The Heat Of Solution Formula calculates the enthalpy change when one mole of a solute dissolves completely in a solvent at constant pressure, expressed as \( q = m imes c imes \Delta T \) for calorimetry or \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \) for thermodynamic analysis. This concept is covered in NCERT Class 11 Chemistry, Chapter 6 (Thermodynamics), and is equally important for Class 12 Physical Chemistry revision. For JEE Main and NEET aspirants, heat of solution problems appear regularly in the thermodynamics and solutions sections. This article covers the formula, derivation, a complete formula sheet, three solved examples, CBSE exam tips, and FAQs.
Key Heat Of Solution Formulas at a Glance
Quick reference for the most important heat of solution formulas used in CBSE and competitive exams.
- Calorimetry heat formula: \( q = m imes c imes \Delta T \)
- Enthalpy of solution: \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \)
- Heat released/absorbed: \( q_{solution} = -q_{calorimeter} \)
- Molar enthalpy of solution: \( \Delta H_{sol} = rac{q}{n} \)
- Hess's Law application: \( \Delta H_{sol} = \sum \Delta H_{products} – \sum \Delta H_{reactants} \)
- Temperature change: \( \Delta T = T_{final} – T_{initial} \)
- Heat capacity of solution: \( q = C_{cal} imes \Delta T \)
What is the Heat Of Solution Formula?
The Heat Of Solution Formula defines the enthalpy change that occurs when one mole of a solute dissolves in a large excess of solvent at constant pressure, reaching infinite dilution. This quantity is also called the molar enthalpy of solution and is denoted by \( \Delta H_{sol} \). The unit is kilojoules per mole (kJ/mol).
When a solute dissolves in a solvent, two competing processes occur simultaneously. First, the lattice energy must be overcome to break the solute apart. Second, hydration (or solvation) energy is released as solvent molecules surround the solute ions or molecules. The net heat of solution depends on which process dominates.
If \( \Delta H_{sol} \) is negative, the dissolution is exothermic and the solution warms up. If \( \Delta H_{sol} \) is positive, the dissolution is endothermic and the solution cools down. This is covered in NCERT Class 11 Chemistry, Chapter 6 (Thermodynamics) and Chapter 2 (Solutions) in Class 12. A calorimeter is the standard instrument used to measure the heat of solution experimentally.
Heat Of Solution Formula — Expression and Variables
The two principal expressions for the heat of solution are given below.
Calorimetric Method:
\[ q = m \times c \times \Delta T \]
Thermodynamic (Born-Haber / Hess's Law) Method:
\[ \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \]
Molar Enthalpy of Solution:
\[ \Delta H_{sol} = \frac{q}{n} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( q \) | Heat absorbed or released by solution | Joules (J) or kJ |
| \( m \) | Mass of solution (solvent + solute) | grams (g) or kg |
| \( c \) | Specific heat capacity of solution | J g⁻¹ K⁻¹ |
| \( \Delta T \) | Change in temperature \( (T_{final} – T_{initial}) \) | Kelvin (K) or °C |
| \( \Delta H_{sol} \) | Molar enthalpy (heat) of solution | kJ mol⁻¹ |
| \( \Delta H_{lattice} \) | Lattice dissociation enthalpy (endothermic, positive) | kJ mol⁻¹ |
| \( \Delta H_{hydration} \) | Enthalpy of hydration (exothermic, negative) | kJ mol⁻¹ |
| \( n \) | Number of moles of solute dissolved | mol |
| \( C_{cal} \) | Heat capacity of calorimeter | J K⁻¹ |
Derivation of the Heat Of Solution Formula
The calorimetric formula derives from the first law of thermodynamics. At constant pressure, heat exchanged equals the enthalpy change. When a solute dissolves in a calorimeter, the heat released or absorbed by the dissolution process equals the heat gained or lost by the solution.
Step 1: The heat gained by the solution is \( q_{solution} = m imes c imes \Delta T \).
Step 2: By conservation of energy, \( q_{dissolution} = -q_{solution} \). The negative sign accounts for the sign convention: if the solution heats up, the process is exothermic.
Step 3: To find the molar enthalpy, divide by the number of moles of solute: \( \Delta H_{sol} = \frac{-q_{solution}}{n} \).
Step 4: Using Hess's Law, \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \), where lattice energy is endothermic (positive) and hydration energy is exothermic (negative). The algebraic sum gives the net heat of solution.
Complete Thermodynamics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat of Solution (Calorimetry) | \( q = m c \Delta T \) | m=mass, c=specific heat, ΔT=temp change | J or kJ | Class 11, Ch 6 |
| Molar Enthalpy of Solution | \( \Delta H_{sol} = q/n \) | q=heat, n=moles | kJ mol⁻¹ | Class 11, Ch 6 |
| Thermodynamic Heat of Solution | \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hyd} \) | ΔHₗₐₜₜᵢᶜₑ=lattice, ΔHₕᵧᵈ=hydration | kJ mol⁻¹ | Class 11, Ch 6 |
| First Law of Thermodynamics | \( \Delta U = q + w \) | ΔU=internal energy, q=heat, w=work | J or kJ | Class 11, Ch 6 |
| Enthalpy Change at Constant Pressure | \( \Delta H = q_p \) | ΔH=enthalpy change, qₚ=heat at const. pressure | kJ mol⁻¹ | Class 11, Ch 6 |
| Hess's Law | \( \Delta H_{rxn} = \sum \Delta H_{products} – \sum \Delta H_{reactants} \) | ΔH=enthalpy of each species | kJ mol⁻¹ | Class 11, Ch 6 |
| Heat of Combustion | \( \Delta H_c = \frac{-q}{n} \) | q=calorimeter heat, n=moles burned | kJ mol⁻¹ | Class 11, Ch 6 |
| Gibbs Free Energy | \( \Delta G = \Delta H – T \Delta S \) | ΔH=enthalpy, T=temperature, ΔS=entropy | kJ mol⁻¹ | Class 11, Ch 6 |
| Entropy Change | \( \Delta S = \frac{q_{rev}}{T} \) | qᵣₑᵥ=reversible heat, T=temperature | J K⁻¹ mol⁻¹ | Class 11, Ch 6 |
| Calorimeter Heat Capacity | \( q = C_{cal} \times \Delta T \) | Cᶜₐₗ=calorimeter heat capacity, ΔT=temp change | J K⁻¹ | Class 11, Ch 6 |
Heat Of Solution Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: When 5.0 g of NaOH dissolves in 100 g of water in a calorimeter, the temperature rises from 25.0°C to 31.5°C. Calculate the heat released by the dissolution process. (Specific heat capacity of solution = 4.18 J g⁻¹ °C⁻¹)
Given:
- Mass of solute = 5.0 g
- Mass of water = 100 g
- Total mass of solution, m = 100 + 5.0 = 105.0 g
- Tᵢₙᵢₜᵢₐₗ = 25.0°C, Tᶠᵢₙₐₗ = 31.5°C
- c = 4.18 J g⁻¹ °C⁻¹
Step 1: Find \( \Delta T \).
\( \Delta T = T_{final} – T_{initial} = 31.5 – 25.0 = 6.5 \, ^\circ C \)
Step 2: Apply the heat formula \( q = m \times c \times \Delta T \).
\( q = 105.0 \times 4.18 \times 6.5 = 2852.85 \, \text{J} \approx 2.85 \, \text{kJ} \)
Step 3: Since the temperature rose, the dissolution is exothermic. The heat released by dissolution equals \( -q_{solution} \).
\( q_{dissolution} = -2852.85 \, \text{J} = -2.85 \, \text{kJ} \)
Answer
Heat released by dissolution = 2.85 kJ (exothermic process).
Example 2 (Class 11-12 Level)
Problem: Calculate the molar enthalpy of solution of NH₄NO₃. When 4.0 g of NH₄NO₃ (molar mass = 80 g/mol) dissolves in 50 g of water, the temperature drops from 22.0°C to 18.6°C. Assume the specific heat of the solution is 4.18 J g⁻¹ °C⁻¹ and ignore the heat capacity of the calorimeter.
Given:
- Mass of NH₄NO₃ = 4.0 g, molar mass = 80 g/mol
- Mass of solution, m = 50 + 4.0 = 54.0 g
- Tᵢₙᵢₜᵢₐₗ = 22.0°C, Tᶠᵢₙₐₗ = 18.6°C
- c = 4.18 J g⁻¹ °C⁻¹
Step 1: Calculate \( \Delta T \).
\( \Delta T = 18.6 – 22.0 = -3.4 \, ^\circ C \)
Step 2: Calculate heat absorbed by the solution.
\( q_{solution} = m \times c \times \Delta T = 54.0 \times 4.18 \times (-3.4) = -767.59 \, \text{J} \)
Step 3: Heat of dissolution = \( -q_{solution} \).
\( q_{dissolution} = +767.59 \, \text{J} = +0.7676 \, \text{kJ} \)
Step 4: Calculate moles of NH₄NO₃ dissolved.
\( n = \frac{4.0}{80} = 0.05 \, \text{mol} \)
Step 5: Calculate molar enthalpy of solution.
\( \Delta H_{sol} = \frac{q_{dissolution}}{n} = \frac{+0.7676}{0.05} = +15.35 \, \text{kJ/mol} \)
Answer
Molar enthalpy of solution of NH₄NO₃ = +15.35 kJ/mol (endothermic; temperature dropped).
Example 3 (JEE/NEET Level)
Problem: The lattice dissociation enthalpy of KCl is +699 kJ/mol and the enthalpy of hydration of K⁺ and Cl⁻ ions are −322 kJ/mol and −363 kJ/mol respectively. (a) Calculate the heat of solution of KCl. (b) Predict whether the dissolution will be endothermic or exothermic. (c) If 7.45 g of KCl (molar mass = 74.5 g/mol) dissolves in 200 g of water, estimate the temperature change. (c = 4.18 J g⁻¹ °C⁻¹)
Given:
- \( \Delta H_{lattice} = +699 \, \text{kJ/mol} \)
- \( \Delta H_{hyd}(K^+) = -322 \, \text{kJ/mol} \)
- \( \Delta H_{hyd}(Cl^-) = -363 \, \text{kJ/mol} \)
- Mass of KCl = 7.45 g, molar mass = 74.5 g/mol
- Mass of water = 200 g
Step 1: Calculate total enthalpy of hydration.
\( \Delta H_{hydration} = (-322) + (-363) = -685 \, \text{kJ/mol} \)
Step 2: Apply the heat of solution formula.
\( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} = +699 + (-685) = +14 \, \text{kJ/mol} \)
Step 3: Since \( \Delta H_{sol} \) is positive, the dissolution is endothermic. The solution temperature will drop.
Step 4: Calculate moles of KCl.
\( n = \frac{7.45}{74.5} = 0.1 \, \text{mol} \)
Step 5: Calculate total heat absorbed.
\( q_{dissolution} = n \times \Delta H_{sol} = 0.1 \times 14000 = 1400 \, \text{J} \)
Step 6: Find temperature change. Total mass of solution = 200 + 7.45 = 207.45 g.
\( \Delta T = \frac{-q_{dissolution}}{m \times c} = \frac{-1400}{207.45 \times 4.18} = \frac{-1400}{867.1} \approx -1.61 \, ^\circ C \)
Answer
(a) \( \Delta H_{sol} \) of KCl = +14 kJ/mol (endothermic). (b) The dissolution is endothermic; the solution cools. (c) Temperature drops by approximately 1.61°C.
CBSE Exam Tips 2025-26
- State the sign convention clearly: In CBSE 2025-26 board exams, always mention whether \( \Delta H_{sol} \) is positive (endothermic) or negative (exothermic). Examiners award marks for this statement.
- Use both formulas: We recommend practising both the calorimetric formula \( q = mc\Delta T \) and the thermodynamic formula \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \). Questions can appear from either approach.
- Include units in every step: CBSE marking schemes deduct marks for missing units. Always write kJ/mol for molar enthalpy of solution.
- Remember the negative sign: \( q_{dissolution} = -q_{solution} \). This sign reversal is a common source of errors in board exams. Write it explicitly in your solution.
- Relate to real-life examples: CBSE 2025-26 papers often include application-based questions. Dissolution of NH₄NO₃ (endothermic, used in cold packs) and NaOH (exothermic) are favourite examples.
- Hess's Law questions: In Class 11 exams, Hess's Law-based calculations of \( \Delta H_{sol} \) carry 3-5 marks. Our experts suggest practising at least five such problems before the board exam.
Common Mistakes to Avoid
- Forgetting to include the solute mass: Many students use only the mass of the solvent in \( q = mc\Delta T \). Always use the total mass of the solution (solvent + solute) unless the question specifies otherwise.
- Ignoring the sign of \( \Delta T \): If the temperature drops, \( \Delta T \) is negative. Substituting a positive value here gives the wrong sign for \( q \) and leads to an incorrect conclusion about exothermic or endothermic nature.
- Confusing lattice energy and lattice enthalpy: Lattice dissociation enthalpy is always positive (endothermic) because energy is required to break the ionic lattice. Hydration enthalpy is always negative (exothermic). Mixing up the signs is a very common JEE error.
- Not converting units: The formula \( q = mc\Delta T \) gives the answer in Joules. Always convert to kJ (divide by 1000) before calculating \( \Delta H_{sol} \) in kJ/mol. Leaving the answer in J/mol is incorrect.
- Using wrong specific heat: The specific heat of water is 4.18 J g⁻¹ K⁻¹. Some students use 4.2 or 1.0 (specific heat of ice). Use the value given in the question; if not given, use 4.18 J g⁻¹ K⁻¹ for dilute aqueous solutions.
JEE/NEET Application of Heat Of Solution Formula
In our experience, JEE aspirants encounter the Heat Of Solution Formula in two distinct contexts: calorimetry-based numerical problems and Hess's Law / Born-Haber cycle questions. NEET questions tend to focus more on the conceptual aspect, asking students to predict whether a dissolution is endothermic or exothermic based on given lattice and hydration enthalpies.
Application Pattern 1 — Calorimetry Numericals: JEE Main frequently provides calorimeter data (mass, temperature change, specific heat) and asks for the molar enthalpy of solution. The key skill is correctly applying \( q_{dissolution} = -q_{solution} \) and then dividing by moles. These questions carry 4 marks each in JEE Main 2025.
Application Pattern 2 — Born-Haber Cycle: JEE Advanced tests the thermodynamic formula \( \Delta H_{sol} = \Delta H_{lattice} + \Delta H_{hydration} \) in multi-step Hess's Law problems. Students must correctly assign signs to each step. Lattice dissociation is endothermic; hydration is exothermic. The net sign of \( \Delta H_{sol} \) determines solubility trends, which is also tested in JEE Advanced physical chemistry.
Application Pattern 3 — NEET Conceptual Questions: NEET often asks: “Which of the following dissolves with a decrease in temperature?” The correct answer is always the salt with a positive (endothermic) heat of solution. Examples include KNO₃, NH₄Cl, and NH₄NO₃. Salts with negative (exothermic) heat of solution, like NaOH and H₂SO₄, cause temperature rise. Memorising a short list of common examples is highly effective for NEET preparation.
In our experience, students who master the sign conventions and unit conversions in the Heat Of Solution Formula score full marks on these questions in both JEE and NEET.
FAQs on Heat Of Solution Formula
For more related formulas and study resources, explore our complete Physics Formulas hub on ncertbooks.net. You may also find the Electric Current Formula and the de Broglie Wavelength Formula useful for your JEE and NEET preparation. For official NCERT syllabus details, visit the NCERT official website.