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આ MCQ મોડ્યુલ આના પર આધારિત છે: Thermochemistry
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Thermochemistry
🎓 Class 11
Chemistry
CBSE
Theory
Ch 5 – Thermodynamics
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Thermochemistry
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Thermochemistry: Heat Capacity, Calorimetry and Hess's Law
5.10 Heat Capacity
Different substances need different amounts of heat to raise their temperature by the same amount. The heat capacity C is defined by:
\[q = C\,\Delta T\]| Quantity | Symbol | Definition | Units |
|---|---|---|---|
| Heat capacity | C | q / ΔT for a body | J K⁻¹ |
| Specific heat capacity | c (or s) | q / (m · ΔT) | J g⁻¹ K⁻¹ |
| Molar heat capacity | C_m | q / (n · ΔT) | J K⁻¹ mol⁻¹ |
5.10.1 C_p and C_v for Gases
For gases, two values are defined depending on what is held constant:
- C_v = molar heat capacity at constant volume → q_v = nC_v ΔT = ΔU
- C_p = molar heat capacity at constant pressure → q_p = nC_p ΔT = ΔH
Mayer's Relation (ideal gas):
\[C_p - C_v = R\]
At constant pressure, extra energy is needed to do PV-work as the gas expands; hence C_p > C_v. For monatomic ideal gas: C_v = (3/2)R, C_p = (5/2)R. For diatomic: C_v = (5/2)R, C_p = (7/2)R.
| Substance | Specific heat (J g⁻¹ K⁻¹) | Molar heat (J K⁻¹ mol⁻¹) |
|---|---|---|
| Water (liquid) | 4.18 | 75.3 |
| Aluminium | 0.90 | 24.3 |
| Iron | 0.45 | 25.1 |
| Copper | 0.39 | 24.5 |
| Lead | 0.13 | 26.4 |
| Air (1 atm, 298 K) | 1.01 | 29.1 |
5.11 Calorimetry — Measuring Heat Changes
Calorimetry measures heat changes in a device called a calorimeter. Two main types:
5.11.1 Coffee-Cup Calorimeter (Constant P → ΔH)
A simple polystyrene cup with a thermometer and stirrer. Reaction occurs at atmospheric pressure. Heat absorbed/released = q_p = ΔH.
\[q = m \cdot c \cdot \Delta T \quad \text{(for water)}\]5.11.2 Bomb Calorimeter (Constant V → ΔU)
A sealed steel "bomb" submerged in a water bath, used for combustion reactions. Constant volume → q_v = ΔU. Then ΔH = ΔU + Δn_g RT.
5.12 Standard Enthalpy of Reaction (Δ_rH°)
The standard state of a substance is its pure form at 1 bar pressure (and a stated temperature, usually 298 K). Standard enthalpies are denoted with a superscript ° (or ⊖).
5.12.1 Standard Enthalpy of Formation, Δ_fH°
Definition: The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states.
Convention: Δ_fH° of any element in its most stable form = 0 (e.g., O₂(g), C(graphite), H₂(g)).
Convention: Δ_fH° of any element in its most stable form = 0 (e.g., O₂(g), C(graphite), H₂(g)).
| Substance | Δ_fH° (kJ mol⁻¹) | Substance | Δ_fH° (kJ mol⁻¹) |
|---|---|---|---|
| H₂O(l) | −285.8 | CO₂(g) | −393.5 |
| H₂O(g) | −241.8 | CO(g) | −110.5 |
| NH₃(g) | −46.1 | CH₄(g) | −74.8 |
| HCl(g) | −92.3 | C₂H₅OH(l) | −277.7 |
| NO(g) | +90.3 | C₆H₆(l) | +49.0 |
For any reaction:
\[\Delta_r H^\circ = \sum n_p\,\Delta_f H^\circ(\text{products}) - \sum n_r\,\Delta_f H^\circ(\text{reactants})\]5.12.2 Other Standard Enthalpies
| Type | Definition | Sign |
|---|---|---|
| Δ_fusH° | 1 mol solid → 1 mol liquid (at melting point) | + (endo) |
| Δ_vapH° | 1 mol liquid → 1 mol vapour (at boiling point) | + (endo) |
| Δ_subH° | 1 mol solid → 1 mol vapour (sublimation) | + (endo) |
| Δ_cH° | 1 mol substance burnt completely in O₂ | − (exo) |
| Δ_neutH° | 1 mol H⁺ + 1 mol OH⁻ → 1 mol H₂O | −57.1 kJ (strong-strong) |
| Δ_aH° | 1 mol gaseous atoms from element in standard state | + (endo) |
| Δ_bondH° | 1 mol gaseous bonds broken (homolytic) | + (endo) |
| Δ_solH° | 1 mol substance dissolved (infinite dilution) | varies |
| Δ_latticeH° | 1 mol ionic solid → gaseous ions | + (endo) |
5.13 Hess's Law of Constant Heat Summation
Hess's Law: The standard enthalpy change of a reaction is the same whether the reaction is carried out in one step or in several steps. Corollary: Thermochemical equations can be added, subtracted and multiplied algebraically along with their ΔH values.
5.14 Bond Enthalpies
Mean bond enthalpies let us estimate Δ_rH° from bonds broken/formed:
\[\Delta_r H \approx \sum \Delta_\text{bond}H(\text{bonds broken}) - \sum \Delta_\text{bond}H(\text{bonds formed})\]| Bond | Bond enthalpy (kJ mol⁻¹) | Bond | Bond enthalpy (kJ mol⁻¹) |
|---|---|---|---|
| H–H | 436 | C=C | 614 |
| O=O | 498 | C≡C | 839 |
| N≡N | 946 | O–H | 463 |
| Cl–Cl | 242 | C–H | 414 |
| H–Cl | 431 | C=O | 740 |
🎯 Interactive: Hess's-Law Calculator
Suppose you have three thermochemical equations with known ΔH values. Pick how to combine them and compute the target ΔH.
Target ΔH = c₁·ΔH₁ + c₂·ΔH₂ + c₃·ΔH₃ = —
🧪 Activity 5.3 — Heat of Neutralization (kitchen-version)
Setup: In two foam cups, take 50 mL of 1 M HCl and 50 mL of 1 M NaOH. Note initial T. Mix them, stir, and record final T.
Predict: What temperature change ΔT do you expect? Calculate the molar enthalpy of neutralization assuming the heat capacity is the same as water.
Typical observation: ΔT ≈ +6.8 °C (rises from 25 °C to ~31.8 °C).
Heat absorbed by solution: q = m·c·ΔT = (100 g)(4.18 J g⁻¹ K⁻¹)(6.8) = 2842 J ≈ 2.84 kJ.
Moles of H⁺ = 0.050 mol. Δ_neutH = −2.84/0.050 = −56.8 kJ/mol ≈ −57 kJ/mol (NCERT value).
Why the same value for any strong acid + strong base? The reaction is just H⁺(aq) + OH⁻(aq) → H₂O(l). The acid and base anions/cations stay as spectators!
Worked Example 5.5: Δ_fH from Hess's Law
Calculate Δ_fH° of CH₄(g) given:
(i) C(s) + O₂(g) → CO₂(g); ΔH₁ = −393.5 kJ
(ii) H₂(g) + ½O₂(g) → H₂O(l); ΔH₂ = −285.8 kJ
(iii) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH₃ = −890.4 kJ
Target: C(s) + 2H₂(g) → CH₄(g) (Δ_fH° = ?)
Multiply (ii) by 2: 2H₂ + O₂ → 2H₂O(l); 2ΔH₂ = −571.6 kJ
Add (i) + 2(ii): C + 2H₂ + 2O₂ → CO₂ + 2H₂O; ΔH = −393.5 + (−571.6) = −965.1 kJ
Reverse (iii): CO₂ + 2H₂O → CH₄ + 2O₂; −ΔH₃ = +890.4 kJ
Sum: C + 2H₂ → CH₄; Δ_fH°(CH₄) = −965.1 + 890.4 = −74.7 kJ/mol. ✓ matches NCERT value.
Multiply (ii) by 2: 2H₂ + O₂ → 2H₂O(l); 2ΔH₂ = −571.6 kJ
Add (i) + 2(ii): C + 2H₂ + 2O₂ → CO₂ + 2H₂O; ΔH = −393.5 + (−571.6) = −965.1 kJ
Reverse (iii): CO₂ + 2H₂O → CH₄ + 2O₂; −ΔH₃ = +890.4 kJ
Sum: C + 2H₂ → CH₄; Δ_fH°(CH₄) = −965.1 + 890.4 = −74.7 kJ/mol. ✓ matches NCERT value.
Worked Example 5.6: Bond Enthalpy Estimation
Estimate the enthalpy of formation of HCl(g) from H₂(g) + Cl₂(g) → 2HCl(g) using bond enthalpies: H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ/mol.
Bonds broken: 1×(H–H) + 1×(Cl–Cl) = 436 + 242 = 678 kJ (absorbed)
Bonds formed: 2×(H–Cl) = 2×431 = 862 kJ (released)
Δ_rH = 678 − 862 = −184 kJ (for 2 mol HCl)
Per mole HCl: Δ_fH°(HCl) = −92 kJ/mol. ✓ Matches the NCERT value of −92.3 kJ/mol.
Bonds formed: 2×(H–Cl) = 2×431 = 862 kJ (released)
Δ_rH = 678 − 862 = −184 kJ (for 2 mol HCl)
Per mole HCl: Δ_fH°(HCl) = −92 kJ/mol. ✓ Matches the NCERT value of −92.3 kJ/mol.
🎯 Competency-Based Questions
Q1. The standard enthalpy of formation of any element in its most stable form is: L1 Remember
Answer: (c) Zero. By convention, Δ_fH°(O₂, g) = 0, Δ_fH°(C, graphite) = 0, etc.
Q2. Which is more useful for measuring ΔH directly: bomb calorimeter or coffee-cup calorimeter? Explain. L2 Understand
Answer: Coffee-cup calorimeter — operates at constant atmospheric pressure, so q_p = ΔH directly. Bomb calorimeter is at constant V, giving q_v = ΔU; ΔH must be calculated using ΔH = ΔU + Δn_g RT.
Q3. The molar heat of fusion of ice is 6.0 kJ/mol. How much heat is required to melt 36 g of ice at 0 °C? L3 Apply
Answer: Moles = 36/18 = 2 mol. Heat required = 2 × 6.0 = 12 kJ.
Q4. Strong acid + strong base neutralization gives ΔH ≈ −57 kJ/mol regardless of identity. Why? L4 Analyse
Answer: Strong acids and strong bases dissociate completely in water — the actual reaction is simply H⁺(aq) + OH⁻(aq) → H₂O(l). The cations and anions are spectator ions. So the same enthalpy is released whether it is HCl + NaOH, HNO₃ + KOH, or HBr + NaOH.
For weak acid/base, |ΔH| is less than 57 kJ because some heat goes into ionizing the weak species first.
For weak acid/base, |ΔH| is less than 57 kJ because some heat goes into ionizing the weak species first.
Q5. HOT (Apply): Δ_fH°(CO₂, g) = −393.5 kJ/mol; Δ_fH°(H₂O, l) = −285.8 kJ/mol; Δ_fH°(C₂H₆, g) = −84.7 kJ/mol. Calculate the standard enthalpy of combustion of ethane: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l). L3 Apply
Answer: Δ_rH° = [4(−393.5) + 6(−285.8)] − [2(−84.7) + 7(0)]
= [−1574 + (−1714.8)] − [−169.4]
= −3288.8 + 169.4 = −3119.4 kJ (for 2 mol C₂H₆)
Per mole ethane: Δ_cH° = −1559.7 kJ/mol.
= [−1574 + (−1714.8)] − [−169.4]
= −3288.8 + 169.4 = −3119.4 kJ (for 2 mol C₂H₆)
Per mole ethane: Δ_cH° = −1559.7 kJ/mol.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Hess's law is a corollary of the first law of thermodynamics.
R: Enthalpy is a state function and depends only on initial and final states.
Answer: (A). Both true; R explains A. The path-independence of state functions guarantees that the total ΔH is the same whichever route the reaction follows.
A: C_p is greater than C_v for an ideal gas.
R: At constant pressure, part of the absorbed heat is used to do PV work as the gas expands.
Answer: (A). Both true; R explains A. Mayer's relation: C_p − C_v = R for one mole of an ideal gas.
A: The enthalpy of neutralization of HCl by NaOH is the same as that of HNO₃ by KOH.
R: All strong acid–strong base reactions involve the same net ionic equation.
Answer: (A). Both true; R explains A. The net reaction in every case is H⁺(aq) + OH⁻(aq) → H₂O(l), Δ_neutH ≈ −57 kJ/mol.
Frequently Asked Questions — Thermochemistry: Heat Capacity, Calorimetry and Hess's Law
What is calorimetry?
Calorimetry is the experimental measurement of heat exchanged during a physical or chemical process. It uses a device called a calorimeter, which is well-insulated to prevent heat loss to the surroundings. In NCERT Class 11 Chemistry Chapter 5, two main types are described: (1) the bomb calorimeter measures heat at constant volume (ΔU) for combustion reactions; (2) the coffee-cup calorimeter measures heat at constant pressure (ΔH) for solution-phase reactions like neutralisation. The basic equation is q = m × c × ΔT, where m is mass, c is specific heat capacity and ΔT is temperature change.
What is the standard enthalpy of formation?
The standard enthalpy of formation (Δ_f H°) is the enthalpy change when one mole of a substance is formed from its constituent elements in their standard states (most stable form at 1 bar and 298 K). For elements in their standard states, Δ_f H° = 0 by definition. NCERT Class 11 Chemistry Chapter 5 provides a table of Δ_f H° values used to calculate the standard enthalpy of any reaction: ΔH°_reaction = Σ Δ_f H°(products) − Σ Δ_f H°(reactants). Examples: Δ_f H°(H₂O, l) = −285.8 kJ/mol; Δ_f H°(CO₂, g) = −393.5 kJ/mol.
How is the enthalpy of combustion measured?
The standard enthalpy of combustion (Δ_c H°) is the enthalpy change when one mole of a substance undergoes complete combustion in oxygen at standard conditions. It is measured experimentally using a bomb calorimeter where the substance is burned in excess O₂ at constant volume. The heat released raises the temperature of the calorimeter and the water surrounding it. From q = C_cal × ΔT, ΔU is calculated and then ΔH = ΔU + Δn_g RT. NCERT Class 11 Chemistry Chapter 5 examples: combustion of glucose (Δ_c H° = −2802 kJ/mol), methane (Δ_c H° = −890.4 kJ/mol).
What is Hess's law of constant heat summation?
Hess's law of constant heat summation states that the enthalpy change of an overall chemical reaction is the sum of the enthalpy changes of the individual steps, regardless of the actual pathway taken. This is a direct consequence of enthalpy being a state function. NCERT Class 11 Chemistry Chapter 5 uses Hess's law to calculate enthalpies of reactions that cannot be measured directly. Example: ΔH for C(s) + ½O₂(g) → CO(g) is computed using the ΔH values of C + O₂ → CO₂ and CO + ½O₂ → CO₂. Hess's law also underlies the Born-Haber cycle for lattice energies.
What is bond enthalpy and how is it used?
Bond enthalpy (or bond dissociation enthalpy) is the average energy required to break one mole of a particular type of bond in the gaseous state. For diatomic molecules like H₂, ΔH°(H−H) = 435.0 kJ/mol exactly. For polyatomic molecules, average bond enthalpies are used (e.g., C−H = 414 kJ/mol). NCERT Class 11 Chemistry Chapter 5 uses bond enthalpies to estimate enthalpies of reactions: ΔH_reaction ≈ Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed). This method gives approximate values useful when Δ_f H° data is unavailable.
What are the standard states used in thermochemistry?
Standard states in NCERT Class 11 Chemistry Chapter 5 thermochemistry are defined at 1 bar pressure and a specified temperature (usually 298.15 K = 25°C). For a pure gas, the standard state is the pure gas at 1 bar behaving ideally. For a pure liquid or solid, the standard state is the most stable form at 1 bar. For a solute in solution, the standard state is the hypothetical 1 mol/L solution behaving ideally. Standard enthalpies (Δ_f H°, Δ_c H°, etc.) are tabulated for these reference conditions. The superscript ° denotes standard state.
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