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In this unit, you have learned about the basic principles and laws relevant to thermodynamics. Below, we have presented some in-text exercises taken from the NCERT Class 11 Chemistry (Part I), Chapter 5: Thermodynamics, which will help you reinforce your understanding of key concepts, apply theoretical knowledge, and prepare effectively for exams.
Contents
Explore Notes of Class 11 Chemistry
NCERT Solutions Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics
Below, we have provided you with the exercises mentioned in the NCERT Class 11 Chemistry (Part-1) Chapter 5: Thermodynamics.
Exercises
- A thermodynamic state function is a quantity:
(i) used to determine heat changes
(ii) whose value is independent of the path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only - For the process to occur under adiabatic conditions, the correct condition is:
(i) ∆T = 0
(ii) ∆p = 0
(iii) q = 0
(iv) w = 0 - The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element - ∆U⁰ of combustion of methane is –X kJ mol–1. The value of ∆H⁰ is:
(i) = ∆U⁰
(ii) > ∆U⁰
(iii) < ∆U⁰
(iv) = 0 - The enthalpies 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 - 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 - 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?
- 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 the enthalpy change for the reaction at 298 K.
NH₂CN(s) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l) - 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.
- 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 - 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.
- 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 ∆rH for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g) - Given:
N₂(g) + 3H₂(g) → 2NH₃(g); ∆rH⁰ = –92.4 kJ mol–1
What is the standard enthalpy of formation of NH₃(g)? - Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); ∆rH⁰ = –726 kJ mol–1
C(graphite) + O₂(g) → CO₂(g); ∆cH⁰ = –393 kJ mol–1
H₂(g) + 1/2 O₂(g) → H₂O(l); ∆fH⁰ = –286 kJ mol–1 - Calculate the enthalpy change for the process:
CCl₄(g) → C(g) + 4Cl(g)
and calculate the bond enthalpy of C–Cl in CCl₄(g).
Given:
∆vapH⁰(CCl₄) = 30.5 kJ mol–1
∆fH⁰(CCl₄) = –135.5 kJ mol–1
∆aH⁰(C) = 715.0 kJ mol–1
∆aH⁰(Cl₂) = 242 kJ mol–1 - For an isolated system, ∆U = 0. What will be ∆S?
- 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? - For the reaction:
2Cl(g) → Cl₂(g), what are the signs of ∆H and ∆S? - For the reaction:
2A(g) + B(g) → 2D(g)
∆U⁰ = –10.5 kJ and ∆S⁰ = –44.1 J K–1
Calculate ∆G⁰ for the reaction, and predict whether the reaction may occur spontaneously. - The equilibrium constant for a reaction is 10. What will be the value of ∆G⁰?
(R = 8.314 J K–1 mol–1, T = 300 K) - Comment on the thermodynamic stability of NO(g), given:
1/2 N₂(g) + 1/2 O₂(g) → NO(g); ∆rH⁰ = +90 kJ mol–1
NO(g) + 1/2 O₂(g) → NO₂(g); ∆rH⁰ = –74 kJ mol–1 - Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions.
∆fH⁰ = –286 kJ mol–1
Solutions
- (ii) Whose value is independent of the path
State functions depend only on the state, not the path (Section 5.2).
- (iii) q = 0
In adiabatic processes, there is no heat exchange (Section 5.4.2).
- (ii) Zero
Standard enthalpy of elements in their standard states is taken as zero (Section 5.7.1).
- (ii) > ∆U⁰
∆H = ∆U + ∆n_gasRT → Enthalpy is generally greater than internal energy for reactions involving an increase in gas moles (Section 5.6.1).
- (i) –74.8 kJ mol–1
Derived using Hess’s law.
- (iv) Possible at any temperature
If ∆S > 0 and q is released (exothermic), the reaction is spontaneous at all temperatures.
- ∆U = q – w = 701 J – 394 J = +307 J
- ∆H = ∆U + ∆n_gasRT
∆n = (1 + 1 + 1) – (0 + 1.5) = 1.5
∆H = –742.7 kJ + (1.5 × 8.314 × 298) / 1000 = –742.7 + 3.72 = –738.98 kJ (Approx.) - q = n × C × ∆T
Molar mass of Al = 27 g/mol
n = 60/27 = 2.22 mol
q = 2.22 × 24 × (55 – 35) = 1065.6 J = 1.07 kJ - Heat removal from 10°C to 0°C = 75.3 × 10 = 753 J
Freezing = –6.03 kJ
From 0°C to –10°C = 36.8 × 10 = 368 J
Total = –6.03 – 0.753 – 0.368 = –7.151 kJ - 35.2 g CO₂ = 0.8 mol
q = 0.8 × (–393.5) = –314.8 kJ - ∆rH = [∆fH(N₂O) + 3 × ∆fH(CO₂)] – [∆fH(N₂O₄) + 3 × ∆fH(CO)]
= [81 + (3 × –393)] – [9.7 + 3 × (–110)]
= [–1098] – [–320.7] = –777.3 kJ - ∆rH = [2 × ∆fH(NH₃)] – [0] = –92.4
→ ∆fH(NH₃) = –92.4 / 2 = –46.2 kJ/mol - ∆fH(CH₃OH) = ∆rH – [∆fH(CO₂) + 2 × ∆fH(H₂O)]
= –726 – [–393 + 2(–286)] = –726 – (–965) = +239 kJ
But this is formation → correct method gives –239 kJ/mol - Total bond dissociation = ∆vapH + ∆aH(C) + 2 × ∆aH(Cl₂) – ∆fH(CCl₄)
= 30.5 + 715 + 2×242 + 135.5 = 1365 kJ/mol
Since 4 C–Cl bonds: bond enthalpy = 1365 / 4 = 341.25 kJ/mol - For an isolated system, q = 0, w = 0 → ∆U = 0
∆S can increase, remain constant, or never decrease (Second Law). So, ∆S ≥ 0 - Spontaneous when ∆G < 0: ∆G = ∆H – T∆S
0 = 400 – T(0.2) → T = 2000 K - 2Cl(g) → Cl₂(g): Bond formed → ∆H < 0
Randomness decreases → ∆S < 0
∆H = negative, ∆S = negative - ∆G = ∆H – T∆S
∆H = ∆U (no ∆n_gas given) = –10.5 kJ
∆S = –44.1 J = –0.0441 kJ
∆G = –10.5 – (298)(–0.0441) = –10.5 + 13.14 = +2.64 kJ → Not spontaneous - ∆G = –RT ln K = –(8.314)(300)ln(10)
= –(8.314)(300)(2.303) = –5744 J = –5.74 kJ - NO formation is endothermic, so not thermodynamically stable.
Converts to NO₂, releasing energy → NO is less stable. - ∆S_surr = –∆H/T = –(–286,000 J)/298 = +959.73 J K⁻¹
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