Thermodynamics — Class 11 Chemistry NCERT Solutions (Free)

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TL;DR: Free step-by-step NCERT solutions for Class 11 Chemistry chapter "Thermodynamics" — 4 important questions with detailed answers for CBSE board exam pr…

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NCERT Solutions for Class 11 Chemistry — Thermodynamics. Step-by-step answers to all 4 textbook questions from this chapter, written for CBSE board preparation and free to use.

Thermodynamics — All 4 Questions Solved

Q1. Define system, surroundings, and universe in thermodynamics. Explain the types of thermodynamic systems with examples.

Basic Thermodynamic Concepts:

System:
Definition: The part of the universe that we focus on or study.
Examples:
→ A beaker containing water and salt dissolving
→ A reaction vessel in a laboratory
→ A battery
→ A combustion engine

Surroundings:
Definition: Everything else in the universe except the system; the environment outside the system.
Examples:
→ Air around a beaker
→ The walls of a reaction vessel
→ Heat dissipating to the environment

Universe:
Definition: The combination of the system and surroundings.
Formula: Universe = System + Surroundings

Types of Thermodynamic Systems:

1. OPEN SYSTEM
Definition: A system that can exchange both matter and energy (heat, work) with the surroundings.

Characteristics:
→ Mass can enter or leave the system
→ Energy can enter or leave the system
→ System is not isolated from surroundings

Examples:
• A boiling pot of water (steam escapes, heat enters from stove)
• Human body (takes in food and oxygen, releases heat and CO₂)
• A cup of tea (exchanges heat with air, water evaporates)
• A burning candle (takes oxygen, releases CO₂ and heat)
• An open container with a chemical reaction (gases can escape)
• A growing plant (takes in water, CO₂, light; releases O₂)

2. CLOSED SYSTEM
Definition: A system that can exchange energy (heat, work) with the surroundings, but NOT matter.

Characteristics:
→ No mass can enter or leave (matter is conserved)
→ Energy can enter or leave (heat and work are exchanged)
→ System is partially isolated (matter, not energy)

Examples:
• A sealed container with liquid in equilibrium with its vapor (energy exchange, no mass loss)
• A sealed syringe with gas (can compress/expand, exchange energy, but gas stays inside)
• A closed car door (heat can enter/leave, but matter inside stays)
• A sealed flask with reacting chemicals (reaction occurs, energy released, but substances stay inside)
• A closed battery (chemical reaction, energy released as electricity, but no matter lost)
• A sealed jar with plants (gas exchange within system, but no external matter enters/leaves)

3. ISOLATED SYSTEM
Definition: A system that cannot exchange either matter or energy with the surroundings.

Characteristics:
→ No mass can enter or leave
→ No energy can enter or leave (no heat, no work)
→ Completely isolated from environment
→ Theoretical concept (difficult to achieve in practice)

Examples:
• Universe (by definition, nothing outside it)
• An insulated, sealed container (in theory; in practice, some heat transfer occurs)
• Thermally insulated flask with a reaction (ideal case)
• Adiabatic system with rigid, impermeable walls
• A perfectly insulated chamber (theoretical)

Note: True isolated systems are idealized; in practice, some energy transfer always occurs.

Comparison Table:

Property Open System Closed System Isolated System
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Matter exchange Yes (in/out) No No
Energy exchange Yes (in/out) Yes (in/out) No
Mass of system Changes Constant Constant
Common in nature Very common Common Rare (ideal)
Examples Boiling water Sealed flask Universe (approx.)
Living organisms Sealed syringe Perfect thermos
Burning candle Car interior Adiabatic container

Energy Exchange Mechanisms:

Heat (Q):
→ Transfer of thermal energy due to temperature difference
→ Positive Q: heat enters system (endothermic)
→ Negative Q: heat leaves system (exothermic)
→ Symbol: Q (units: J, kJ)

Work (W):
→ Transfer of energy due to mechanical force
→ Positive W: work done on system (compression)
→ Negative W: work done by system (expansion)
→ Symbol: W (units: J, kJ)
→ Formula: W = -P × ΔV (for constant pressure)

Internal Energy (U or E):
→ Total energy of particles in the system
→ Includes kinetic + potential energy of molecules
→ Change in internal energy: ΔU = Q - W (First Law)

Significance of System Classification:

→ Open systems: Most real-world phenomena (cooking, breathing, combustion)
→ Closed systems: Laboratory experiments, sealed containers
→ Isolated systems: Theoretical reference for calculating maximum efficiency
→ Understanding system type is crucial for applying thermodynamic laws correctly

Conclusion: Systems are classified by their ability to exchange matter and energy with surroundings. Most real systems are open, while closed and isolated systems are useful conceptual models in thermodynamics.

Q2. State the first law of thermodynamics. Explain its significance and application with a numerical example.

First Law of Thermodynamics Statement:
Energy can neither be created nor destroyed; it can only be converted from one form to another.

Alternative Statement:
The increase in internal energy of a system equals the heat added to the system minus the work done by the system.

Mathematical Expression:
ΔU = Q - W

Where:
ΔU = Change in internal energy of the system (J or kJ)
Q = Heat absorbed by the system (J or kJ)
W = Work done by the system (J or kJ)

Alternative forms:
ΔU = Q + W' (where W' is work done ON the system, opposite sign convention)
Q = ΔU + W (rearranged form)

Sign Convention (Important):

For ΔU (Change in internal energy):
→ ΔU > 0: Internal energy increases (positive change)
→ ΔU < 0: Internal energy decreases (negative change)

For Q (Heat):
→ Q > 0: Heat is absorbed by system (endothermic, heat flows in)
→ Q < 0: Heat is released by system (exothermic, heat flows out)

For W (Work done by system):
→ W > 0: System does work on surroundings (expansion)
→ W < 0: Work is done on system (compression)

For constant pressure processes:
W = -P × ΔV = -P(Vfinal - Vinitial)
Or: W = -P(V₂ - V₁)

Note: W = -PΔV because when volume increases (ΔV > 0), the system does work (W is negative in our convention)

Significance of First Law:

1. Conservation Principle:
→ Energy is conserved in all processes
→ Total energy of an isolated system remains constant
→ Energy cannot disappear; it is transformed

2. Connects Microscopic and Macroscopic:
→ Links internal energy (microscopic motion of particles) with heat and work (macroscopic observations)
→ Bridges molecular scale and observable phenomena

3. Foundation of Thermodynamics:
→ First principle upon which all other thermodynamic laws are built
→ Allows calculation of energy changes in chemical and physical processes

4. Predictive Power:
→ Given any two of (ΔU, Q, W), the third can be calculated
→ Enables prediction of energy transformations

5. Applications:
→ Chemical reactions (heat released or absorbed)
→ Phase transitions (melting, boiling, sublimation)
→ Expansion/compression of gases
→ Mechanical work and heat

Numerical Example 1: Combustion Reaction

Problem:
During combustion of 1 mole of methane (CH₄) in a rigid container (constant volume):
→ Heat released (Q) = -890 kJ (negative because heat is released)
→ No volume change (ΔV = 0), so no work is done by or on the gas (W = 0)

Calculate the change in internal energy.

Solution:
Given:
Q = -890 kJ (heat released, exothermic)
W = 0 (rigid container, constant volume, ΔV = 0)

Using First Law:
ΔU = Q - W
ΔU = (-890) - 0
ΔU = -890 kJ

Interpretation:
→ Internal energy decreases by 890 kJ
→ All the heat released goes into decreasing the internal energy
→ No work is done because volume is constant
→ The chemical energy stored in bonds is converted to heat

Numerical Example 2: Expansion of Gas at Constant Pressure

Problem:
A gas expands at constant pressure (1 atm) from 2 L to 5 L. Heat absorbed by the gas is 30 kJ. Calculate the change in internal energy.

Given:
Initial volume (V₁) = 2 L
Final volume (V₂) = 5 L
Pressure (P) = 1 atm = 101,325 Pa
Heat absorbed (Q) = 30 kJ = 30,000 J

Step 1: Calculate change in volume.
ΔV = V₂ - V₁ = 5 - 2 = 3 L = 3 × 10⁻³ m³ = 0.003 m³

Step 2: Calculate work done by system.
W = -P × ΔV
W = -(101,325 Pa) × (0.003 m³)
W = -303.975 J ≈ -304 J ≈ -0.304 kJ

Note: Gas expands (ΔV > 0), so system does work (W is negative)
The negative sign indicates work is done BY the system (energy leaves)

Step 3: Apply First Law.
ΔU = Q - W
ΔU = 30 kJ - (-0.304 kJ)
ΔU = 30 + 0.304
ΔU = 30.304 kJ ≈ 30.3 kJ

Interpretation:
→ Internal energy increases by 30.3 kJ
→ Heat absorbed by system = 30 kJ
→ Work done by system = 0.304 kJ
→ Most of the heat goes into increasing internal energy; small amount goes into doing expansion work
→ ΔU = Q - W (positive Q adds energy, negative W removes energy, net is increase)

Numerical Example 3: Compression of Gas

Problem:
A gas is compressed isothermally (constant temperature) from 10 L to 2 L. Work of 5 kJ is done on the gas. Calculate the heat change if internal energy remains constant.

Given:
Work done ON system (W_on) = 5 kJ
ΔU = 0 (isothermal for ideal gas with no temperature change)

Note: W_on is work done ON system (positive), but our formula uses W (work done BY system)
So: W = -W_on = -5 kJ (work done by system is negative)

Step 1: Apply First Law.
ΔU = Q - W
0 = Q - (-5)
0 = Q + 5
Q = -5 kJ

Interpretation:
→ No change in internal energy (isothermal, T constant)
→ 5 kJ of work is done ON the gas (compression)
→ 5 kJ of heat is released by the gas
→ Energy flows: Work energy (5 kJ in) → Heat energy (5 kJ out)
→ All mechanical work is converted to heat and released

Application to Real Processes:

1. Heating at constant volume (rigid container):
W = 0 (no expansion)
ΔU = Q (all heat goes into internal energy)
Example: Heating a sealed flask

2. Heating at constant pressure (open container):
W = -PΔV (expansion does work)
ΔU = Q + PΔV (some heat increases internal energy, some does work)
Example: Boiling water in open pot

3. Adiabatic process (thermally insulated):
Q = 0 (no heat exchange)
ΔU = -W (work change equals internal energy change)
Example: Rapid expansion in insulated system

4. Isothermal expansion of ideal gas:
ΔU = 0 (T constant, internal energy only depends on T)
Q = W (heat absorbed = work done by system)
Example: Slow expansion at constant T

Conclusion:
The First Law of Thermodynamics is a statement of energy conservation. It allows us to track energy flow between heat, work, and internal energy in any thermodynamic process. ΔU = Q - W is the fundamental equation that governs all energy transformations in thermodynamics.

Q3. Define enthalpy (H) and explain the difference between ΔU and ΔH. When is ΔH = ΔU?

Enthalpy Definition:
Enthalpy (H) is the total heat content of a system at constant pressure. It is a thermodynamic property that combines internal energy and the pressure-volume work.

Mathematical Definition:
H = U + PV

Where:
H = Enthalpy (J or kJ)
U = Internal energy (J or kJ)
P = Pressure (Pa or atm)
V = Volume (m³ or L)
PV = Pressure-volume work term

For changes in enthalpy:
ΔH = ΔU + Δ(PV)
ΔH = ΔU + (PV)final - (PV)initial
ΔH = ΔU + P(Vfinal - Vinitial)
ΔH = ΔU + P × ΔV (at constant pressure)

Difference Between ΔU and ΔH:

1. Definition:
ΔU = Change in internal energy (energy stored in molecules)
ΔH = Change in enthalpy (heat absorbed/released at constant pressure)

2. Conditions:
ΔU = applies to ALL processes (any pressure, any volume)
ΔH = specifically defined for constant pressure processes

3. Relationship:
At constant pressure: ΔH = ΔU + P × ΔV
Or: ΔH = ΔU + Δ(PV)

4. Heat and Work:
At constant pressure:
Q_constant pressure = ΔH
Q_constant volume = ΔU
Therefore: Q_p = ΔH and Q_v = ΔU

5. Which is Larger?
The difference between ΔH and ΔU is the work term (P × ΔV)
If gas expands (ΔV > 0): P × ΔV > 0, so ΔH > ΔU
If gas contracts (ΔV < 0): P × ΔV < 0, so ΔH < ΔU
If no volume change (ΔV = 0): ΔH = ΔU

Comparison Table:

Property ΔU (Internal Energy) ΔH (Enthalpy)
━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━
Definition Change in total energy Heat at constant pressure
of molecules
Condition Any pressure, any volume Constant pressure
Formula ΔU = Q - W ΔH = ΔU + P × ΔV
Relation at const P ΔU = Q_v - W ΔH = Q_p
Work included Yes (W is explicit) Yes (P×ΔV term)
Measurement Difficult Easy (at const P)
Application Thermodynamics Chemistry (most reactions)
Usual condition Constant volume Constant pressure
Example Bomb calorimeter (Cv) Typical lab reaction (Cp)

When is ΔH = ΔU?

Since: ΔH = ΔU + P × ΔV

For ΔH = ΔU, we need: P × ΔV = 0

This occurs when:

1. Constant Volume (ΔV = 0):
ΔV = 0 → P × ΔV = 0
Therefore: ΔH = ΔU
Examples:
• Rigid container (rigid walls, no volume change)
• Solid or liquid undergoing reaction (volume change negligible)
• Bomb calorimeter (rigid container)
• Heating/cooling a sealed, rigid vessel

2. Incompressible Liquids and Solids:
→ Liquids and solids have very small volume changes
→ ΔV ≈ 0
→ Therefore: ΔH ≈ ΔU
Examples:
• Dissolution of salt in water
• Phase changes of liquids and solids
• Most organic reactions in solution
• Melting, freezing, crystallization

3. Reactions with No Change in Moles of Gas:
For reactions involving gases:
Δn = (moles of gaseous products) - (moles of gaseous reactants)

If Δn = 0 (no net change in gas moles):
ΔH ≈ ΔU (approximately; small difference due to liquid/solid volume changes)

For ideal gases at constant pressure:
ΔH = ΔU + Δ(nRT)
If Δn = 0: ΔH = ΔU

Examples:
• N₂ + 3H₂ → 2NH₃ (Δn = 2 - 4 = -2, so ΔH ≠ ΔU significantly)
• H₂ + Cl₂ → 2HCl (Δn = 2 - 2 = 0, so ΔH ≈ ΔU)
• 2H₂O → 2H₂ + O₂ (Δn = 3 - 0 = 3, so ΔH ≠ ΔU)

Numerical Example 1: Constant Volume Process

Problem:
Combustion of 1 mol of methane in a rigid container at 25°C:
ΔU = -890 kJ

Since ΔV = 0 (rigid container, constant volume)

Calculate ΔH.

Solution:
At constant volume: ΔV = 0
Using: ΔH = ΔU + P × ΔV
ΔH = -890 + P × 0
ΔH = -890 kJ

Answer: ΔH = ΔU = -890 kJ (they are equal)

Numerical Example 2: Combustion at Constant Pressure

Problem:
Combustion of 1 mol of methane at 25°C and 1 atm (constant pressure):
ΔU = -890 kJ
Volume change: ΔV = -3 L (volume decreases as products are formed)

Calculate ΔH.

Solution:
Given:
ΔU = -890 kJ = -890,000 J
P = 1 atm = 101,325 Pa
ΔV = -3 L = -0.003 m³

Calculate P × ΔV:
P × ΔV = 101,325 Pa × (-0.003 m³) = -303.975 J ≈ -304 J ≈ -0.304 kJ

Using: ΔH = ΔU + P × ΔV
ΔH = -890 + (-0.304)
ΔH = -890.304 kJ ≈ -890.3 kJ

Answer: ΔH ≈ -890.3 kJ (slightly different from ΔU, but the difference is small for this reaction)

Note: The difference (0.3 kJ out of 890 kJ) is typically negligible and ΔH ≈ ΔU for condensed phase reactions.

Numerical Example 3: Gas Phase Reaction

Problem:
For the reaction: H₂(g) + Cl₂(g) → 2HCl(g) at 25°C and constant pressure
ΔU = -184 kJ
Δn = moles of gaseous products - moles of gaseous reactants
Δn = 2 - 2 = 0

Calculate ΔH.

Solution:
For ideal gases at constant pressure:
ΔH = ΔU + Δ(nRT)
ΔH = ΔU + ΔnRT
ΔH = -184 + (0) × R × T
ΔH = -184 + 0
ΔH = -184 kJ

Answer: ΔH = ΔU = -184 kJ (because Δn = 0)

Note: When moles of gas don't change, ΔH = ΔU exactly.

Numerical Example 4: Gas Production

Problem:
For the reaction: 2H₂O(l) → 2H₂(g) + O₂(g) at 25°C and constant pressure
ΔU = 572 kJ (endothermic, energy required)
Δn = 3 - 0 = 3 mol
T = 298 K
R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)

Calculate ΔH and the difference from ΔU.

Solution:
For ideal gases at constant pressure:
ΔH = ΔU + ΔnRT
ΔH = 572 + (3) × (8.314/1000) × 298
ΔH = 572 + (3) × (0.008314) × 298
ΔH = 572 + 7.43
ΔH = 579.43 kJ ≈ 579 kJ

Difference:
ΔH - ΔU = P × ΔV = ΔnRT = 7.43 kJ

Answer: ΔH ≈ 579 kJ (ΔH > ΔU because gas is produced, Δn > 0)

Interpretation:
→ More energy (579 kJ) is required at constant pressure than at constant volume (572 kJ)
→ Extra 7.43 kJ is needed because the system must do work to push the atmosphere as it expands
→ At constant volume, no expansion work, so less energy needed

Practical Significance:

1. In Chemistry:
→ Most reactions occur at constant pressure (open vessels)
→ ΔH is the relevant measure (heat of reaction at constant P)
→ ΔH values are tabulated for reactions

2. In Thermodynamics:
→ ΔU is the fundamental quantity (from First Law)
→ ΔH is convenient for constant pressure processes
→ ΔH is easier to measure experimentally

3. Standard States:
→ ΔH° = Standard enthalpy change at 25°C, 1 atm
→ ΔU° = Standard internal energy change
→ For liquids and solids: ΔH° ≈ ΔU° (small volume change)
→ For gases: ΔH° ≠ ΔU° (unless Δn = 0)

4. Calorimetry:
→ At constant pressure: measure ΔH (q_p in bomb calorimeter is ΔU)
→ At constant volume: measure ΔU (q_v in reaction calorimeter)

Conclusion: ΔH = ΔU when there is no volume change (ΔV = 0) or when the work term P×ΔV is negligible. This occurs for processes in rigid containers or for reactions of liquids and solids where volume changes are minimal.

Q4. Calculate the enthalpy change (ΔH) for the reaction: 2H₂(g) + O₂(g) → 2H₂O(l) using standard enthalpies of formation given: ΔH°f[H₂O(l)] = -285.8 kJ/mol, ΔH°f[H₂(g)] = 0 kJ/mol, ΔH°f[O₂(g)] = 0 kJ/mol.

Standard Enthalpy of Formation Definition:
Standard enthalpy of formation (ΔH°f) is the enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions (25°C, 1 atm).

Convention:
→ Elements in their standard state have ΔH°f = 0 kJ/mol
→ Compounds have ΔH°f values that are tabulated
→ Standard state indicated by ° (degree symbol)

Formula for ΔH°rxn (Enthalpy of Reaction):

ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants)

Where:
Σ = summation (sum of all terms)
ΔH°f(products) = sum of [mole coefficients × ΔH°f of each product]
ΔH°f(reactants) = sum of [mole coefficients × ΔH°f of each reactant]

Alternative form:
ΔH°rxn = [ΔH°f(products) with coefficients] - [ΔH°f(reactants) with coefficients]

Given Reaction:
2H₂(g) + O₂(g) → 2H₂O(l)

Given Data:
ΔH°f[H₂(g)] = 0 kJ/mol (element in standard state)
ΔH°f[O₂(g)] = 0 kJ/mol (element in standard state)
ΔH°f[H₂O(l)] = -285.8 kJ/mol (compound, tabulated value)

Step 1: Identify reactants and products.
Reactants: 2 mol H₂(g) + 1 mol O₂(g)
Products: 2 mol H₂O(l)

Step 2: Calculate sum for products.
Σ ΔH°f(products) = 2 mol × ΔH°f[H₂O(l)]
Σ ΔH°f(products) = 2 mol × (-285.8 kJ/mol)
Σ ΔH°f(products) = -571.6 kJ

Step 3: Calculate sum for reactants.
Σ ΔH°f(reactants) = 2 mol × ΔH°f[H₂(g)] + 1 mol × ΔH°f[O₂(g)]
Σ ΔH°f(reactants) = 2 mol × (0 kJ/mol) + 1 mol × (0 kJ/mol)
Σ ΔH°f(reactants) = 0 + 0
Σ ΔH°f(reactants) = 0 kJ

Step 4: Apply formula.
ΔH°rxn = Σ ΔH°f(products) - Σ ΔH°f(reactants)
ΔH°rxn = (-571.6 kJ) - (0 kJ)
ΔH°rxn = -571.6 kJ

Answer: ΔH°rxn = -571.6 kJ

Interpretation:

1. Negative value (ΔH < 0):
→ Reaction is exothermic (releases heat)
→ Burning hydrogen releases energy
→ Reaction is thermodynamically favorable

2. Magnitude:
→ 571.6 kJ of heat is released
→ Per mole of reaction as written (2 mol H₂ + 1 mol O₂ → 2 mol H₂O)

3. Per mole of H₂O:
→ Heat released per mole of H₂O = -571.6 kJ / 2 mol = -285.8 kJ/mol
→ This equals ΔH°f[H₂O(l)], confirming our calculation

4. Physical significance:
→ Hydrogen-oxygen combustion is highly exothermic
→ Used as rocket fuel (high energy release)
→ Reaction goes essentially to completion
→ Large amount of heat liberated makes it useful for propulsion

5. Equation with heat term:
2H₂(g) + O₂(g) → 2H₂O(l) ΔH° = -571.6 kJ
Or: 2H₂(g) + O₂(g) → 2H₂O(l) + 571.6 kJ
(Writing +571.6 kJ on product side emphasizes heat release)

Verification using Alternative Formula:

ΔH°rxn = [Σ n × ΔH°f(products)] - [Σ n × ΔH°f(reactants)]

Where n = stoichiometric coefficients

Products: 2 × (-285.8) = -571.6 kJ
Reactants: 2 × (0) + 1 × (0) = 0 kJ
ΔH°rxn = -571.6 - 0 = -571.6 kJ ✓

Key Points for Using ΔH°f:

1. Elements have ΔH°f = 0 in their standard state:
→ H₂(g), O₂(g), N₂(g), C(graphite), etc.
→ Different forms of element: C(graphite) = 0, C(diamond) ≠ 0

2. Compounds have non-zero ΔH°f:
→ Values are given in tables
→ Negative ΔH°f: formation is exothermic (stable compound)
→ Positive ΔH°f: formation is endothermic (less stable)

3. States of matter matter:
→ H₂O(l) ≠ H₂O(g) in terms of ΔH°f values
→ Different ΔH°f values for different physical states

4. Stoichiometric coefficients are crucial:
→ Must multiply ΔH°f by the coefficient in the equation
→ Different equations give different ΔH values

5. Sign and magnitude:
→ Negative ΔH: exothermic, heat released
→ Positive ΔH: endothermic, heat absorbed
→ Large |ΔH|: large energy change

Related Reactions (Same Overall Process, Different Stoichiometry):

Reaction A: H₂(g) + ½O₂(g) → H₂O(l)
ΔH°rxn = 1 × (-285.8) - [0 + 0] = -285.8 kJ
(Heat per mole of H₂O formed)

Reaction B: 2H₂(g) + O₂(g) → 2H₂O(l)
ΔH°rxn = 2 × (-285.8) - [0 + 0] = -571.6 kJ
(Same reaction, doubled equation, doubled heat)

Conclusion: The enthalpy change for the combustion of hydrogen is -571.6 kJ for the reaction as written, indicating a highly exothermic process. Using standard enthalpies of formation allows calculation of reaction enthalpy changes without experimental measurement.

Thermodynamics — Practice MCQs with Answers

Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.

MCQ 1. Which of the following is an intensive property?

• A. Mass
• B. Volume
• C. Density
• D. Enthalpy

Answer: C. Density — Intensive properties do not depend on the amount of matter in the system, while extensive properties do. Density, being a ratio of two extensive properties (mass/volume), is an intensive property.

MCQ 2. An isolated system is one in which:

• A. Neither heat nor matter can be exchanged with the surroundings
• B. Only heat can be exchanged with the surroundings
• C. Only matter can be exchanged with the surroundings
• D. Both heat and matter can be exchanged with the surroundings

Answer: A. Neither heat nor matter can be exchanged with the surroundings — An isolated system is completely sealed off from its surroundings, preventing the exchange of both energy (heat) and matter.

MCQ 3. If q is the heat added to a system and w is the work done by the system, what is the change in internal energy (ΔU) according to the First Law of Thermodynamics?

• A. ΔU = q + w
• B. ΔU = q - w
• C. ΔU = -q + w
• D. ΔU = -q - w

Answer: B. ΔU = q - w — The First Law of Thermodynamics states ΔU = q + w, where q is heat absorbed *by* the system and w is work done *on* the system. If w is work done *by* the system, its sign convention is negative, hence ΔU = q + (-w) = q - w.

MCQ 4. When work is done *by* the system on the surroundings, the sign of work (w) is conventionally:

• A. Positive
• B. Negative
• C. Zero
• D. Depends on the process

Answer: B. Negative — According to the IUPAC convention, work done *by* the system (expansion) is negative, as the system expends its energy. Work done *on* the system (compression) is positive.

MCQ 5. For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at constant temperature and pressure, which of the following is true?

• A. ΔH = ΔU
• B. ΔH > ΔU
• C. ΔH < ΔU
• D. ΔH = 0

Answer: C. ΔH < ΔU — The relationship between ΔH and ΔU is ΔH = ΔU + Δn_g RT. For this reaction, Δn_g = (2 - (1+3)) = -2. Since Δn_g is negative, Δn_g RT is a negative value, meaning ΔH = ΔU - |Δn_g RT|, so ΔH < ΔU.

MCQ 6. Given the following reactions: 1. C(s) + O₂(g) → CO₂(g) ; ΔH₁ = -393.5 kJ 2. H₂(g) + ½O₂(g) → H₂O(l) ; ΔH₂ = -285.8 kJ 3. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) ; ΔH₃ = -890.3 kJ Calculate the enthalpy of formation of methane (CH₄(g)).

• A. -74.8 kJ
• B. 74.8 kJ
• C. -170.6 kJ
• D. 170.6 kJ

Answer: A. -74.8 kJ — The target reaction is C(s) + 2H₂(g) → CH₄(g). Using Hess's Law: ΔH_f(CH₄) = ΔH₁ + 2ΔH₂ - ΔH₃ = -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -74.8 kJ.

MCQ 7. For a spontaneous process in an isolated system, which of the following is true?

• A. ΔS_total > 0
• B. ΔG < 0
• C. ΔH < 0
• D. ΔU = 0

Answer: A. ΔS_total > 0 — According to the Second Law of Thermodynamics, for a spontaneous process in an isolated system (or the universe), the total entropy (system + surroundings) must increase (ΔS_total > 0).

MCQ 8. A reaction has ΔH = +50 kJ/mol and ΔS = +150 J/K·mol. At what temperature will the reaction become spontaneous?

• A. Above 333.3 K
• B. Below 333.3 K
• C. At 333.3 K
• D. Never

Answer: A. Above 333.3 K — For spontaneity, ΔG = ΔH - TΔS < 0. Set ΔG = 0 to find the equilibrium temperature: 0 = 50,000 J - T(150 J/K). This gives T = 50,000/150 = 333.3 K. Since ΔH is positive and ΔS is positive, the reaction becomes spontaneous at higher temperatures, so T > 333.3 K.

MCQ 9. A process that can be reversed by an infinitesimal change in a variable is called a(n):

• A. Irreversible process
• B. Adiabatic process
• C. Reversible process
• D. Isobaric process

Answer: C. Reversible process — A reversible process is defined as one that can be reversed by an infinitesimal change in the external conditions, restoring both the system and surroundings to their initial states.

MCQ 10. 1 mole of an ideal gas expands isothermally from 10 L to 20 L at 300 K against a constant external pressure of 1 atm. What is the work done (in Joules)? (1 L·atm = 101.3 J)

• A. -1013 J
• B. -2494 J
• C. -10 L·atm
• D. -2026 J

Answer: A. -1013 J — This is an irreversible expansion against a constant external pressure, so w = -P_ext * ΔV. w = -1 atm * (20 L - 10 L) = -10 L·atm. Converting to Joules: -10 L·atm * 101.3 J/L·atm = -1013 J.

MCQ 11. Which statement is true regarding molar heat capacity at constant volume (C_v)?

• A. C_v = R
• B. C_v = (dU/dT)_V
• C. C_v = (dH/dT)_P
• D. C_v = C_p + R

Answer: B. C_v = (dU/dT)_V — Molar heat capacity at constant volume (C_v) is defined as the change in internal energy with respect to temperature at constant volume, i.e., C_v = (dU/dT)_V.

MCQ 12. The standard enthalpy of vaporization of water at 100°C is 40.66 kJ/mol. What is the standard entropy of vaporization of water at 100°C?

• A. 109.0 J/K·mol
• B. -109.0 J/K·mol
• C. 406.6 J/K·mol
• D. 0 J/K·mol

Answer: A. 109.0 J/K·mol — At a phase transition (like boiling), the process is reversible and ΔG = 0, so ΔS = ΔH / T. ΔS_vap = 40.66 kJ/mol / (100 + 273.15) K = 40660 J/mol / 373.15 K ≈ 109.0 J/K·mol.

MCQ 13. The Zeroth Law of Thermodynamics deals with:

• A. Conservation of energy
• B. Direction of heat flow
• C. Absolute zero temperature
• D. Thermal equilibrium

Answer: D. Thermal equilibrium — The Zeroth Law states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other, establishing the concept of temperature and thermal equilibrium.

MCQ 14. In an adiabatic process, which of the following is true?

• A. q = 0
• B. ΔU = 0
• C. w = 0
• D. ΔH = 0

Answer: A. q = 0 — An adiabatic process is defined as one in which no heat exchange occurs between the system and its surroundings, meaning q = 0.

MCQ 15. For a reaction, ΔH = -30 kJ/mol and ΔS = -100 J/K·mol. At what temperature will the reaction be spontaneous?

• A. Below 300 K
• B. Above 300 K
• C. At 300 K
• D. The reaction is always spontaneous

Answer: A. Below 300 K — For spontaneity, ΔG = ΔH - TΔS < 0. Set ΔG = 0 to find the equilibrium temperature: 0 = -30,000 J - T(-100 J/K), which gives T = 300 K. Since both ΔH and ΔS are negative, the reaction is spontaneous at low temperatures (where the -TΔS term is less positive than the negative ΔH), so T < 300 K.

MCQ 16. The standard state for a substance refers to:

• A. 0°C and 1 atm pressure
• B. 25°C and 1 atm pressure
• C. Any temperature with 1 atm pressure
• D. 25°C and 1 bar pressure

Answer: D. 25°C and 1 bar pressure — The standard state in thermodynamics is typically defined as 298.15 K (25°C) and 1 bar (10^5 Pa) pressure for gases, and 1 M concentration for solutions. For liquids/solids, it's the pure substance at 1 bar.

MCQ 17. Given average bond enthalpies: C-H = 413 kJ/mol, Cl-Cl = 242 kJ/mol, C-Cl = 328 kJ/mol, H-Cl = 431 kJ/mol. Calculate ΔH for the reaction: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g).

• A. -104 kJ
• B. +104 kJ
• C. -100 kJ
• D. +100 kJ

Answer: A. -104 kJ — ΔH = Σ(Bond enthalpies of reactants broken) - Σ(Bond enthalpies of products formed). Bonds broken: 1 C-H (413) + 1 Cl-Cl (242) = 655 kJ. Bonds formed: 1 C-Cl (328) + 1 H-Cl (431) = 759 kJ. ΔH = 655 - 759 = -104 kJ.

MCQ 18. For a reaction, the standard Gibbs free energy change (ΔG°) is -13.8 kJ/mol at 298 K. Calculate the equilibrium constant (K_eq) for this reaction. (R = 8.314 J/K·mol)

• A. 250
• B. 100
• C. 50
• D. 20

Answer: A. 250 — The relationship is ΔG° = -RT ln K_eq. So, -13800 J/mol = -(8.314 J/K·mol)(298 K) ln K_eq. This yields ln K_eq = 13800 / (8.314 * 298) ≈ 5.57. Therefore, K_eq = e^(5.57) ≈ 262, which is closest to 250.

MCQ 19. The Third Law of Thermodynamics states that:

• A. Energy cannot be created or destroyed
• B. The entropy of an isolated system increases in a spontaneous process
• C. The entropy of a perfect crystalline substance is zero at absolute zero temperature
• D. Heat cannot spontaneously flow from a colder body to a hotter body

Answer: C. The entropy of a perfect crystalline substance is zero at absolute zero temperature — The Third Law of Thermodynamics provides a reference point for entropy, stating that the entropy of a perfect crystal at absolute zero (0 K) is zero.

MCQ 20. A reaction is endothermic (ΔH > 0) and results in a decrease in entropy (ΔS < 0). This reaction will be:

• A. Always spontaneous
• B. Never spontaneous
• C. Spontaneous at high temperatures
• D. Spontaneous at low temperatures

Answer: B. Never spontaneous — ΔG = ΔH - TΔS. If ΔH > 0 and ΔS < 0, then the -TΔS term will be positive. Thus, ΔG will always be positive (positive + positive), meaning the reaction is never spontaneous under these conditions.

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