be:
(i) – 74.8 kJ mol–1
(ii) – 52.27 kJ mol–1
(iii) + 74.8 kJ mol–1
(iv) + 52.26 kJ mol–1
Ans: (i) – 74.8 kJ mol–1
6. A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be:
(i) possible at high temperature.
(ii) possible only at low temperature.
(iii) not possible at any temperature.
(iv) possible at any temperature.
Ans: (iv) possible at any temperature.
7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Ans: According to 1st law of thermodynamic
△U = q + w
Heat absorbed by the system,
Q = 701J
Work done by the system = – 304 J
Change in internal energy,
△U = + W = 7o1 J – 394J
= 307J.
8. The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb calorimeter, and ∆U was found to be – 742.7 kJ mol–1 at 298 K. Calculate enthalpy change for the reaction at 298 K.
NH2CN(g) + 3 2 O2(g) → N2(g) + CO2(g) + H2O(l)
Ans: To calculate the enthalpy change (ΔH) for the reaction, we can use the relationship between internal energy change (ΔU) and enthalpy change (ΔH):
ΔH=ΔU+ΔngRT
Where:
ΔU is the change in internal energy, given as −742.7 kJ mol−1
Δng is the change in the number of moles of gas during the reaction,
R is the gas constant (8.314 J mol−1K−1)
T is the temperature, given as 298 K.
ΔU = 742.7kJmol– 1
ΔH = ΔU + (Δng)RT
Δng = (1 + 1) – 3/2 = 1/2 mol
R = 8.314 10- 3 kjk– 1 mol-1,
T = 298 K
According to the relation,
ΔH = ΔU + Δng Rt
ΔH = -742.7 + (1/2) × 8.314 × 10-3 kj-1 + 298 K
= -742.7 – 1.239
= – 741.5kj mol-1
9. Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol–1 K–1.
Ans: Number of moles of Aluminum (Al) 60/27 moles
Specific heat capacity CP = 24 J/mol∘C
Temperature change ΔT = 55−53 = 2∘C
No. of moles of Al (m)
= 60g / 27g mol-1
= 2.22 mol
Molar heat capacity (C)
= 24J Mol-1 K-1
Rise in Temperature (ΔT)
= 55 – 35 = 20∘C
= 20 K
Heat Evolved (Q) = C × m × T
= (24J mol-1 k-1) × (2.22mol) × (20K)
= 1065.6 J
= 1.067kJ.
10. Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0° C to ice at –10.0° C. ∆fusH = 6.03 kJ mol–1 at 0° C. Cp [H2O(l)] = 75.3 J mol–1 K–1 Cp [H2O(s)] = 36.8 J mol–1 K–1.
Ans: Total enthalpy change involved in the transformation is the sum of the following changes and can be represented as:
According to Hess’s law:
ΔH = ΔH1 + ΔH2 + ΔH3
ΔH1 = 75.3 J mol-1 K-1 (10K)
= 753 l Mol-1
ΔH2 = -6.03 KJ mol-1
= – 6030 j mol-1
ΔH3 = 36.8 J mol-1 K-1 (-10K)
= – 36.8 J mol-1 K-1 (-10K)
= – 36.8 J mol-1
Total enthalpy change:
ΔH = (753 – 6030 – 368) J mol-1
= 7645 J mol-1
= 5.645 kj mol-1.
11. Enthalpy of combustion of carbon to CO2 is –393.5 kJ mol–1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas.
Ans: Heat of combustion for 1 mole of CO2 is −393.5kJ/mol
Number of moles in 35.2 g of CO2:
Moles of CO2 = 35.2g /44g/mol = 0.8 mol
Heat of combustion for 0.8 mole of CO2:
Heat released = − 393.5kJ / mol × 0.8mol
= −314.8kJ.
12. Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, – 393, 81 and 9.7 kJ mol–1 respectively. Find the value of ∆r H for the reaction:
N2O4(g) + 3CO(g) → N2O(g) + 3CO2(g)
Ans: Enthalpy of reaction ΔrH
ΔrH = [81+3(−393)]−[9.7+3(−110)]
ΔrH = [81−1179]−[9.7−330]
= 777.7 kJ/ mol.
13. Given N2(g) + 3H2(g) → 2NH3(g); ∆r H° = –92.4 kJ mol–1 What is the standard enthalpy of formation of NH3 gas?
Ans: ΔfHӨNH3(g) = -(92.4)/2
= -46.2kJ Mol-1
The standard enthalpy of formation of NH3 gas is -46.2 kJ/mol.
14. Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
(i) CH3OH (l) + 3/2 O2(g) → CO2(g) + 2H2O(l); ∆rH° = –726 kJ mol–1
(ii) C(g) + O2(g) → CO2(g); ∆cH° = –393 kJ mol–1
(iii) H2(g) + 1/2 O2(g) → H2O(l); ∆fH° = –286 kJ mol–1.
Ans: The standard enthalpy of formation of CH3OH (l) is calculated using Hess’s Law:
ΔfH∘(CH3OH (l)) = ΔrH + ΔcH +2 × ΔfH (H2O)
ΔfH∘(CH3OH (l)) = −726kJ/mol + 393kJ/mol + 572kJ/mol
= −239kJ/mol
∴ ΔfH∘(CH3OH (l)) = −239 kJ/mol.
15. Calculate the enthalpy change for the process CCl4(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C – Cl in CCl4(g).
∆vapH°(CCl4) = 30.5 kJ mol–1
∆f H° (CCl4) = –135.5 kJ mol–1.
∆aH° (C) = 715.0 kJ mol–1, where ∆aH° is enthalpy of atomisation.
∆aH° (Cl2) = 242 kJ mol–1
Ans: Given data:
(i) CCI4(l) → CCl4(g);
∆vapH∘ = 30.5 kJ mol-1
(ii) C(S) + 2CI2 (g) → CCI4(l),
∆f H∘ = – 135.5 kJ mol-1
(iii) C(s) → C(g),
∆aH∘ = 715.8 kJ mol-1
(iv) Cl2(g) → 2Cl-(g),
∆aH∘ = 242 kJ mol-1
The equation we aim at is:
CCI4 → C(g) + 4 Cl(g); ΔH = ?
ΔH = ΔaH(C) + 4 × 1/2 × ΔaH (Cl2) − ΔfH(CCl4)
= 715.0kJ/ mol l 4 × 121.0kJ/mol + 135.5kJ/mol
= 715.0 + 484.0 + 135.5
=1334.5kJ/mol
Bond enthalpy calculation:
Bond enthalpy of C-Cl = (1334.5−30.5) / 4
= 1304/4
= 326 kJ/mol
So, the enthalpy change is 1334.5 kJ/mol, and the bond enthalpy of C-Cl in CCl4 is 326 kJ/mol.
16. For an isolated system, ∆U = O, what will be ∆S?
Ans: In an isolated system, where ∆U = O, the internal energy remains constant. For such a system, the change in entropy (∆S) must reflect the tendency towards increased disorder. According to the second law of thermodynamics, entropy tends to increase in an isolated system. Thus, for any spontaneous process occurring within the system, ∆S will be greater than zero (∆S > O). example of two gases contained separately in two bulbs connected by a stop-cock and isolated from the surrounding as an example of an isolated system. On opening the stop-cock, the two gases mix up, i.e.; the system becomes more disordered. This implies that ∆S > O but ∆U = O for the process. change some words.
17. For the reaction at 298 K, 2A + B → C ∆H = 400 kJ mol–1 and ∆S = 0.2 kJ K–1 mol–1 At what temperature will the reaction become spontaneous considering ∆H and ∆S to be constant over the temperature range.
Ans: To determine the temperature at which the reaction becomes spontaneous, use the condition that ΔGmust be zero for spontaneity:
ΔG = ΔH − TΔS
For spontaneity, ΔG should be less than zero. Set ΔG = 0
0 = ΔH − TΔS0
Rearrange to find the temperature T:
T = ΔH/ΔST
= 2000 K
Thus reaction will be in a state of equilibrium at 2000 K and will be spontaneous above this temperature.
18. For the reaction, 2 Cl(g) → Cl2(g), what are the signs of ∆H and ∆S?
Ans: ∆H: negative because energy is released in bond formation.
The entropy (∆S) is negative because the disorder decreases when atoms combine to form molecules.
19. For the reaction
2 A(g) + B(g) → 2D(g)
∆U° = –10.5 kJ and
∆SӨ = –44.1 JK–1.
Calculate ∆G for the reaction, and predict whether the reaction may occur spontaneously.
Ans: Calculator of ∆G, use
ΔG = ΔU + Δ(PV)
Given ∆U = -10.5 kJ, and for gases, Δ(PV) = ΔnRT.
Here, Δn = -1Δn = −1, R = 8.314 J/K·mol,
T = 298 K:
Δ(PV) = −1 × 8.314 × 298 = −2478 J
ΔG = −10,500−2478 = −12,978 J
Since ΔG is negative, the reaction is spontaneous.
According to Gibbs Helmholtz equation:
ΔGо = ΔHо – TΔо
DGо = (-12.978 kJ) – (298K) × (-0.0441 KJK -1)
= (12.978 + 13.112)
= -12.978 + 13.142
= 0.164 KJ
20. The equilibrium constant for a reaction is 10. What will be the value of ∆G?
R = 8.314 JK–1 mol–1, T = 300 K.
Ans: To find ∆G, use the equation ∆G = -RT ln K. Given R
= 8.314 J/K·mol, T
= 300 K, and K = 10:
ΔG = − 8.314 × 300 × ln(10)
= − 8.314 × 300 × 2.3026
= − 5740 J/mol
So, ∆G = -5740 J/mol.
= 5.740 KJ mol-1
21. Comment on the thermodynamic stability of NO(g), given
1/2 N2(g) + 1/2 O2(g) = NO(g); ∆rHӨ = 90 kJ mol–1
NO(g) + 1/2 O2(g) = NO2(g); ∆rHӨ = –74 kJ mol–1
Ans: The positive enthalpy of formation for NO indicates it is less stable. Energy is released during the formation of NO₂, making NO₂ more stable and favouring its formation over NO.
22. Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions.
∆H° = –286 kJ mol–1
Ans: qrev = (-∆HӨ) = -286 kJ mol-1
= – 286000 J mol-1
∆S(surrounding) = qrev/T = (286000 J mol-1)/ 298 K
= 959 JK-1 Mol-1.
Hence the entropy change in surroundings will be 0.96KJ/mol.
