What is the First Law of Thermodynamics and its mathematical expression?

The First Law states that energy can neither be created nor destroyed, only transformed. Mathematically: ΔU = q + w, where ΔU is change in internal energy, q is heat supplied, and w is work done on the system.

What is Hess’s Law of Constant Heat Summation?

Hess’s Law states that the overall enthalpy change of a chemical reaction is identical whether the reaction takes place in a single step or in a series of multiple steps, because enthalpy is a state function.

What is the condition for spontaneity in terms of Gibbs free energy (ΔG)?

For a process at constant temperature and pressure, ΔG = ΔH - TΔS. A reaction is spontaneous if ΔG 0, and in equilibrium when ΔG = 0.

What is the relationship between ΔH and ΔU in chemical reactions?

For gaseous reactions, ΔH = ΔU + Δn_g · RT, where Δn_g is (moles of gaseous products - moles of gaseous reactants), R is gas constant (8.314 J/mol·K), and T is temperature in Kelvin.

ALL UNITS

Unit 05 • Physical Chemistry

Thermodynamics

Class 11 Chemistry Chapter 5 comprehensive notes, First, Second & Third Laws, state functions ($U, H, S, G$), $C_p - C_v = R$, Hess's Law, calorimetry, Born-Haber cycle, spontaneity criteria ($\Delta G < 0$), in-text problems (5.15.14), NCERT exercises (5.1 to 5.22), revision sheets, and 3-level chapter tests.

Introduction: The Science of Energy Transformations

"Thermodynamics is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown." — Albert Einstein.

Chemical energy stored in molecules is released as heat or work during chemical transformations (e.g., combustion of fuels, galvanic cells). Thermodynamics deals with the quantitative energy transformations of macroscopic systems in equilibrium — independent of the microscopic pathway or rate of reaction.

Three Core Questions Thermodynamics Answers

Feasibility (Spontaneity): Will a chemical reaction occur on its own under specified conditions ($\Delta G < 0$)?
Energy Changes: How much heat is released/absorbed ($\Delta H$) or useful work produced?
Extent of Reaction: To what extent does the reaction proceed before reaching equilibrium ($K_{eq}$)?

5.1 Basic Thermodynamic Terms

Term	Scientific Definition	Everyday Laboratory Example
System	The specific part of the universe chosen for thermodynamic observation and study.	The reacting chemicals inside a beaker or test tube.
Surroundings	The entire remaining universe outside the system (specifically the immediate interacting environment).	The laboratory air, beaker walls, and water bath surrounding the system.
Boundary	The real or imaginary surface separating the system from its surroundings (rigid, non-rigid, conducting, adiabatic).	The glass wall of the beaker.

$$\text{Universe} = \text{System} + \text{Surroundings}$$

System, Surroundings, and Thermodynamic Boundary

UNIVERSE (System + Surroundings) SURROUNDINGS Immediate thermal & mechanical environment SURROUNDINGS SYSTEM Reactants & Products (Under Observation) Boundary Wall Energy (q, w) Universe = System + Surroundings

5.1.2 Types of Systems

Open System: Can exchange both matter and energy with surroundings (e.g., reactants in an open beaker).
Closed System: Can exchange energy (heat/work) but NOT matter with surroundings (e.g., liquid in a sealed metallic container).
Isolated System: Can exchange neither matter nor energy with surroundings (e.g., hot liquid inside an ideal thermos flask with adiabatic walls).

Classification of Thermodynamic Systems

5.1.3 State Variables & State Functions

A state function is a thermodynamic property whose value depends solely on the state of the system (initial and final states) and is completely independent of the path or mechanism taken to reach that state.

State Functions: Pressure ($p$), Volume ($V$), Temperature ($T$), Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$).
Path Functions: Heat ($q$) and Work ($w$) — their values depend on the specific path followed.

Extensive vs Intensive Thermodynamic Properties (Partition Model)

5.1.4 Internal Energy ($U$) and the First Law

Internal Energy ($U$): The total microscopic energy contained within a system — sum of kinetic energies (translational, rotational, vibrational) and potential energies (intermolecular, chemical bonding, electronic, nuclear).

The First Law of Thermodynamics:
"Energy can neither be created nor destroyed, although it may be converted from one form to another."
$$\Delta U = q + w$$

IUPAC Sign Conventions (Crucial for Numericals)

Heat absorbed by system ($q > 0$): Positive ($+$).
Heat released by system ($q < 0$): Negative ($-$).
Work done ON the system (compression, $w > 0$): Positive ($+$).
Work done BY the system (expansion, $w < 0$): Negative ($-$).

5.1

Express the change in internal energy of a system when: (i) No heat is absorbed, but work (w) is done on the system; (ii) No work is done, but heat (q) is taken out; (iii) Work (w) is done by the system and heat (q) is supplied to the system.

Using $\Delta U = q + w$:
(i) $q = 0 \implies \mathbf{\Delta U = w_{\text{ad}}}$ (wall is adiabatic).
(ii) $w = 0 \implies \mathbf{\Delta U = -q}$ (wall is thermally conducting / diathermic).
(iii) Heat supplied $= +q$, work done by system $= -w \implies \mathbf{\Delta U = q - w}$ (closed system).

(i) ΔU = w_ad; (ii) ΔU = −q; (iii) ΔU = q − w.

5.2 Pressure-Volume Work & Free Expansion

When a gas in a cylinder fitted with a frictionless piston changes volume against an external pressure $p_{ex}$: $$w = - p_{ex} \Delta V = - p_{ex} (V_f - V_i)$$

Gas Expansion ($V_f > V_i$): $\Delta V$ is positive $\implies w$ is negative (work is done by the gas on surroundings).
Gas Compression ($V_f < V_i$): $\Delta V$ is negative $\implies w$ is positive (work is done on the gas by surroundings).
Free Expansion in Vacuum ($p_{ex} = 0$): $$w = - 0 \times \Delta V = \mathbf{0}$$ No work is done during free expansion of an ideal gas into a vacuum, whether reversible or irreversible!
Isothermal Reversible Expansion of an Ideal Gas: $$w_{\text{rev}} = - \int_{V_i}^{V_f} p \, dV = - nRT \ln \frac{V_f}{V_i} = \mathbf{- 2.303 \, nRT \log \frac{V_f}{V_i}} = \mathbf{- 2.303 \, nRT \log \frac{p_i}{p_f}}$$

PV Indicator Diagram: Irreversible vs Reversible Expansion Work

5.2

Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 °C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?

Expansion is into a vacuum $\implies p_{ex} = 0$.
$w = - p_{ex} (V_f - V_i) = - 0 \times (10 - 2) = \mathbf{0}$.
Since process is isothermal ($T = \text{constant}$), for an ideal gas $\Delta U = 0$.
$\Delta U = q + w \implies q = -w = \mathbf{0}$.

Work done = 0; Heat absorbed = 0.

5.3

Consider the same expansion (2 L to 10 L), but this time against a constant external pressure of 1 atm. How much work is done and heat absorbed?

$w = - p_{ex} (V_f - V_i) = - 1\text{ atm} \times (10 - 2)\text{ L} = \mathbf{-8\text{ L-atm}}$.
Converting to Joules ($1\text{ L-atm} = 101.325\text{ J}$):
$w = -8 \times 101.325\text{ J} = \mathbf{-810.6\text{ J}}$.
Since $\Delta U = 0$, $q = -w = \mathbf{+8\text{ L-atm}} = \mathbf{+810.6\text{ J}}$.

w = −8 L-atm (−810.6 J); q = +8 L-atm (+810.6 J).

5.4

Calculate the work done for the expansion from 2 L to 10 L of 1 mol of an ideal gas at 25 °C conducted reversibly.

$w_{\text{rev}} = - 2.303 \, nRT \log \frac{V_f}{V_i}$
$= - 2.303 \times 1\text{ mol} \times 8.314\text{ J K}^{-1}\text{mol}^{-1} \times 298\text{ K} \times \log \frac{10}{2}$
$= - 2.303 \times 8.314 \times 298 \times 0.6990 = \mathbf{-3988\text{ J}} = \mathbf{-3.988\text{ kJ}}$.
In L-atm units: $w_{\text{rev}} = -39.37\text{ L-atm}$.

w_rev = −3.988 kJ (−39.37 L-atm); q = +3.988 kJ.

5.3 Enthalpy ($H$), $\Delta H$ vs $\Delta U$, and Heat Capacities

Most laboratory reactions are conducted at constant atmospheric pressure, not constant volume. To represent the heat change at constant pressure ($q_p$), we define Enthalpy ($H$):

$$H = U + pV$$ $$\Delta H = \Delta U + p\Delta V = q_p \qquad (\text{at constant pressure})$$ $$\Delta U = q_v \qquad (\text{at constant volume})$$

Relation between $\Delta H$ and $\Delta U$ for Gaseous Reactions

For reactions involving ideal gases ($pV = nRT \implies p\Delta V = \Delta n_g RT$):

$$\Delta H = \Delta U + \Delta n_g RT$$ Where $\Delta n_g = \sum n_{\text{products(g)}} - \sum n_{\text{reactants(g)}}$.

If $\Delta n_g = 0 \implies \mathbf{\Delta H = \Delta U}$ (e.g., $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}$).
If $\Delta n_g > 0 \implies \mathbf{\Delta H > \Delta U}$ (e.g., $PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}$).
If $\Delta n_g < 0 \implies \mathbf{\Delta H < \Delta U}$ (e.g., $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$).

5.5

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100 °C is 41 kJ mol⁻¹. Calculate the internal energy change, when 1 mol of water is vapourised.

Reaction: $H_2O_{(l)} \longrightarrow H_2O_{(g)}$
$\Delta n_g = 1 - 0 = 1\text{ mol}$.
$\Delta U = \Delta H - \Delta n_g RT$
$= 41.00\text{ kJ mol}^{-1} - (1 \times 8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1} \times 373\text{ K})$
$= 41.00 - 3.10 = \mathbf{37.90\text{ kJ mol}^{-1}}$.

ΔU = 37.90 kJ mol⁻¹.

Heat Capacity & The $C_p - C_v = R$ Relation

Heat Capacity ($C$): $q = C \Delta T$.
Specific Heat ($c$): Heat required to raise temperature of $1\text{ g}$ of substance by $1\text{ K}$ ($q = m \cdot c \cdot \Delta T$).
Molar Heat Capacity ($C_m$): Heat required for $1\text{ mole}$ ($q = n \cdot C_m \cdot \Delta T$).
At Constant Volume: $q_v = \Delta U = C_v \Delta T \implies C_v = \left(\frac{\partial U}{\partial T}\right)_V$.
At Constant Pressure: $q_p = \Delta H = C_p \Delta T \implies C_p = \left(\frac{\partial H}{\partial T}\right)_p$.

$$\mathbf{C_p - C_v = R} \qquad (\text{for 1 mole of an ideal gas})$$ Why? At constant pressure, heat supplied not only increases temperature (internal energy) but also performs expansion work against external pressure.

5.4 Calorimetry, Enthalpy of Reactions & Hess's Law

Calorimeter Type	Operating Condition	What It Measures	Key Formula
Bomb Calorimeter	Constant Volume ($\Delta V = 0, w = 0$)	Internal Energy of Combustion ($\Delta U_c$)	$q_v = \Delta U = - C_{\text{cal}} \Delta T$
Coffee-Cup Calorimeter	Constant Atmospheric Pressure	Enthalpy of Reaction ($\Delta_r H$)	$q_p = \Delta H = m \cdot c \cdot \Delta T$

Bomb Calorimeter Setup (Measurement of ΔU at Constant Volume)

Hess's Law of Constant Heat Summation

Enthalpy Profile Diagrams: Exothermic (ΔH < 0) vs Endothermic (ΔH > 0)

Because enthalpy is a state function, the total enthalpy change for a reaction is identical whether it occurs in a single step or through a series of steps: $$\Delta_r H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots$$

Hess's Law Enthalpy Cycle Representation

Standard Enthalpies of Different Reactions

Standard Enthalpy of Formation ($\Delta_f H^\circ$): Enthalpy change when 1 mole of a compound is formed from its elements in their standard reference states. By convention, $\mathbf{\Delta_f H^\circ = 0}$ for all pure elements in their reference state (e.g., $O_{2(g)}, C_{\text{graphite}}, S_{\text{rhombic}}, Fe_{(s)}$). $$\Delta_r H^\circ = \sum a_i \Delta_f H^\circ(\text{products}) - \sum b_i \Delta_f H^\circ(\text{reactants})$$
Standard Enthalpy of Combustion ($\Delta_c H^\circ$): Always exothermic (negative). Heat released when 1 mole of substance is completely burnt in excess oxygen.
Bond Dissociation Enthalpy vs Mean Bond Enthalpy: For diatomic molecules, atomization equals bond enthalpy. For polyatomics (like $CH_4$), breaking 4 successive $C-H$ bonds requires $427, 439, 452, 347\text{ kJ mol}^{-1}$ $\implies$ Mean Bond Enthalpy $= \frac{1665}{4} = 416\text{ kJ mol}^{-1}$. $$\Delta_r H^\circ = \sum \text{Bond Enthalpies (Reactants)} - \sum \text{Bond Enthalpies (Products)}$$
Lattice Enthalpy & Born-Haber Cycle: For $NaCl_{(s)} \rightarrow Na^+_{(g)} + Cl^-_{(g)}$ ($\Delta_{\text{lattice}}H = +788\text{ kJ mol}^{-1}$): $$\Delta_f H^\circ = \Delta_{\text{sub}}H + \Delta_i H + \frac{1}{2}\Delta_{\text{bond}}H + \Delta_{eg}H - \Delta_{\text{lattice}}H$$

Born-Haber Cycle: Thermochemical Determination of Lattice Enthalpy of NaCl

5.6

1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm. Temperature rises from 298 K to 299 K. Heat capacity of calorimeter is 20.7 kJ/K. What is the enthalpy change for the reaction per mole of graphite?

Heat absorbed by calorimeter $= C_v \Delta T = 20.7\text{ kJ/K} \times (299 - 298)\text{ K} = 20.7\text{ kJ}$.
Heat evolved by 1 g graphite $= -20.7\text{ kJ}$.
Molar mass of carbon $= 12.01\text{ g/mol}$.
$\Delta U = -20.7 \times 12.01 = \mathbf{-2.48 \times 10^2\text{ kJ mol}^{-1}}$.
Reaction: $C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} \implies \Delta n_g = 1 - 1 = 0 \implies \mathbf{\Delta H = \Delta U = -2.48 \times 10^2\text{ kJ mol}^{-1}}$.

ΔH = ΔU = −2.48 × 10² kJ mol⁻¹.

5.7

A swimmer is covered with 18 g of water. How much heat must be supplied to evaporate this water at 298 K? Calculate ΔU of vapourisation (Δ_vap H = 44.01 kJ/mol at 298 K).

Moles of water $= \frac{18\text{ g}}{18\text{ g/mol}} = 1\text{ mol}$.
Heat supplied $q_p = \Delta_{\text{vap}}H = \mathbf{44.01\text{ kJ}}$.
$\Delta_{\text{vap}}U = \Delta_{\text{vap}}H - \Delta n_g RT = 44.01 - (1 \times 8.314 \times 10^{-3} \times 298) = 44.01 - 2.48 = \mathbf{41.53\text{ kJ}}$.

q = 44.01 kJ; Δ_vap U = 41.53 kJ.

5.8

Calculate the internal energy change when 1 mol of water at 100 °C and 1 bar is converted to ice at 0 °C. (Δ_fus H = 6.00 kJ/mol, Cp of water = 4.2 J/g·°C).

Step 1: Cooling liquid water: $\Delta H_1 = - (18\text{ g} \times 4.2\text{ J/g·°C} \times 100\text{ °C}) = -7560\text{ J} = -7.56\text{ kJ}$.
Step 2: Freezing water to ice: $\Delta H_2 = -\Delta_{\text{fus}}H = -6.00\text{ kJ}$.
Total $\Delta H = -7.56 + (-6.00) = \mathbf{-13.56\text{ kJ mol}^{-1}}$.
Since only condensed phases are involved, $p\Delta V \approx 0 \implies \mathbf{\Delta U = \Delta H = -13.56\text{ kJ mol}^{-1}}$.

ΔH = ΔU = −13.56 kJ mol⁻¹.

5.9

Calculate the standard enthalpy of formation of benzene from the enthalpies of combustion of benzene (−3267 kJ/mol), carbon (−393.5 kJ/mol), and hydrogen (−285.83 kJ/mol).

Desired reaction: $6C_{\text{graphite}} + 3H_{2(g)} \longrightarrow C_6H_{6(l)}$
$\Delta_f H^\circ = 6 \Delta_c H^\circ(C) + 3 \Delta_c H^\circ(H_2) - \Delta_c H^\circ(C_6H_6)$
$= 6(-393.5) + 3(-285.83) - (-3267)$
$= -2361.0 - 857.49 + 3267.0 = \mathbf{+48.51\text{ kJ mol}^{-1}}$.

Δ_f H°(benzene) = +48.51 kJ mol⁻¹.

5.5 Spontaneity, Entropy ($S$) & Gibbs Free Energy ($G$)

A spontaneous process is an irreversible process that has an intrinsic tendency to occur on its own without requiring an external continuous driving force.

Enthalpy alone is NOT the sole criterion: While many exothermic reactions ($\Delta H < 0$) are spontaneous, many endothermic processes are also spontaneous (e.g., evaporation of water, melting of ice above $0^\circ\text{C}$, dissolution of $NH_4Cl$).
Entropy ($S$): A state function measuring the degree of randomness or disorder in the system: $$\Delta S = \frac{q_{\text{rev}}}{T} \qquad \text{Units: J K}^{-1}\text{mol}^{-1}$$
Second Law of Thermodynamics: In any spontaneous process, the total entropy of the universe increases: $$\mathbf{\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0}$$ $$\text{At equilibrium: } \mathbf{\Delta S_{\text{total}} = 0}$$
Third Law of Thermodynamics: The entropy of a perfectly crystalline solid approaches zero as the temperature approaches absolute zero ($0\text{ K}$): $$\lim_{T \to 0} S = 0$$

Gibbs Energy ($G$) and Reaction Spontaneity

To determine spontaneity under laboratory conditions (constant $T$ and $p$), we use the Gibbs Free Energy equation:

$$\mathbf{\Delta G = \Delta H - T \Delta S}$$

$\mathbf{\Delta G < 0}$ (Negative): Spontaneous process (feasible).
$\mathbf{\Delta G = 0}$ (Zero): System is at dynamic equilibrium.
$\mathbf{\Delta G > 0}$ (Positive): Non-spontaneous process (reverse reaction is spontaneous).

$\Delta H$	$\Delta S$	$\Delta G = \Delta H - T\Delta S$	Spontaneity Conditions
−	+	Always negative	Spontaneous at all temperatures
−	−	Negative at low $T$; Positive at high $T$	Spontaneous only at low temperature ($T < \frac{\Delta H}{\Delta S}$)
+	+	Positive at low $T$; Negative at high $T$	Spontaneous only at high temperature ($T > \frac{\Delta H}{\Delta S}$)
+	−	Always positive	Non-spontaneous at all temperatures

Gibbs Free Energy Spontaneity Map: Interplay of Enthalpy (ΔH) and Entropy (ΔS)

Relation between $\Delta_r G^\circ$ and Equilibrium Constant ($K$)

$$\mathbf{\Delta_r G^\circ = - RT \ln K = - 2.303 \, RT \log K}$$ If $\Delta_r G^\circ < 0 \implies K > 1$ (products favoured at equilibrium); if $\Delta_r G^\circ > 0 \implies K < 1$ (reactants favoured).

5.10

Predict in which of the following, entropy increases/decreases: (i) A liquid crystallizes into a solid; (ii) Temperature of a crystalline solid is raised from 0 K to 115 K; (iii) 2NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g); (iv) H₂(g) → 2H(g).

(i) Decreases (solid has ordered lattice).
(ii) Increases (particles oscillate and gain vibrational disorder).
(iii) Increases (solid reactant produces two moles of gaseous products).
(iv) Increases (1 molecule dissociates into 2 free atoms).

(i) Decreases; (ii) Increases; (iii) Increases; (iv) Increases.

5.11

For the oxidation of iron: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s), entropy change is −549.4 J K⁻¹ mol⁻¹ at 298 K. In spite of negative entropy change, why is the reaction spontaneous? (Δ_r H° = −1648 × 10³ J mol⁻¹).

Spontaneity is determined by $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$.
$\Delta S_{\text{surr}} = \frac{-\Delta_r H^\circ}{T} = \frac{-(-1648 \times 10^3\text{ J})}{298\text{ K}} = \mathbf{+5530\text{ J K}^{-1}\text{mol}^{-1}}$.
$\Delta S_{\text{total}} = -549.4 + 5530 = \mathbf{+4980.6\text{ J K}^{-1}\text{mol}^{-1}} > 0$.
Because $\Delta S_{\text{total}} > 0$, the reaction is spontaneous!

ΔS_total = +4980.6 J K⁻¹ mol⁻¹ > 0. Huge heat released to surroundings increases surroundings' entropy enormously.

5.12

Calculate Δ_r G° for the conversion of oxygen to ozone, 3/2 O₂(g) → O₃(g) at 298 K, if K_p for this conversion is 2.47 × 10⁻²⁹.

$\Delta_r G^\circ = - 2.303 \, RT \log K_p$
$= - 2.303 \times 8.314\text{ J K}^{-1}\text{mol}^{-1} \times 298\text{ K} \times \log(2.47 \times 10^{-29})$
$= - 5705.85 \times (-28.607) = \mathbf{+163,000\text{ J mol}^{-1}} = \mathbf{+163\text{ kJ mol}^{-1}}$.

Δ_r G° = +163 kJ mol⁻¹ (Large positive value explains why ozone decomposes back into oxygen).

5.13

Find the value of the equilibrium constant for a reaction at 298 K if Δ_r G° = −13.6 kJ mol⁻¹.

$\log K = \frac{-\Delta_r G^\circ}{2.303 RT} = \frac{-(-13.6 \times 10^3\text{ J})}{2.303 \times 8.314 \times 298} = \frac{13600}{5705.85} = 2.38$.
$K = \text{antilog}(2.38) = \mathbf{2.4 \times 10^2}$.

K = 2.4 × 10² (K > 1 indicates products predominate).

5.14

At 60 °C, dinitrogen tetroxide (N₂O₄) is 50% dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

Reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.
For 50% dissociation ($\alpha = 0.5$):
$p_{N_2O_4} = \frac{1 - 0.5}{1 + 0.5} \times 1 = \frac{0.5}{1.5} = \frac{1}{3}\text{ atm}$.
$p_{NO_2} = \frac{2 \times 0.5}{1 + 0.5} \times 1 = \frac{1.0}{1.5} = \frac{2}{3}\text{ atm}$.
$K_p = \frac{p_{NO_2}^2}{p_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \mathbf{1.33\text{ atm}}$.
$\Delta_r G^\circ = - 2.303 \, RT \log K_p = - 2.303 \times 8.314 \times 333\text{ K} \times \log(1.33) = \mathbf{-796\text{ J mol}^{-1}} = \mathbf{-0.796\text{ kJ mol}^{-1}}$.

K_p = 1.33 atm; Δ_r G° = −796 J mol⁻¹ (−0.796 kJ mol⁻¹).

Chapter Summary (Unit 5)

Thermodynamics: Macroscopic study of energy transformations between system and surroundings.
First Law: Conservation of energy: $\Delta U = q + w$. State functions depend only on initial/final state ($U, H, S, G$).
PV Work: $w = -p_{ex}\Delta V$. For isothermal reversible expansion: $w_{\text{rev}} = -2.303 nRT \log(V_f/V_i)$.
Enthalpy: $H = U + pV \implies \Delta H = \Delta U + \Delta n_g RT$. For ideal gases: $C_p - C_v = R$.
Hess's Law: Total enthalpy change is independent of the path ($\Delta_r H = \sum \Delta H_i$).
Entropy & Second Law: Measure of disorder. For spontaneous change: $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$.
Gibbs Energy: $\Delta G = \Delta H - T\Delta S$. Spontaneous if $\Delta G < 0$; equilibrium if $\Delta G = 0$. $\Delta_r G^\circ = -2.303 RT \log K$.

Thermodynamics Conceptual Quiz (25 MCQs)

Test your conceptual understanding of thermodynamic laws, state functions, work, enthalpy, calorimetry, entropy, and Gibbs free energy. Select an option to see instant feedback and explanation.

1 Which of the following is an intensive property?

Incorrect: Mass depends on quantity of matter (extensive).

Incorrect: Volume is extensive.

Correct: Density is mass per unit volume and is independent of sample size (intensive).

Incorrect: Internal energy is extensive.

2 For an adiabatic process, which of the following is true?

Incorrect: Temperature may change in an adiabatic process.

Correct: By definition, an adiabatic system permits no heat transfer through its boundary (q = 0).

Incorrect: Work can be done adiabatically (ΔU = w_ad).

Incorrect: Pressure may vary.

3 In an isolated system, which of the following statements is correct?

Incorrect: Isolated system exchanges neither.

Incorrect: That describes a closed system.

Correct: An isolated system is bounded by adiabatic and rigid walls allowing no transfer of matter or energy.

Incorrect: That describes an open system.

4 The mathematical expression of the First Law of Thermodynamics is:

Incorrect: Under IUPAC conventions, w is positive when work is done on the system.

Correct: ΔU = q + w under IUPAC convention.

Incorrect: That defines enthalpy change.

Incorrect: That is the Gibbs-Helmholtz equation.

5 Work done during the free expansion of an ideal gas into a vacuum (pex = 0) is:

Incorrect: Reversible expansion gives maximum work.

Incorrect: Expansion work against resistance is negative.

Correct: Because p_ex = 0, w = −p_ex·ΔV = 0.

Incorrect: Free expansion does zero work.

6 The relationship between ΔH and ΔU for a reaction involving gases is given by:

Incorrect: The sign before ΔngRT should be positive.

Correct: Since H = U + pV and pΔV = ΔngRT, ΔH = ΔU + ΔngRT.

Incorrect: Rearrangement gives ΔU = ΔH − ΔngRT.

Incorrect: Work is pΔV, not pΔT.

7 For the reaction C(graphite) + O₂(g) → CO₂(g), which relation holds true?

Incorrect: Only when Δng > 0.

Incorrect: Only when Δng < 0.

Correct: Δng = 1 (CO₂) − 1 (O₂) = 0. Therefore ΔH = ΔU + 0 = ΔU.

Incorrect: The combustion reaction is exothermic (ΔH = −393.5 kJ/mol).

8 For an ideal gas, the difference between molar heat capacity at constant pressure (Cp) and at constant volume (Cv) is:

Incorrect: Cp is always greater than Cv.

Correct: Mayer's relation states Cp − Cv = R for one mole of an ideal gas.

Incorrect: Cp > Cv, not Cv > Cp.

Incorrect: Cp / Cv = γ (heat capacity ratio).

9 In a bomb calorimeter, reactions are carried out under which condition?

Incorrect: Constant pressure is used in coffee-cup calorimeter.

Correct: A bomb calorimeter is a rigid sealed steel vessel where ΔV = 0 (constant volume, w = 0, qv = ΔU).

Incorrect: Temperature changes during combustion.

Incorrect: Heat is exchanged between bomb and water bath.

10 By IUPAC convention, the standard enthalpy of formation (Δf H°) is taken as zero for:

Incorrect: Compounds have non-zero enthalpies of formation.

Correct: By convention, Δf H° = 0 for elements in their standard reference states (e.g., O₂(g), C(graphite), S(rhombic)).

Incorrect: Water has Δf H° = −285.8 kJ/mol.

Incorrect: Gaseous C is an atomized state (ΔaH = +715 kJ/mol).

11 Hess's Law of constant heat summation is a direct consequence of:

Correct: Hess's Law follows from enthalpy being a state function, rooted in the First Law (conservation of energy).

Incorrect: Second law deals with entropy and spontaneity.

Incorrect: Third law deals with absolute entropy at 0 K.

Incorrect: Le Chatelier deals with equilibrium shifts.

12 The enthalpy of combustion (Δc H°) of any combustible substance is always:

Incorrect: Combustion releases energy.

Incorrect: Endothermic combustion does not sustain itself.

Correct: Combustion reactions always release heat to the surroundings (Δc H° < 0).

Incorrect: Bond enthalpy can be positive or negative depending on context.

13 For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), Δr H° = −92.4 kJ. The standard enthalpy of formation of NH₃(g) is:

Incorrect: That is for forming 2 moles of NH₃.

Incorrect: Formation is exothermic.

Correct: Δf H° is per mole of substance formed: Δf H° = −92.4 / 2 = −46.2 kJ mol⁻¹.

Incorrect: Wrong sign and magnitude.

14 What is the lattice enthalpy of an ionic crystal like NaCl?

Incorrect: That is enthalpy of fusion/solidification.

Correct: Lattice enthalpy is the enthalpy required to dissociate 1 mole of solid ionic compound into free gaseous ions (NaCl(s) → Na⁺(g) + Cl⁻(g), +788 kJ/mol).

Incorrect: Melting produces liquid ions, not gaseous ions.

Incorrect: It involves sublimation and dissociation too.

15 In which of the following physical changes does entropy decrease?

Incorrect: Melting increases molecular randomness (ΔS > 0).

Incorrect: Vaporisation produces highly disordered gas (ΔS > 0).

Correct: Freezing/crystallisation locks molecules into an ordered lattice, decreasing entropy (ΔS < 0).

Incorrect: Sublimation creates gas from solid (ΔS > 0).

16 For a reversible process, the entropy change (ΔS) is calculated as:

Correct: ΔS = q_rev / T (units: J K⁻¹ mol⁻¹).

Incorrect: Entropy is inversely proportional to temperature.

Incorrect: Work is not the randomising thermal term.

Incorrect: Dimensionally incorrect.

17 According to the Second Law of Thermodynamics, for any spontaneous process:

Incorrect: System entropy alone can be negative (e.g., freezing) if surroundings increase more.

Incorrect: Surroundings alone is not the sole requirement.

Correct: Total entropy of the universe must strictly increase (ΔS_total > 0).

Incorrect: ΔS_total = 0 occurs only at equilibrium.

18 The Third Law of Thermodynamics states that:

Incorrect: That is the First Law.

Correct: At 0 K, constituent particles in a pure crystal are in perfect order with zero randomness (lim T→0 S = 0).

Incorrect: That is the Second Law.

Incorrect: Absolute zero is unattainable.

19 Under constant temperature and pressure, the criterion for a process to be spontaneous is:

Incorrect: ΔG > 0 means non-spontaneous.

Incorrect: ΔG = 0 indicates equilibrium.

Correct: A process is spontaneous if Gibbs free energy decreases (ΔG < 0).

Incorrect: Many endothermic reactions are spontaneous if TΔS > ΔH.

20 If a reaction has ΔH > 0 and ΔS > 0, it will be spontaneous at:

Incorrect: At low T, ΔH dominates making ΔG positive.

Correct: At high temperatures, the −TΔS term becomes large and negative, overcoming positive ΔH so ΔG < 0.

Incorrect: It is non-spontaneous at low T.

Incorrect: It can be spontaneous when T > ΔH/ΔS.

21 A reaction with ΔH < 0 and ΔS > 0 is:

Incorrect: It doesn't need high temperature.

Incorrect: It is spontaneous at high temperatures too.

Correct: ΔG = ΔH − TΔS will always be negative regardless of temperature (negative minus positive).

Incorrect: It is always spontaneous.

22 The standard Gibbs energy change (Δr G°) is related to the equilibrium constant (K) by:

Incorrect: Missing negative sign.

Correct: Δr G° = −RT ln K = −2.303 RT log K.

Incorrect: Natural log requires 2.303 multiplier for base-10 log.

Incorrect: Incorrect functional form.

23 If the equilibrium constant K for a reaction is much greater than 1 (K >> 1), then Δr G° must be:

Incorrect: Δr G° = 0 when K = 1.

Incorrect: Positive Δr G° gives K < 1.

Correct: When K > 1, log K is positive, so Δr G° = −2.303 RT log K is negative.

Incorrect: Δr G° depends on entropy as well.

24 For the dissociation reaction Cl₂(g) → 2Cl(g), the signs of ΔH and ΔS are:

Correct: Bond breaking absorbs energy (ΔH > 0) and 1 molecule gives 2 free atoms, increasing disorder (ΔS > 0).

Incorrect: Reverse reaction (bond formation) has negative ΔH and ΔS.

Incorrect: Entropy increases.

Incorrect: Bond breaking is never exothermic.

25 A system absorbs 701 J of heat and does 394 J of work. The change in internal energy (ΔU) is:

Incorrect: Work done BY system is negative.

Correct: ΔU = q + w = (+701 J) + (−394 J) = +307 J.

Incorrect: Heat absorbed is positive.

Incorrect: Neither value is completely subtracted.

Quiz Results

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NCERT Textbook Exercises (5.1 to 5.22 Solved)

Complete step-by-step solutions for all 22 end-of-chapter questions from CBSE / NCERT Class 11 Chemistry Chapter 5 (Thermodynamics).

5.1

Choose the correct answer. A thermodynamic state function is a quantity: (i) used to determine heat changes, (ii) whose value is independent of path, (iii) used to determine pressure volume work, (iv) whose value depends on temperature only.

A state function is a thermodynamic property whose numerical value depends solely upon the initial and final states of the system and is completely independent of the path or mechanism taken to reach that state (e.g., $U, H, S, G, p, V, T$).

Correct Answer: (ii) whose value is independent of path.

5.2

For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT = 0, (ii) Δp = 0, (iii) q = 0, (iv) w = 0.

An adiabatic process is defined as a process in which no heat enters or leaves the system across its boundary ($q = 0$). Under these conditions, any work done changes internal energy directly ($\Delta U = w_{\text{ad}}$).

Correct Answer: (iii) q = 0.

5.3

The enthalpies of all elements in their standard states are: (i) unity, (ii) zero, (iii) < 0, (iv) different for each element.

By international thermodynamic convention, the standard molar enthalpy of formation ($\Delta_f H^\circ$) of an element in its most stable reference state of aggregation at $298.15\text{ K}$ and $1\text{ bar}$ pressure is arbitrarily assigned as zero (e.g., $O_{2(g)}, C_{\text{graphite}}, S_{\text{rhombic}}, Fe_{(s)}$).

Correct Answer: (ii) zero.

5.4

ΔU° of combustion of methane is −X kJ mol⁻¹. The value of ΔH° is: (i) = ΔU°, (ii) > ΔU°, (iii) < ΔU°, (iv) = 0.

Write the balanced thermochemical equation for the combustion of methane: $$CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$$ Calculate the change in moles of gaseous species ($\Delta n_g$): $$\Delta n_g = n_p(g) - n_r(g) = 1 - (1 + 2) = 1 - 3 = \mathbf{-2}$$ Now relate $\Delta H^\circ$ and $\Delta U^\circ$: $$\Delta H^\circ = \Delta U^\circ + \Delta n_g RT = \Delta U^\circ + (-2)RT = \Delta U^\circ - 2RT$$ Since $2RT > 0$, subtracting a positive term makes $\Delta H^\circ$ more negative than $\Delta U^\circ$: $$\mathbf{\Delta H^\circ < \Delta U^\circ}$$

Correct Answer: (iii) < ΔU°.

5.5

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are −890.3 kJ mol⁻¹, −393.5 kJ mol⁻¹, and −285.8 kJ mol⁻¹ respectively. Enthalpy of formation of CH₄(g) will be: (i) −74.8 kJ mol⁻¹, (ii) −52.27 kJ mol⁻¹, (iii) +74.8 kJ mol⁻¹, (iv) +52.26 kJ mol⁻¹.

The formation equation of methane is: $$C_{\text{graphite}} + 2H_{2(g)} \longrightarrow CH_{4(g)} \qquad \Delta_f H^\circ = ?$$ Using standard enthalpies of combustion: $$\Delta_f H^\circ(CH_4) = \Delta_c H^\circ(C) + 2\Delta_c H^\circ(H_2) - \Delta_c H^\circ(CH_4)$$ $$= (-393.5) + 2(-285.8) - (-890.3)$$ $$= -393.5 - 571.6 + 890.3 = -965.1 + 890.3 = \mathbf{-74.8\text{ kJ mol}^{-1}}$$

Correct Answer: (i) −74.8 kJ mol⁻¹.

5.6

A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be: (i) possible at high temperature, (ii) possible only at low temperature, (iii) not possible at any temperature, (iv) possible at any temperature.

1. The reaction releases heat ($+q$) $\implies$ it is exothermic ($\mathbf{\Delta H < 0}$).
2. The problem states that entropy change is positive ($\mathbf{\Delta S > 0}$).
Using the Gibbs equation: $$\Delta G = \Delta H - T \Delta S$$ Since $\Delta H$ is negative and $(-T\Delta S)$ is also negative at any absolute temperature ($T > 0\text{ K}$), the sum $\Delta G$ is strictly negative ($\Delta G < 0$) at all temperatures.

Correct Answer: (iv) possible at any temperature (spontaneous at all T).

5.7

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

According to the First Law of Thermodynamics: $$\Delta U = q + w$$ Applying IUPAC sign conventions:

Heat absorbed by the system: $q = \mathbf{+701\text{ J}}$
Work done by the system: $w = \mathbf{-394\text{ J}}$

$$\Delta U = (+701\text{ J}) + (-394\text{ J}) = \mathbf{+307\text{ J}}$$

ΔU = +307 J (Internal energy of the system increases by 307 J).

5.8

The reaction of cyanamide, NH₂CN(s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be −742.7 kJ mol⁻¹ at 298 K. Calculate enthalpy change for the reaction at 298 K: NH₂CN(s) + 3/2 O₂(g) → N₂(g) + CO₂(g) + H₂O(l).

Balanced reaction: $$NH_2CN_{(s)} + \frac{3}{2}O_{2(g)} \longrightarrow N_{2(g)} + CO_{2(g)} + H_2O_{(l)}$$ Find $\Delta n_g$ (only counting gaseous species): $$\Delta n_g = n_p(g) - n_r(g) = (1 + 1) - \frac{3}{2} = 2 - 1.5 = \mathbf{+0.5\text{ mol}}$$ Given: $\Delta U = -742.7\text{ kJ mol}^{-1}$, $T = 298\text{ K}$, $R = 8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1}$.
Now calculate $\Delta H$: $$\Delta H = \Delta U + \Delta n_g RT$$ $$= -742.7\text{ kJ} + [0.5 \times (8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1}) \times 298\text{ K}]$$ $$= -742.7 + 1.239 = \mathbf{-741.46\text{ kJ mol}^{-1}} \approx \mathbf{-741.5\text{ kJ mol}^{-1}}$$

ΔH = −741.5 kJ mol⁻¹.

5.9

Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35 °C to 55 °C. Molar heat capacity of Al is 24 J mol⁻¹ K⁻¹.

1. Moles of aluminium ($n$): $$n = \frac{\text{Mass}}{\text{Molar mass of Al}} = \frac{60.0\text{ g}}{27.0\text{ g mol}^{-1}} = 2.222\text{ mol}$$ 2. Temperature change ($\Delta T$): $$\Delta T = 55^\circ\text{C} - 35^\circ\text{C} = 20\text{ K}$$ 3. Heat required ($q$): $$q = n \cdot C_m \cdot \Delta T = 2.222\text{ mol} \times 24\text{ J mol}^{-1}\text{K}^{-1} \times 20\text{ K} = 1066.67\text{ J} = \mathbf{1.07\text{ kJ}}$$

q = 1.07 kJ (1067 J).

5.10

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 °C to ice at −10.0 °C. Δ_fus H = 6.03 kJ mol⁻¹ at 0 °C. Cp[H₂O(l)] = 75.3 J mol⁻¹ K⁻¹, Cp[H₂O(s)] = 36.8 J mol⁻¹ K⁻¹.

The transformation occurs in 3 consecutive thermodynamic steps:

Cooling liquid water from $10.0^\circ\text{C}$ to $0^\circ\text{C}$ ($\Delta T = -10\text{ K}$): $$\Delta H_1 = n \cdot C_p(l) \cdot \Delta T = 1\text{ mol} \times 75.3\text{ J K}^{-1}\text{mol}^{-1} \times (-10\text{ K}) = -753\text{ J} = \mathbf{-0.753\text{ kJ}}$$
Freezing water to ice at $0^\circ\text{C}$ (Phase change): $$\Delta H_2 = - \Delta_{\text{fus}}H = \mathbf{-6.03\text{ kJ}}$$
Cooling ice from $0^\circ\text{C}$ to $-10.0^\circ\text{C}$ ($\Delta T = -10\text{ K}$): $$\Delta H_3 = n \cdot C_p(s) \cdot \Delta T = 1\text{ mol} \times 36.8\text{ J K}^{-1}\text{mol}^{-1} \times (-10\text{ K}) = -368\text{ J} = \mathbf{-0.368\text{ kJ}}$$

Total enthalpy change: $$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = (-0.753) + (-6.03) + (-0.368) = \mathbf{-7.151\text{ kJ mol}^{-1}}$$

ΔH = −7.151 kJ mol⁻¹ (Heat released = 7.151 kJ).

5.11

Enthalpy of combustion of carbon to CO₂ is −393.5 kJ mol⁻¹. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

Formation equation: $$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} \qquad \Delta_r H = -393.5\text{ kJ mol}^{-1}$$ Molar mass of $CO_2 = 12 + 2(16) = 44\text{ g mol}^{-1}$.
Moles of $CO_2$ formed: $$n = \frac{35.2\text{ g}}{44\text{ g mol}^{-1}} = \mathbf{0.80\text{ mol}}$$ Heat released upon formation of $0.80\text{ mol}$ of $CO_2$: $$q = 0.80\text{ mol} \times 393.5\text{ kJ mol}^{-1} = \mathbf{314.8\text{ kJ}}$$

Heat released = 314.8 kJ (or ΔH = −314.8 kJ).

5.12

Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are −110, −393, 81 and 9.7 kJ mol⁻¹ respectively. Find the value of Δ_r H for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g).

Reaction: $$N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$$ Using formula: $\Delta_r H = \sum \Delta_f H(\text{products}) - \sum \Delta_f H(\text{reactants})$: $$\Delta_r H = [\Delta_f H(N_2O) + 3\Delta_f H(CO_2)] - [\Delta_f H(N_2O_4) + 3\Delta_f H(CO)]$$ $$= [81 + 3(-393)] - [9.7 + 3(-110)]$$ $$= [81 - 1179] - [9.7 - 330]$$ $$= -1098 - (-320.3) = -1098 + 320.3 = \mathbf{-777.7\text{ kJ mol}^{-1}}$$

Δ_r H = −777.7 kJ mol⁻¹.

5.13

Given N₂(g) + 3H₂(g) → 2NH₃(g); Δ_r H° = −92.4 kJ mol⁻¹. What is the standard enthalpy of formation of NH₃ gas?

By definition, the standard enthalpy of formation ($\Delta_f H^\circ$) is the enthalpy change for the formation of 1 mole of the compound from its constituent elements in their reference states: $$\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \longrightarrow NH_{3(g)} \qquad \Delta_f H^\circ = \frac{\Delta_r H^\circ}{2}$$ $$\Delta_f H^\circ(NH_3, g) = \frac{-92.4\text{ kJ mol}^{-1}}{2} = \mathbf{-46.2\text{ kJ mol}^{-1}}$$

Δ_f H°[NH₃(g)] = −46.2 kJ mol⁻¹.

5.14

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: (1) CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l); Δ_r H° = −726 kJ mol⁻¹; (2) C(graphite) + O₂(g) → CO₂(g); Δ_c H° = −393 kJ mol⁻¹; (3) H₂(g) + 1/2 O₂(g) → H₂O(l); Δ_f H° = −286 kJ mol⁻¹.

Target formation reaction: $$C_{\text{graphite}} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow CH_3OH_{(l)} \qquad \Delta_f H^\circ = ?$$ Using Hess's Law: $$\Delta_f H^\circ(CH_3OH) = \text{Eq}(2) + 2 \times \text{Eq}(3) - \text{Eq}(1)$$ $$= (-393) + 2(-286) - (-726)$$ $$= -393 - 572 + 726 = -965 + 726 = \mathbf{-239\text{ kJ mol}^{-1}}$$

Δ_f H°[CH₃OH(l)] = −239 kJ mol⁻¹.

5.15

Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4Cl(g) and calculate bond enthalpy of CCl in CCl₄(g). Given: Δ_vap H(CCl₄) = 30.5 kJ mol⁻¹; Δ_f H(CCl₄) = −135.5 kJ mol⁻¹; Δ_a H(C) = 715.0 kJ mol⁻¹; Δ_a H(Cl₂) = 242 kJ mol⁻¹.

The atomization of gaseous $CCl_4$ is: $$CCl_{4(g)} \longrightarrow C_{(g)} + 4Cl_{(g)} \qquad \Delta_a H = ?$$ Formation of $CCl_{4(l)}$: $$C_{(s)} + 2Cl_{2(g)} \longrightarrow CCl_{4(l)} \qquad \Delta_f H^\circ = -135.5\text{ kJ mol}^{-1}$$ Enthalpy of formation of $CCl_{4(g)}$: $$\Delta_f H^\circ(CCl_{4, g}) = \Delta_f H^\circ(CCl_{4, l}) + \Delta_{\text{vap}}H = -135.5 + 30.5 = \mathbf{-105.0\text{ kJ mol}^{-1}}$$ Now express $\Delta_f H^\circ(CCl_{4, g})$ in terms of atomization energies: $$\Delta_f H^\circ(CCl_{4, g}) = \Delta_a H(C) + 2\Delta_a H(Cl_2) - \Delta_a H(CCl_{4, g})$$ $$-105.0 = 715.0 + 2(242) - \Delta_a H(CCl_{4, g})$$ $$-105.0 = 715.0 + 484.0 - \Delta_a H(CCl_{4, g}) = 1199.0 - \Delta_a H(CCl_{4, g})$$ $$\mathbf{\Delta_a H(CCl_{4, g}) = 1199.0 + 105.0 = 1304.0\text{ kJ mol}^{-1}}$$ Since $CCl_4$ has 4 identical $C-Cl$ bonds: $$\text{Bond Enthalpy of } C-Cl = \frac{\Delta_a H}{4} = \frac{1304.0}{4} = \mathbf{326.0\text{ kJ mol}^{-1}}$$

Enthalpy change for atomization = 1304 kJ mol⁻¹; CCl Bond Enthalpy = 326 kJ mol⁻¹.

5.16

For an isolated system, ΔU = 0, what will be ΔS?

In an isolated system, there is no exchange of energy or matter with the surroundings ($q = 0, w = 0 \implies \Delta U = 0$).
According to the Second Law of Thermodynamics, any spontaneous change within an isolated system is accompanied by an increase in randomness/disorder: $$\mathbf{\Delta S > 0} \qquad (\text{for a spontaneous process})$$ At thermodynamic equilibrium: $$\mathbf{\Delta S = 0}$$

ΔS > 0 for a spontaneous process (and ΔS = 0 at equilibrium).

5.17

For the reaction at 298 K, 2A + B → C, ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?

For a reaction to become spontaneous: $$\Delta G = \Delta H - T \Delta S < 0 \implies T \Delta S > \Delta H \implies T > \frac{\Delta H}{\Delta S}$$ At the equilibrium transition temperature ($T_{eq}$ where $\Delta G = 0$): $$T_{eq} = \frac{\Delta H}{\Delta S} = \frac{400\text{ kJ mol}^{-1}}{0.2\text{ kJ K}^{-1}\text{mol}^{-1}} = \mathbf{2000\text{ K}}$$ Since both $\Delta H$ and $\Delta S$ are positive, the reaction will be spontaneous at all temperatures above 2000 K ($T > 2000\text{ K}$).

The reaction becomes spontaneous at temperatures above 2000 K (T > 2000 K).

5.18

For the reaction 2Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?

1. Sign of $\Delta H$: Bond formation is always an exothermic process because the bonded state is at a lower potential energy than isolated free atoms $\implies \mathbf{\Delta H < 0}$ (negative).
2. Sign of $\Delta S$: Two moles of separate gaseous atoms combine to form one mole of diatomic molecules ($2\text{ mol} \rightarrow 1\text{ mol}$). The number of independently moving particles decreases, reducing randomness $\implies \mathbf{\Delta S < 0}$ (negative).

Both ΔH and ΔS are negative (ΔH < 0 and ΔS < 0).

5.19

For the reaction 2A(g) + B(g) → 2D(g), ΔU° = −10.5 kJ and ΔS° = −44.1 J K⁻¹. Calculate ΔG° for the reaction at 298 K, and predict whether the reaction may occur spontaneously.

1. Find $\Delta n_g$: $$\Delta n_g = 2 - (2 + 1) = 2 - 3 = \mathbf{-1}$$ 2. Calculate $\Delta H^\circ$: $$\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$$ $$= -10.5\text{ kJ} + [(-1) \times 8.314 \times 10^{-3}\text{ kJ K}^{-1}\text{mol}^{-1} \times 298\text{ K}]$$ $$= -10.5 - 2.478 = \mathbf{-12.98\text{ kJ}}$$ 3. Calculate $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$: $$\Delta G^\circ = -12.98\text{ kJ} - [298\text{ K} \times (-44.1 \times 10^{-3}\text{ kJ K}^{-1})]$$ $$= -12.98\text{ kJ} - (-13.14\text{ kJ}) = -12.98 + 13.14 = \mathbf{+0.16\text{ kJ}} \quad (\mathbf{+160\text{ J}})$$ Since $\Delta G^\circ$ is positive ($\Delta G^\circ > 0$), the reaction is non-spontaneous at $298\text{ K}$.

ΔG° = +0.16 kJ (+160 J). The reaction is non-spontaneous at 298 K.

5.20

The equilibrium constant for a reaction is 10. What will be the value of ΔG°? (R = 8.314 J K⁻¹ mol⁻¹, T = 300 K).

Use the relation between standard free energy and equilibrium constant: $$\Delta_r G^\circ = - 2.303 \, RT \log K$$ Substitute the given values: $$\Delta_r G^\circ = - 2.303 \times 8.314\text{ J K}^{-1}\text{mol}^{-1} \times 300\text{ K} \times \log(10)$$ Since $\log(10) = 1$: $$\Delta_r G^\circ = - 2.303 \times 8.314 \times 300 \times 1 = \mathbf{-5744.14\text{ J mol}^{-1}} = \mathbf{-5.744\text{ kJ mol}^{-1}}$$

ΔG° = −5.744 kJ mol⁻¹ (−5744 J mol⁻¹).

5.21

Comment on the thermodynamic stability of NO(g), given: 1/2 N₂(g) + 1/2 O₂(g) → NO(g); Δ_r H° = 90 kJ mol⁻¹ and NO(g) + 1/2 O₂(g) → NO₂(g); Δ_r H° = −74 kJ mol⁻¹.

1. Formation of $NO_{(g)}$: The enthalpy of formation of $NO_{(g)}$ is positive ($\Delta_f H^\circ = +90\text{ kJ mol}^{-1}$). This means $NO_{(g)}$ is an endothermic compound and is thermodynamically unstable relative to its constituent elements ($N_2$ and $O_2$), tending to decompose back into elements under standard conditions.
2. Oxidation to $NO_{2(g)}$: The reaction of $NO_{(g)}$ with $O_2$ to form $NO_2$ is exothermic ($\Delta_r H^\circ = -74\text{ kJ mol}^{-1}$). This indicates that $NO_{(g)}$ readily undergoes further oxidation to form $NO_2$, showing that $NO_2$ is more stable than $NO$.

NO(g) is thermodynamically unstable with respect to its elements (positive Δ_f H = +90 kJ/mol), but NO₂(g) is relatively more stable.

5.22

Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δ_f H° = −286 kJ mol⁻¹.

The formation of water is exothermic: $$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)} \qquad \Delta_r H^\circ = -286\text{ kJ mol}^{-1} = -286 \times 10^3\text{ J mol}^{-1}$$ Heat given off by the system is absorbed by the surroundings under reversible conditions at $T = 298\text{ K}$: $$q_{\text{surr}} = - \Delta_r H^\circ = -(-286 \times 10^3\text{ J}) = \mathbf{+286 \times 10^3\text{ J}}$$ Entropy change of the surroundings ($\Delta S_{\text{surr}}$): $$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = \frac{+286 \times 10^3\text{ J mol}^{-1}}{298\text{ K}} = \mathbf{+959.73\text{ J K}^{-1}\text{mol}^{-1}} \approx \mathbf{+960\text{ J K}^{-1}\text{mol}^{-1}}$$

ΔS_surr = +959.7 J K⁻¹ mol⁻¹ (or +960 J K⁻¹ mol⁻¹).

Revision Cheat-Sheets & Core Formulations

Quick-reference formula cards, spontaneity criteria tables, thermodynamic laws summaries, and high-yield mnemonics for Thermodynamics.

Master Formula Cheat-Sheet: Thermodynamics

Physical Quantity / Law	Mathematical Formula	Conditions & Conventions
First Law of Thermodynamics	$$\Delta U = q + w$$	$q>0$ (absorbed), $w>0$ (work done on system)
Pressure-Volume Work	$$w = - p_{\text{ex}} \Delta V$$	Single-step irreversible expansion/compression
Isothermal Reversible Work	$$w_{\text{rev}} = - 2.303 \, nRT \log \frac{V_f}{V_i} = - 2.303 \, nRT \log \frac{p_i}{p_f}$$	Ideal gas, constant $T$
Free Expansion (in vacuum)	$$w = 0, \quad q = 0, \quad \Delta U = 0$$	$p_{\text{ex}} = 0$ (vacuum), isothermal ideal gas
Enthalpy Definition	$$H = U + pV \implies \Delta H = q_p$$	$q_p = \Delta H$ (at constant pressure)
Relation between $\Delta H$ and $\Delta U$	$$\Delta H = \Delta U + \Delta n_g RT$$	$\Delta n_g = n_p(g) - n_r(g)$
Heat Capacity at Const. Vol.	$$C_v = \left(\frac{\partial U}{\partial T}\right)_V \implies q_v = \Delta U = n C_v \Delta T$$	Constant volume ($w=0$)
Heat Capacity at Const. Press.	$$C_p = \left(\frac{\partial H}{\partial T}\right)_p \implies q_p = \Delta H = n C_p \Delta T$$	Constant pressure
Mayer's Relation	$$C_p - C_v = R$$	Per mole of an ideal gas
Hess's Law of Heat Summation	$$\Delta_r H^\circ = \sum a_i \Delta_f H^\circ(\text{products}) - \sum b_i \Delta_f H^\circ(\text{reactants})$$	Standard enthalpies of formation
Reaction Enthalpy from Bonds	$$\Delta_r H^\circ = \sum \text{B.E.}(\text{reactants}) - \sum \text{B.E.}(\text{products})$$	Valid strictly for gaseous species
Lattice Enthalpy (Born-Haber)	$$\Delta_f H^\circ = \Delta_{\text{sub}}H + \Delta_i H + \frac{1}{2}\Delta_{\text{bond}}H + \Delta_{eg}H - \Delta_{\text{lattice}}H$$	Cycle sums to zero
Entropy Change (Reversible)	$$\Delta S = \frac{q_{\text{rev}}}{T}$$	Units: $\text{J K}^{-1}\text{mol}^{-1}$
Second Law (Spontaneity)	$$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$$	$\Delta S_{\text{total}} = 0$ at equilibrium
Surroundings Entropy Change	$$\Delta S_{\text{surr}} = \frac{-\Delta_r H}{T}$$	Thermal equilibrium with surroundings
Gibbs Free Energy	$$\Delta G = \Delta H - T \Delta S$$	Constant $T$ and $p$
Gibbs Energy & Equilibrium	$$\Delta_r G^\circ = - RT \ln K = - 2.303 \, RT \log K$$	$\Delta G = 0$ at equilibrium

Spontaneity Decision Matrix ($\Delta G = \Delta H - T\Delta S$)

$\Delta H$	$\Delta S$	Sign of $\Delta G$	Reaction Feasibility & Effect of Temperature	Key Physical Example
Negative (−)	Positive (+)	Always Negative (−)	Spontaneous at all temperatures	$2H_2O_{2(l)} \rightarrow 2H_2O_{(l)} + O_{2(g)}$
Negative (−)	Negative (−)	Negative at low $T$; Positive at high $T$	Spontaneous at low $T$ ($T < \Delta H/\Delta S$)	Freezing of water ($T < 273\text{ K}$), synthesis of $NH_3$
Positive (+)	Positive (+)	Positive at low $T$; Negative at high $T$	Spontaneous at high $T$ ($T > \Delta H/\Delta S$)	Melting of ice ($T > 273\text{ K}$), $CaCO_3$ decomposition
Positive (+)	Negative (−)	Always Positive (+)	Non-spontaneous at all temperatures	$3O_{2(g)} \rightarrow 2O_{3(g)}$ at standard conditions

🧠 Thermodynamics Mnemonics & Memory Hooks

Concept	Mnemonic / Memory Hook	Core Scientific Fact
First Law Signs	"Positive In, Negative Out" (PINO)	Heat added ($+q$) or work on ($+w$) increases internal energy ($U$)
Intensive Properties	"P-D-T are Free of Quantity"	Pressure, Density, and Temperature do NOT depend on size/mass
Extensive Properties	"Big Volume, Big Mass, Big Energy"	Volume, Mass, $U, H, S, G$ double when the system size doubles
Mayer's Equation	"Pressure has More Room"	$C_p > C_v$ because gas does work when expanding at constant pressure
Hess's Law Bonds	"Reactants Break, Products Make" (RB − PM)	$\Delta_r H = \sum \text{B.E.}(\text{reactants}) - \sum \text{B.E.}(\text{products})$
Bomb Calorimeter	"Bomb is Rigid, Volume is Frigid"	$\Delta V = 0 \implies w = 0 \implies q_v = \Delta U$
Spontaneity Rule	"Down with Energy, Up with Entropy"	Nature favors minimum $\Delta H$ (exothermic) and maximum $\Delta S$ (chaos)
Equilibrium Criterion	"Zero Free Energy at Peace"	At dynamic equilibrium, $\Delta G = 0$ and $\Delta S_{\text{total}} = 0$

Chapter Tests (3 Difficulty Levels)

Benchmark your mastery across Foundation, Intermediate, and Advanced tiers with instant scoring and detailed feedback.

L1.1 Which of the following represents a closed system?

Correct: Heat can be conducted through copper, but matter cannot escape the sealed flask.

Incorrect: That is an open system (steam escapes).

Incorrect: That approximates an isolated system.

Incorrect: Humans exchange matter and energy (open system).

L1.2 Which of the following is NOT a state function?

Incorrect: U is a state function.

Incorrect: H is a state function.

Correct: Work and heat are path functions; their values depend on the pathway taken.

Incorrect: S is a state function.

L1.3 The enthalpy change of an exothermic reaction is always:

Incorrect: Positive enthalpy change indicates an endothermic reaction.

Correct: Heat is evolved to the surroundings, so ΔH < 0.

Incorrect: Zero means no thermal change.

Incorrect: ΔH and ΔS have different physical dimensions.

L1.4 What is the value of standard enthalpy of formation for pure O₂(g) at 298 K?

Incorrect: That is the bond enthalpy of O=O.

Incorrect: That is Δ_f H° of liquid water.

Incorrect: Formation enthalpy of elements is not unity.

Correct: By convention, Δ_f H° = 0 for any pure element in its standard reference state.

L1.5 At absolute zero temperature (0 K), the entropy of a perfectly crystalline solid is:

Correct: By the Third Law of Thermodynamics, perfect order exists at 0 K so S = 0.

Incorrect: High temperature increases entropy towards large values.

Incorrect: Absolute entropy cannot be negative.

Incorrect: S and H are distinct physical entities.

Result

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L2.1 For which of the following gaseous reactions is ΔH strictly equal to ΔU?

Incorrect: Δng = 2 − 4 = −2.

Correct: Δng = 2 − (1 + 1) = 0. Therefore ΔH = ΔU + 0 = ΔU.

Incorrect: Δng = 2 − 1 = +1.

Incorrect: Δng = 2 − 3 = −1.

L2.2 A gas expands from 10 L to 20 L against a constant external pressure of 2 atm. The work done on the gas is:

Incorrect: Expansion work is negative.

Incorrect: p_ex is 2 atm, not vacuum.

Incorrect: ΔV = 10 L, p = 2 atm.

Correct: w = −p_ex·ΔV = −2 atm × (20 − 10) L = −20 L-atm = −20 × 101.325 J = −2026.5 J.

L2.3 In the reaction C(s) + H₂O(g) → CO(g) + H₂(g), what is the sign of ΔS?

Correct: 1 mole of gas reacts to form 2 moles of gas (Δng = +1), substantially increasing randomness.

Incorrect: Gaseous moles increase, not decrease.

Incorrect: Gas phase production creates high entropy change.

Incorrect: Physical state analysis predicts ΔS cleanly.

L2.4 What is the temperature at which a reaction with ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹ reaches equilibrium?

Incorrect: Watch the unit conversion between kJ and J.

Incorrect: ΔH must be in Joules (30,000 J).

Correct: At equilibrium ΔG = 0 ⇒ T = ΔH / ΔS = 30,000 J / 100 J K⁻¹ = 300 K.

Incorrect: 30,000 / 100 = 300 K.

L2.5 If the equilibrium constant K for a reaction is equal to 1, then the standard Gibbs energy change (Δ_r G°) is:

Incorrect: Positive when K < 1.

Correct: Since Δ_r G° = −RT ln K and ln(1) = 0, Δ_r G° = 0.

Incorrect: Negative when K > 1.

Incorrect: Standard free energy is finite.

Result

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L3.1 Why is the rusting of iron (4Fe + 3O₂ → 2Fe₂O₃) spontaneous at room temperature even though ΔS_sys is −549.4 J K⁻¹ mol⁻¹?

Correct: ΔS_surr = −ΔH/T = +5530 J/K. Total entropy change is positive (+4980.6 J/K), driving spontaneity.

Incorrect: Energy changes are substantial.

Incorrect: Rusting is strongly exothermic.

Incorrect: System entropy is always included in ΔS_total.

L3.2 For 1 mole of an ideal gas undergoing isothermal reversible expansion, which set of conditions is correct?

Incorrect: That is free expansion. In reversible expansion work is done.

Incorrect: Isothermal means ΔU = 0.

Correct: ΔU = 0 for isothermal ideal gas, so heat absorbed exactly equals work done by the gas.

Incorrect: For isothermal ideal gas ΔH = ΔU = 0.

L3.3 Using Born-Haber cycle data: Δ_sub H(Na) = +108.4, Δi H(Na) = +496, ½ Δ_bond H(Cl₂) = +121, Δ_eg H(Cl) = −348.6, and Δ_f H°(NaCl) = −411.2 kJ/mol. What is the lattice enthalpy of NaCl(s)?

Incorrect: That is enthalpy of formation.

Correct: Δ_lattice H = Δ_sub H + Δi H + ½ Δ_bond H + Δ_eg H − Δ_f H = 108.4 + 496 + 121 − 348.6 − (−411.2) = +788 kJ/mol.

Incorrect: Lattice enthalpy for dissociation into gaseous ions is positive.

Incorrect: Arithmetic error.

L3.4 Why is Cp always greater than Cv for one mole of an ideal gas?

Incorrect: Speed depends purely on temperature.

Incorrect: Heat transfer occurs at constant volume (qv = ΔU).

Incorrect: H = U + pV, so H > U for positive pV.

Correct: At constant volume, w = 0 so all heat goes into ΔU. At constant pressure, extra heat is required to do mechanical expansion work.

L3.5 For the dissociation N₂O₄(g) ⇌ 2NO₂(g) at 60 °C (333 K) and 1 atm with 50% dissociation, Kp = 1.33 atm. What is Δ_r G°?

Correct: Δ_r G° = −2.303 RT log(1.33) = −2.303 × 8.314 × 333 × 0.1239 = −796 J/mol.

Incorrect: Kp > 1 leads to negative ΔG°.

Incorrect: Joules were mislabeled as kJ.

Incorrect: Kp = 1.33 ≠ 1.

Result

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