QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 5.2NCERT Class 11 · Chemistry · Chapter 5

Enthalpy, Heat Capacity & Calorimetry

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 8 min read

Most chemical laboratory experiments and industrial reactions occur in open vessels under constant atmospheric pressure. Under constant pressure, heat exchanged is not equal to the change in internal energy (ΔU). To quantify constant-pressure thermal changes, thermodynamics introduces the state function Enthalpy (H).

1. Enthalpy (H) & Constant Pressure Heat (qp)

Enthalpy is defined as the total heat content of a system, mathematically given by:

H = U + PV

Since internal energy U, pressure P, and volume V are state functions, Enthalpy (H) is also an extensive state function.

For a process occurring at constant pressure (P = constant):

ΔH = ΔU + Δ(PV) = ΔU + PΔV

From the First Law (ΔU = qp − PΔV), substituting ΔU + PΔV gives:

ΔH = qp   (Heat exchanged at constant pressure)

Similarly, at constant volume (ΔV = 0, w = 0):

ΔU = qv   (Heat exchanged at constant volume)

2. Fundamental Relation: ΔH = ΔU + ΔngRT

For chemical reactions involving gaseous species, the volume change is primarily governed by the change in the number of gaseous moles. For an ideal gas, PV = nRT. At constant temperature T:

Δ(PV) = Δ(ngRT) = ΔngRT

ΔH = ΔU + ΔngRT

Where Δng = ∑ ng(products) − ∑ ng(reactants) (strictly count ONLY species in the gaseous state (g); ignore solids (s) and liquids (l)).

Condition Physical Meaning Example Reaction
Δng = 0 ΔH = ΔU (No expansion work) H2(g) + I2(g) → 2HI(g) (Δng = 2 − 2 = 0)
Δng > 0 ΔH > ΔU (System expands, work done by system) PCl5(g) → PCl3(g) + Cl2(g) (Δng = 2 − 1 = +1)
Δng < 0 ΔH < ΔU (System contracts, work done on system) N2(g) + 3H2(g) → 2NH3(g) (Δng = 2 − 4 = −2)

3. Heat Capacity & Mayer's Relation (Cp − Cv = R)

The Heat Capacity (C) of a system is the quantity of heat required to raise its temperature by 1 Kelvin (or 1 °C):

C = dq / dT   (in J K−1)
  • Specific Heat Capacity (c): Heat capacity per unit mass (J g−1 K−1 or J kg−1 K−1).
  • Molar Heat Capacity (Cm): Heat capacity per mole (J mol−1 K−1).

Heat Capacity at Constant Volume (Cv) & Constant Pressure (Cp)

At constant volume: qv = ΔU ⇒ Cv = (∂U / ∂T)v

At constant pressure: qp = ΔH ⇒ Cp = (∂H / ∂T)p

Derivation of Mayer's Relation (Cp − Cv = R) for Ideal Gas

From enthalpy definition for 1 mole of ideal gas: H = U + PV = U + RT.

Differentiating with respect to temperature T:

dH/dT = dU/dT + R

Cp − Cv = R   (Mayer's Equation)

The ratio γ = Cp / Cv is the Poisson ratio (adiabatic exponent):

  • Monatomic gas (He, Ne, Ar): Cv = (3/2)R, Cp = (5/2)R ⇒ γ = 5/3 ≈ 1.67
  • Diatomic gas (N2, O2, CO): Cv = (5/2)R, Cp = (7/2)R ⇒ γ = 7/5 = 1.40
  • Non-linear triatomic gas (H2O, SO2): Cv = 3R, Cp = 4R ⇒ γ = 4/3 ≈ 1.33

Calorimetry Apparatus: Bomb Calorimeter vs Coffee-Cup Calorimeter

Schematic comparison of constant volume (ΔU) vs constant pressure (ΔH) calorimeters
BOMB CALORIMETER (Constant Volume, ΔV = 0)     COFFEE-CUP CALORIMETER (Constant Pressure)
+------------------------------------------+    +------------------------------------------+
|  [Thermometer]    [Motor Stirrer]        |    |       [Thermometer]     [Stirrer]        |
|        |                |                |    |             |               |            |
|     +--|----------------|--+             |    |          +--|---------------|--+         |
|     |  v                v  | (Water Bath)|    |          |  v               v  |         |
|     |  ..................  |             |    |     === Styrofoam Insulated Cup ===      |
|     |  . +------------+ .  |             |    |     |                           |        |
|     |  . | Heavy Steel| .  |             |    |     |   Reacting Aqueous Soln   |        |
|     |  . | Bomb Vessel| .  |             |    |     |   q_rxn = -q_calorimeter  |        |
|     |  . | (Ignition) | .  |             |    |     |   q_rxn = -(m * c * ΔT)   |        |
|     |  . +------------+ .  |             |    |     +---------------------------+        |
|     +----------------------+             |    +------------------------------------------+
|  q_v = C_cal * ΔT = ΔU                   |    q_p = (m_sol * c_sol + C_cal) * ΔT = ΔH    |
+------------------------------------------+    +------------------------------------------+
Calorimeter Principle: In the bomb calorimeter, the sample is burned in pure oxygen inside a rigid sealed steel container. Since ΔV = 0, no work is performed (w = 0), and heat measured gives ΔU (qv). In the Styrofoam coffee-cup calorimeter, the cup is open to atmospheric pressure, so heat released measures ΔH (qp) directly.

4. Calorimetric Formulae for Bomb & Solution Calorimetry

Bomb Calorimeter: qv = −Ccal × ΔT
Molar ΔU = (qv / msubstance) × Mmolar
Then calculate ΔH using: ΔH = ΔU + ΔngRT

5. Solved High-Yield Numerical Examples & Exam MCQs

Example 1 (JEE Advanced): The combustion of 1.00 mole of benzene, C6H6(l), at 298 K in a bomb calorimeter produces −3267.0 kJ of heat. Calculate the standard enthalpy of combustion (ΔcH°) of benzene at 298 K. (R = 8.314 J K−1 mol−1)
Step-by-step Solution:
1. Write the balanced combustion equation:
C6H6(l) + 7.5 O2(g) → 6 CO2(g) + 3 H2O(l)
2. Calculate Δng (gaseous species only):
Δng = ng(products) − ng(reactants) = 6 − 7.5 = −1.5 mol
3. Identify ΔU: Bomb calorimeter measures constant volume heat ⇒ ΔU = −3267.0 kJ mol−1
4. Apply ΔH = ΔU + ΔngRT:
ΔngRT = (−1.5 mol) × (8.314 × 10−3 kJ K−1 mol−1) × (298 K) = −3.716 kJ
ΔcH° = −3267.0 + (−3.716) = −3270.72 kJ mol−1.
Notice that ΔH is slightly more negative than ΔU because work is done on the system by the atmosphere as volume contracts (Δng < 0).
Practice MCQ 1 (NEET): For the reaction: N2(g) + 3H2(g) → 2NH3(g), which of the following relations is correct?
(A) ΔH = ΔU − 2RT
(B) ΔH = ΔU + 2RT
(C) ΔH = ΔU − RT
(D) ΔH = ΔU
Show answer
Correct Answer: (A) ΔH = ΔU − 2RT
Explanation:
Δng = moles of gaseous products − moles of gaseous reactants = 2 − (1 + 3) = −2.
Substituting into ΔH = ΔU + ΔngRT gives ΔH = ΔU + (−2)RT = ΔU − 2RT.
Practice MCQ 2 (JEE Main): For an ideal monoatomic gas, the difference between molar heat capacity at constant pressure and molar heat capacity at constant volume (Cp,m − Cv,m) is equal to:
(A) 2.5 R
(B) 1.5 R
(C) R
(D) 0.67 R
Show answer
Correct Answer: (C) R
Explanation:
By Mayer's relation, for any ideal gas regardless of atomicity, Cp,m − Cv,m = R. For monoatomic gas, Cv = 1.5 R and Cp = 2.5 R, whose difference is exactly R.

6. Frequently Asked Questions (FAQs)

1. Why is enthalpy change (ΔH) preferred over internal energy change (ΔU) in chemical laboratories?
Most practical chemistry experiments are carried out in test tubes, beakers, or flasks open to the atmosphere at constant pressure (~1 atm). Because constant pressure heat equals enthalpy change (qp = ΔH), enthalpy directly tracks the measured thermal energy without needing rigid pressure-resistant vessels.
2. What happens to the temperature of a gas when it undergoes adiabatic expansion?
In an adiabatic process, q = 0. Therefore ΔU = w. In an expansion, the gas does work (w < 0), leading to a reduction in its internal energy (ΔU < 0). Because ΔU = nCvΔT, the gas temperature drops, producing cooling.
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