Thermodynamics: CBSE Class 11 Chemistry NCERT Guide

Fundamental Concepts in Thermodynamics

System and Surroundings

The 'system' is the specific part of the universe we are studying (e.g., a chemical reaction in a beaker). Everything else outside the system is the 'surroundings'. The system and surroundings together make up the universe.

Types of Systems

1. Open System: Can exchange both energy and matter with the surroundings (e.g., an open cup of tea). 2. Closed System: Can exchange energy but not matter (e.g., a sealed bottle of water). 3. Isolated System: Cannot exchange either energy or matter (e.g., an ideal thermos flask).

State Function vs. Path Function

A State Function is a property whose value depends only on the current state of the system, not on how it got there (e.g., temperature, pressure, volume, enthalpy, entropy). A Path Function depends on the path taken between states (e.g., heat (q) and work (w)).

Intensive vs. Extensive Properties

Intensive properties are independent of the amount of substance (e.g., temperature, density). Extensive properties depend on the amount of substance (e.g., mass, volume, internal energy).

The First Law of Thermodynamics: Conservation of Energy

The First Law of Thermodynamics is a version of the universal law of conservation of energy. It states that energy can neither be created nor destroyed; it can only be converted from one form to another. For any thermodynamic system, the change in its internal energy (ΔU) is equal to the heat supplied to the system (q) plus the work done on the system (w).

The mathematical expression is:
ΔU = q + w

Let's break this down:

• ΔU (Change in Internal Energy): Internal energy (U) is the sum of all kinetic and potential energies of all particles in the system. We can't measure the absolute value of U, but we can measure the change (ΔU) when the system goes from an initial state to a final state. It's a state function.
• q (Heat): This is the energy transferred due to a temperature difference between the system and its surroundings. By convention, if heat flows into the system from the surroundings, q is positive. If heat flows out of the system, q is negative.
• w (Work): This is the energy transferred when an object is moved by a force. In chemistry, we often consider pressure-volume work. By IUPAC convention, if work is done on the system (e.g., compression of a gas), w is positive. If work is done by the system (e.g., expansion of a gas), w is negative.

Worked Examples: Calculations in Thermodynamics

• Example 1: Calculating Enthalpy of Reaction (ΔH°rxn) Calculate the standard enthalpy of combustion for methane (CH₄) given the standard enthalpies of formation (ΔH°f): CH₄(g) = -74.8 kJ/mol CO₂(g) = -393.5 kJ/mol H₂O(l) = -285.8 kJ/mol Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) Step 1: Write the formula. The enthalpy of reaction is the sum of the enthalpies of formation of products minus the sum of the enthalpies of formation of reactants, each multiplied by their stoichiometric coefficients. ΔH°rxn = Σ [ΔH°f (products)] - Σ [ΔH°f (reactants)] Step 2: Substitute the values. Remember that the standard enthalpy of formation of an element in its most stable state (like O₂(g)) is zero. ΔH°rxn = [ (1 × ΔH°f(CO₂)) + (2 × ΔH°f(H₂O)) ] - [ (1 × ΔH°f(CH₄)) + (2 × ΔH°f(O₂)) ] ΔH°rxn = [ (1 × -393.5) + (2 × -285.8) ] - [ (1 × -74.8) + (2 × 0) ] Step 3: Calculate the result. ΔH°rxn = [ -393.5 - 571.6 ] - [ -74.8 ] ΔH°rxn = -965.1 + 74.8 ΔH°rxn = -890.3 kJ/mol Final Answer: The standard enthalpy of combustion for methane is -890.3 kJ/mol.
• Example 2: Predicting Spontaneity with Gibbs Free Energy (ΔG) A reaction has an enthalpy change (ΔH) of +100 kJ/mol and an entropy change (ΔS) of +250 J/mol·K. At what temperature will the reaction become spontaneous? Step 1: Understand the condition for spontaneity. A reaction is spontaneous when the Gibbs free energy change (ΔG) is negative (ΔG < 0). The formula connecting ΔG, ΔH, and ΔS is: ΔG = ΔH - TΔS. Step 2: Set ΔG to zero to find the equilibrium temperature. The crossover point between non-spontaneous (ΔG > 0) and spontaneous (ΔG < 0) is at equilibrium, where ΔG = 0. 0 = ΔH - TΔS T = ΔH / ΔS Step 3: Convert units to be consistent. ΔH is in kJ/mol, but ΔS is in J/mol·K. Let's convert ΔH to J/mol. ΔH = 100 kJ/mol × 1000 J/kJ = 100000 J/mol Step 4: Calculate the temperature (T). T = (100000 J/mol) / (250 J/mol·K) T = 400 K Step 5: Interpret the result. At 400 K, the reaction is at equilibrium (ΔG = 0). Since both ΔH and ΔS are positive, the term -TΔS will become more negative as temperature increases. Therefore, the reaction will become spontaneous (ΔG < 0) at any temperature above 400 K. Final Answer: The reaction becomes spontaneous above 400 K.

Exam Traps & Key Tips for Thermodynamics

1. Sign Conventions are CRUCIAL: Always remember the IUPAC conventions. Work done on the system is positive (+w). Heat absorbed by the system is positive (+q). A common mistake is to mix up these signs, which will reverse your answer for ΔU. Write them down at the start of your exam!

2. Check Your Units: A frequent trap is mixing kJ and J. In the Gibbs free energy equation (ΔG = ΔH - TΔS), ΔH is usually given in kJ/mol while ΔS is in J/mol·K. Always convert one of them (usually ΔH to J or ΔS to kJ) before calculating. Forgetting this is an easy way to lose marks.

3. Spontaneity vs. Rate: A negative ΔG indicates a reaction is thermodynamically spontaneous, meaning it can happen without external energy input. It says nothing about how fast it will happen. Diamond turning into graphite is spontaneous (negative ΔG) but is so slow it's practically unobservable. Don't confuse thermodynamics with kinetics.

Practice Questions with Solutions

• Q: A gas in a cylinder absorbs 500 J of heat and expands, doing 200 J of work on the surroundings. What is the change in internal energy (ΔU) of the gas? A: Step 1: Identify the given values and their signs according to IUPAC conventions. Heat is absorbed by the system, so q = +500 J. Work is done by the system on the surroundings, so w = -200 J. Step 2: Apply the First Law of Thermodynamics formula: ΔU = q + w. Step 3: Substitute the values into the formula. ΔU = (+500 J) + (-200 J) ΔU = 300 J Final answer: The change in internal energy of the gas is +300 J.
• Q: Using Hess's Law, calculate the enthalpy of formation for acetylene (C₂H₂) from the following data: 1. C(s) + O₂(g) → CO₂(g); ΔH = -393.5 kJ 2. H₂(g) + ½O₂(g) → H₂O(l); ΔH = -285.8 kJ 3. 2C₂H₂(g) + 5O₂(g) → 4CO₂(g) + 2H₂O(l); ΔH = -2598.8 kJ Target reaction: 2C(s) + H₂(g) → C₂H₂(g) A: Step 1: Manipulate the given equations to match the target reaction. - Multiply equation (1) by 2 to get 2C on the reactant side: 2C(s) + 2O₂(g) → 2CO₂(g); ΔH = 2 × (-393.5) = -787.0 kJ - Equation (2) has H₂ on the reactant side, so keep it as is: H₂(g) + ½O₂(g) → H₂O(l); ΔH = -285.8 kJ - Reverse and divide equation (3) by 2 to get C₂H₂ on the product side: 2CO₂(g) + H₂O(l) → C₂H₂(g) + ⁵/₂O₂(g); ΔH = -(-2598.8) / 2 = +1299.4 kJ Step 2: Add the manipulated equations and their ΔH values. The intermediates (CO₂, H₂O, O₂) should cancel out. (2C + 2O₂) + (H₂ + ½O₂) + (2CO₂ + H₂O) → (2CO₂) + (H₂O) + (C₂H₂ + ⁵/₂O₂) This simplifies to: 2C(s) + H₂(g) → C₂H₂(g) Step 3: Sum the enthalpy changes. ΔH_formation = (-787.0 kJ) + (-285.8 kJ) + (+1299.4 kJ) ΔH_formation = -1072.8 kJ + 1299.4 kJ = +226.6 kJ Final answer: The enthalpy of formation for acetylene is +226.6 kJ/mol.
• Q: Predict whether the reaction 2O₃(g) → 3O₂(g) is spontaneous at 298 K. Given: ΔH = -285.4 kJ/mol and ΔS = +137 J/mol·K. A: Step 1: Write the Gibbs free energy equation: ΔG = ΔH - TΔS. Step 2: Ensure units are consistent. Convert ΔS from J to kJ. ΔS = 137 J/mol·K / 1000 J/kJ = 0.137 kJ/mol·K. Step 3: Substitute the given values into the equation. T = 298 K ΔH = -285.4 kJ/mol ΔS = 0.137 kJ/mol·K ΔG = -285.4 - (298 × 0.137) Step 4: Calculate the value of ΔG. ΔG = -285.4 - 40.826 ΔG = -326.226 kJ/mol Step 5: Interpret the sign of ΔG. Since ΔG is negative, the reaction is spontaneous. Final answer: Yes, the reaction is spontaneous at 298 K because ΔG is negative (-326.226 kJ/mol).
• Q: For the process H₂O(l) → H₂O(g) at 100°C and 1 atm pressure, what is the sign of ΔH, ΔS, and ΔG? A: Step 1: Analyze ΔH (Enthalpy). Boiling water is an endothermic process; it requires heat energy to turn liquid into gas. Therefore, ΔH is positive. Step 2: Analyze ΔS (Entropy). A gas is much more disordered than a liquid. The water molecules have more freedom of movement in the gaseous state. Therefore, entropy increases, and ΔS is positive. Step 3: Analyze ΔG (Gibbs Free Energy). The process describes water boiling at its normal boiling point (100°C and 1 atm). At the boiling point, the liquid and gas phases are in equilibrium. By definition, ΔG = 0 for a system at equilibrium. Final answer: ΔH > 0 (positive), ΔS > 0 (positive), and ΔG = 0.

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