✦ Updated March 2026  ·  JEE · NEET · Boards

Thermodynamics — Class 11 Chapter 5 · Complete NCERT Exercise Solutions

Every question fully stated · Every solution fully explained · IUPAC sign convention · Board + JEE + NEET aligned

22 Questions Solved First Law & ΔU Enthalpy & Hess's Law Entropy & Gibbs Energy Bond Enthalpy Spontaneity
Academia Aeternum

Academia Aeternum Editorial Team

Reviewed by Subject Experts · Exam-Oriented Format · Updated March 2026

Exam Relevance

Why NCERT Thermodynamics Exercise Is Exam-Critical

Thermodynamics contributes 3–5 marks in CBSE Boards and 2–3 questions in JEE Main every year. The Laws of Thermodynamics — energy conservation (First Law), entropy (Second Law), and absolute zero (Third Law) — form the backbone of all physical chemistry. The First Law (ΔU = q + w) alone generates a large share of Board and competitive numericals. Every solution on this page is written so that a student can read it independently, without referring back to the textbook — all question data, all concepts required, and every single calculation step are included in full.

Core Topics

Topics Covered in This Exercise

🔥
First Law & Internal Energy (ΔU)

Internal energy, work done on/by system, heat exchange — IUPAC sign convention applied and explained in every numerical.

⚗️
Enthalpy & Hess's Law

ΔH calculations from formation, combustion & neutralisation enthalpies; multi-step thermochemical Hess's cycles explained from first principles.

📐
Entropy & Gibbs Free Energy

ΔS, spontaneity criteria, ΔG = ΔH − TΔS, and ΔG° = −RT ln K — the critical link between thermodynamics and equilibrium.

🔗
Bond Enthalpy

Bonds broken vs bonds formed — a top-scoring topic in NEET & JEE. Full atomisation cycle shown with Hess's Law.

⚠️
Common Exam Traps

Sign convention errors, J vs kJ unit mistakes, isothermal vs adiabatic confusion — flagged and explained wherever they arise.

📊
PYQ Pattern Mapping

Each solution cross-referenced with Board, JEE Main & NEET question trends. Updated March 2026.

How to Study

Maximising Your Marks

📌 CBSE Board Exams

Board Examination Strategy

  • Follow IUPAC sign convention strictly — examiners deduct marks for sign errors
  • Show every unit conversion explicitly as a separate numbered step
  • Write out thermochemical equations in full before applying Hess's Law
  • State the formula first, then substitute values — never skip the formula step
⚡ JEE Main & NEET

Competitive Examination Strategy

  • Memorise the ΔH / ΔS sign matrix for all four spontaneity combinations
  • Link ΔG° = −RT ln K directly with Chapter 7 (Chemical Equilibrium)
  • Practise integer-type reformulations of every numerical given here
  • For MCQs, eliminate options using dimensional analysis before calculating
SOLUTIONS
NCERT Chapter 5 Exercise

All 22 Questions — Fully Solved

Click any question to expand its complete solution. Every question is self-contained — no textbook needed.

01

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
MCQ State Functions Conceptual
Background Concept

What Is a Thermodynamic State Function?

In thermodynamics, we study the properties of a system — its pressure, temperature, volume, internal energy, enthalpy, and so on. These properties can be divided into two types depending on how their values are determined.

A state function (also called a state variable or state property) is a thermodynamic quantity whose value depends only on the current state of the system, and not on the path or history by which the system arrived at that state. In other words, if we know the conditions (temperature, pressure, volume, composition) at any instant, we can determine the value of every state function completely — it does not matter how many steps were taken, what route was followed, or how much work or heat was involved along the way.

Mathematically, if a system moves from an initial state (1) to a final state (2), the change in any state function \(X\) is always given by:

\[ \Delta X = X_2 - X_1 \]

This expression shows clearly that \(\Delta X\) depends only on \(X_2\) (the final value) and \(X_1\) (the initial value). It is completely independent of the path taken between the two states. No matter how many steps, reversals, or detours occur between state 1 and state 2, \(\Delta X\) always comes out the same.

Common examples of state functions include: internal energy (U), enthalpy (H), entropy (S), Gibbs free energy (G), temperature (T), pressure (p), and volume (V).

In contrast, path functions are quantities whose values do depend on the specific route or process followed. Heat \(q\) and work \(w\) are the most important path functions in thermodynamics. The same initial and final states can be connected by different processes, and the heat absorbed and work done will differ depending on which process is chosen.

Analysis of Each Option

Evaluating Statements (i) through (iv)

✗ (i) "Used to determine heat changes" — This is incorrect. Heat (\(q\)) is itself a path function, not a state function. The amount of heat exchanged depends on how the process is carried out (isothermally, adiabatically, at constant pressure, etc.). A state function cannot be defined as something that determines a path function.
✓ (ii) "Whose value is independent of path" — This is the precise and correct definition of a thermodynamic state function. Since its change depends only on the initial and final states, it is inherently path-independent.
✗ (iii) "Used to determine pressure–volume work" — Pressure–volume work (\(w = -p_{\text{ext}}\Delta V\)) also depends on the process (reversible vs irreversible, isothermal vs adiabatic). Work is a path function, not a state function.
✗ (iv) "Whose value depends on temperature only" — This is incorrect. Most state functions depend on multiple variables. For example, internal energy U depends on temperature, pressure, and composition. Volume V depends on both temperature and pressure. No single state function depends on temperature alone (for a real system).
Answer
(ii) whose value is independent of path
02

Choose the correct answer. For the process to occur under adiabatic conditions, the correct condition is:

(i)   \(\Delta T = 0\)
(ii)   \(\Delta p = 0\)
(iii)   \(q = 0\)
(iv)   \(w = 0\)
MCQ Adiabatic Process First Law
Background Concept

Classification of Thermodynamic Processes

Thermodynamic processes are classified according to which variable is held constant during the change. Understanding these classifications is fundamental to applying the First Law correctly in different situations.

The First Law of Thermodynamics expresses the principle of conservation of energy for thermodynamic systems. It states that energy can neither be created nor destroyed, but it can be transformed from one form to another. For any thermodynamic process, the change in internal energy of the system equals the heat absorbed by the system plus the work done on the system:

\[ \Delta U = q + w \]

Here \(\Delta U\) is the change in internal energy, \(q\) is the heat exchanged with the surroundings (positive when absorbed, negative when released), and \(w\) is the work done (positive when done on the system, negative when done by the system).

The four standard process types are:

\[\begin{array}{ll} \textbf{Isothermal} & \Delta T = 0 \text{ (temperature constant)}\\ \textbf{Isobaric} & \Delta p = 0 \text{ (pressure constant)}\\ \textbf{Isochoric} & \Delta V = 0 \text{ (volume constant)}\\ \textbf{Adiabatic} & q = 0 \text{ (no heat exchange)} \end{array}\]

An adiabatic process is one in which the system is thermally isolated from its surroundings — no heat enters or leaves during the process. This can be achieved either by perfect thermal insulation (for slow processes) or by carrying out the process so rapidly that there is not enough time for heat transfer to occur.

Analysis of Each Option

Evaluating Each Condition

✗ (i) \(\Delta T = 0\) defines an isothermal process (temperature does not change), not an adiabatic process. An adiabatic process generally causes the temperature to change — in fact, a compressed gas heats up adiabatically, and an expanding gas cools down.
✗ (ii) \(\Delta p = 0\) defines an isobaric process (pressure does not change). For example, reactions in open beakers occur at constant atmospheric pressure.
✓ (iii) \(q = 0\) is exactly the defining condition of an adiabatic process. With no heat exchange, the First Law reduces to \(\Delta U = w\), meaning all internal energy changes arise from work done.
✗ (iv) \(w = 0\) defines a process in which no work is done — this corresponds to a constant-volume (isochoric) process when only PV work is considered (\(w = -p_{\text{ext}}\Delta V = 0\) when \(\Delta V = 0\)). It does not define an adiabatic process.

Consequence of the Adiabatic Condition

When \(q = 0\) is substituted into the First Law:

\[\begin{aligned} \Delta U &= q + w \\ \Delta U &= 0 + w \\ \Delta U &= w \end{aligned}\]

This means that in an adiabatic process, any change in internal energy of the system is entirely due to work done. If the surroundings do work on the system (compression), the internal energy increases and the temperature rises. If the system does work on the surroundings (expansion), the internal energy decreases and the temperature falls.

Answer
(iii) q = 0
03

Choose the correct answer. The enthalpies of all elements in their standard states are:

(i)   unity
(ii)   zero
(iii)   less than zero
(iv)   different for each element
MCQ Standard Enthalpy Convention
Background Concept

Standard State and Reference Enthalpy

The standard state of a substance is its most stable physical form at a specified temperature (usually 298 K) and at 1 bar pressure. For example, the standard state of carbon is graphite (not diamond), the standard state of oxygen is O₂(g), and the standard state of sodium is Na(s).

The standard enthalpy of formation, \(\Delta_f H^\Theta\), is defined as the enthalpy change when one mole of a compound is formed from its constituent elements, with each element in its standard state. This quantity serves as the thermochemical reference for all enthalpy calculations.

A reference point must be established for all thermodynamic calculations, just as sea level is used as the reference (zero) for altitude measurements. In thermochemistry, the agreed-upon convention is to assign a value of exactly zero to the standard enthalpy of formation of every element in its standard state. This is not measured — it is a defined convention that makes all thermochemical calculations consistent and universally comparable.

\[\Delta_f H^\Theta \bigl[\text{any element in its standard state}\bigr] = 0\]

Some examples of this convention:

\[\begin{aligned} \Delta_f H^\Theta[\mathrm{H_2(g)}] &= 0\\ \Delta_f H^\Theta[\mathrm{O_2(g)}] &= 0\\ \Delta_f H^\Theta[\mathrm{C(graphite)}] &= 0\\ \Delta_f H^\Theta[\mathrm{Na(s)}] &= 0\\ \Delta_f H^\Theta[\mathrm{Fe(s)}] &= 0 \end{aligned}\]
Analysis

Evaluating Each Option

✗ (i) Unity (= 1) — Incorrect. The standard enthalpies of elements are not assigned a value of one. "Unity" has no thermodynamic significance here.
✓ (ii) Zero — Correct. By internationally agreed convention (IUPAC), the standard enthalpy of formation of every element in its most stable standard state is exactly zero. This is the reference baseline for all thermochemical calculations.
✗ (iii) Less than zero — Incorrect. There is no reason why all elemental enthalpies should be negative; the value zero is assigned purely by convention, not by measurement.
✗ (iv) Different for each element — Incorrect. If each element had a different reference enthalpy, thermochemical tables would be inconsistent and calculations would be impossible. The uniform value of zero for all elements is the whole point of the convention.
Answer
(ii) zero
04

\(\Delta U^\Theta\) of combustion of methane is \(-X\ \mathrm{kJ\ mol^{-1}}\). The value of \(\Delta H^\Theta\) is:

(i)   \(= \Delta U^\Theta\)
(ii)   \(> \Delta U^\Theta\)
(iii)   \(< \Delta U^\Theta\)
(iv)   \(= 0\)
MCQ ΔH vs ΔU Gaseous Reactions
Background Concept

Relationship Between ΔH and ΔU for Reactions Involving Gases

Internal energy change (\(\Delta U\)) represents the total energy change of a system at constant volume — this is what a bomb calorimeter measures.

Enthalpy change (\(\Delta H\)) represents the heat change at constant pressure — this is what is measured in most open-system reactions (since they occur at atmospheric pressure).

For reactions in which gases are produced or consumed, the two quantities differ because the system must do extra work to push back the atmosphere when gas volume changes. The relationship between them is derived from the definition of enthalpy \(H = U + pV\):

\[\Delta H = \Delta U + \Delta(pV)\]

For ideal gases, \(pV = nRT\), so \(\Delta(pV) = \Delta n_g \cdot RT\), giving:

\[\boxed{\Delta H = \Delta U + \Delta n_g RT}\]

where \(\Delta n_g = n_g(\text{products}) - n_g(\text{reactants})\) counts only gaseous species (solids and liquids have negligible \(\Delta(pV)\) contribution and are ignored).

Step-by-Step Solution

Step 1: Write the Balanced Combustion Equation for Methane

\[\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}\]
Note: Water is produced as a liquid under standard conditions (298 K), not as a gas. This is critical — liquid water does not count as a gaseous product.

Step 2: Calculate Δn_g

Count only gaseous species on each side:

\[ \begin{aligned} n_g(\text{reactants}) &= n(\mathrm{CH_4}) + n(\mathrm{O_2}) = 1 + 2 = 3\\ n_g(\text{products}) &= n(\mathrm{CO_2}) \quad \text{(}\mathrm{H_2O}\text{ is liquid — excluded)}\\[6pt] \Delta n_g &= n_g(\text{products}) - n_g(\text{reactants}) = 1 - 3 = -2 \end{aligned} \]

Step 3: Apply the ΔH–ΔU Relationship

\[\begin{aligned} \Delta H &= \Delta U + \Delta n_g RT\\ \Delta H &= \Delta U + (-2)RT\\ \Delta H &= \Delta U - 2RT \end{aligned}\]

Step 4: Interpret the Result

Since \(R > 0\), \(T > 0\), and \(\Delta n_g = -2 < 0\), the correction term \(2RT\) is a positive number. We are subtracting it from \(\Delta U\):

\[\Delta H = \Delta U - 2RT \implies \Delta H < \Delta U\]

In numerical terms: \(\Delta U^\Theta = -X\ \mathrm{kJ\ mol^{-1}}\) (a negative number since combustion is exothermic). Subtracting \(2RT \approx 2 \times 0.008314 \times 298 \approx 4.96\ \mathrm{kJ}\) makes the result even more negative, i.e., \(\Delta H^\Theta\) is more negative than \(\Delta U^\Theta\). Therefore \(\Delta H^\Theta < \Delta U^\Theta\).

Answer
(iii) \(\Delta H^\Theta < \Delta U^\Theta\)
05

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, \(-890.3\ \mathrm{kJ\ mol^{-1}}\), \(-393.5\ \mathrm{kJ\ mol^{-1}}\), and \(-285.8\ \mathrm{kJ\ mol^{-1}}\) respectively. Enthalpy of formation of \(\mathrm{CH_4(g)}\) will be:

(i)   \(-74.8\ \mathrm{kJ\ mol^{-1}}\)
(ii)   \(-52.27\ \mathrm{kJ\ mol^{-1}}\)
(iii)   \(+74.8\ \mathrm{kJ\ mol^{-1}}\)
(iv)   \(+52.26\ \mathrm{kJ\ mol^{-1}}\)
MCQHess's LawEnthalpy of Formation
Background Concept

Enthalpy of Formation and Hess's Law

The standard enthalpy of formation, \(\Delta_f H^\Theta\), of a compound is the enthalpy change when one mole of that compound is formed from its constituent elements, each in their standard states at 298 K and 1 bar.

Direct measurement of \(\Delta_f H^\Theta\) is often impossible (e.g., we cannot simply mix carbon and hydrogen in a flask and get methane). Instead, we use Hess's Law, which states that the total enthalpy change of a chemical reaction is independent of the route taken — it depends only on the initial and final states. We can therefore add and subtract known thermochemical equations to derive an unknown one.

The formation reaction we want to establish is:

\[\mathrm{C(graphite) + 2H_2(g) \rightarrow CH_4(g)} \quad \Delta_f H^\Theta = \;?\]

The three combustion reactions we are given are:

\[\begin{aligned} \text{(A) } &\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)} &\Delta H_A = -890.3\ \mathrm{kJ}\\ \text{(B) } &\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)} &\Delta H_B = -393.5\ \mathrm{kJ}\\ \text{(C) } &\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)} &\Delta H_C = -285.8\ \mathrm{kJ} \end{aligned}\]
Step-by-Step Solution Using Hess's Law

Step 1: Identify What We Need

We need: C + 2H₂ → CH₄. We must combine reactions A, B, and C to obtain this.

Step 2: Decide How to Combine the Reactions

To get C on the left: use reaction (B) as written. ✓
To get 2H₂ on the left: use reaction (C) multiplied by 2. ✓
To get CH₄ on the left → but we need it on the right, so reverse reaction (A) and change the sign of its enthalpy.

Step 3: Write the Modified Reactions

\[\begin{aligned} \text{(B)} \quad &\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)} &\Delta H = -393.5\ \mathrm{kJ}\\ \text{2×(C)} \quad &\mathrm{2H_2(g) + O_2(g) \rightarrow 2H_2O(l)} &\Delta H = 2 \times (-285.8) = -571.6\ \mathrm{kJ}\\ \text{−(A)} \quad &\mathrm{CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)} &\Delta H = +890.3\ \mathrm{kJ} \end{aligned}\]

Step 4: Add the Three Equations

When we add these three equations, species appearing on both sides cancel:

\[\mathrm{C + \cancel{O_2} + 2H_2 + \cancel{O_2} + \cancel{CO_2} + \cancel{2H_2O} \rightarrow \cancel{CO_2} + \cancel{2H_2O} + CH_4 + \cancel{2O_2}}\]

The net equation is:

\[\mathrm{C(graphite) + 2H_2(g) \rightarrow CH_4(g)}\]

Step 5: Sum the Enthalpy Changes

\[\begin{aligned} \Delta_f H^\Theta(\mathrm{CH_4}) &= (-393.5) + (-571.6) + (+890.3)\\ &= -965.1 + 890.3\\ &= -74.8\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]
Answer
(i) \(-74.8\ \mathrm{kJ\ mol^{-1}}\)
06

A reaction, \(\mathrm{A + B \rightarrow C + D + q}\) 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
MCQGibbs EnergySpontaneity
Background Concept

Gibbs Free Energy and Spontaneity

The spontaneity of a chemical reaction at constant temperature and pressure is determined by the Gibbs free energy change, \(\Delta G\). This quantity elegantly combines the two thermodynamic driving forces — the tendency to minimise enthalpy (energy) and the tendency to maximise entropy (disorder).

\[\Delta G = \Delta H - T\Delta S\]

The spontaneity rules are:

\[\begin{array}{ll} \Delta G < 0 & \text{Reaction is spontaneous (thermodynamically feasible)}\\ \Delta G = 0 & \text{System is at equilibrium}\\ \Delta G > 0 & \text{Reaction is non-spontaneous (reverse is spontaneous)} \end{array}\]

The four possible combinations of \(\Delta H\) and \(\Delta S\) signs, and their spontaneity consequences, are summarised below — this table is essential for MCQs:

\[\begin{array}{cclc} \Delta H & \Delta S & \Delta G = \Delta H - T\Delta S & \text{Spontaneity}\\ \hline - & + & \text{Always negative} & \text{Spontaneous at ALL T}\\ - & - & \text{Negative at low T} & \text{Spontaneous at LOW T only}\\ + & + & \text{Negative at high T} & \text{Spontaneous at HIGH T only}\\ + & - & \text{Always positive} & \text{Never spontaneous} \end{array}\]
Step-by-Step Solution

Step 1: Identify the Signs of ΔH and ΔS from the Question

The reaction is written as \(\mathrm{A + B \rightarrow C + D + q}\). The symbol \(q\) on the right-hand side of the equation represents heat being released — it is a product of the reaction. This means the reaction is exothermic.

\[\Delta H < 0 \quad (\text{exothermic — heat is released as a product})\]

The question also states explicitly that the entropy change is positive:

\[\Delta S > 0 \quad (\text{given directly in the question})\]

Step 2: Apply the Gibbs Equation

\[\Delta G = \Delta H - T\Delta S = (\text{negative}) - T \times (\text{positive})\]

The first term \(\Delta H\) is negative. The second term \(T\Delta S\) is always positive (since both \(T > 0\) and \(\Delta S > 0\)). Subtracting a positive number from an already negative number gives a result that is even more negative.

\[\Delta G = (\text{negative}) - (\text{positive}) = \text{negative at ALL values of } T\]

Step 3: Conclusion

Since \(\Delta G < 0\) regardless of the temperature \(T\), the reaction is spontaneous at every temperature. This corresponds to the first row of the spontaneity table: \(\Delta H < 0\), \(\Delta S > 0\) always gives \(\Delta G < 0\).

Answer
(iv) possible at any temperature
07

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?

NumericalFirst LawSign Convention
Background Concept

First Law of Thermodynamics — IUPAC Sign Convention

The First Law of Thermodynamics states that the total energy of an isolated system is constant. For any process, the change in internal energy equals the algebraic sum of heat and work:

\[\Delta U = q + w\]

The sign convention used in modern chemistry (IUPAC convention) defines everything from the system's perspective:

\[\begin{array}{ll} q > 0 & \text{Heat absorbed by the system from surroundings}\\ q < 0 & \text{Heat released by the system to surroundings}\\[4pt] w > 0 & \text{Work done on the system by surroundings (compression)}\\ w < 0 & \text{Work done by the system on surroundings (expansion)} \end{array}\]
Exam Trap: The question says the system does 394 J of work — this means work is done by the system, so \(w = -394\ \mathrm{J}\) (negative). Students often forget to apply the negative sign here and get the wrong answer.
Step-by-Step Solution

Step 1: Assign Signs to q and w

\[\begin{aligned} q &= +701\ \mathrm{J} \quad \text{(heat is \emph{absorbed} by system → positive)}\\ w &= -394\ \mathrm{J} \quad \text{(work is done \emph{by} system → negative)} \end{aligned}\]

Step 2: Apply the First Law

\[\begin{aligned} \Delta U &= q + w\\ \Delta U &= 701\ \mathrm{J} + (-394\ \mathrm{J})\\ \Delta U &= 701\ \mathrm{J} - 394\ \mathrm{J}\\ \Delta U &= +307\ \mathrm{J} \end{aligned}\]

Step 3: Physical Interpretation

The positive value of \(\Delta U = +307\ \mathrm{J}\) means the internal energy of the system has increased by 307 J. This makes physical sense: the system gained 701 J as heat but spent 394 J doing work on the surroundings, so the net gain to the system's internal energy is \(701 - 394 = 307\ \mathrm{J}\).

Answer
\(\Delta U = +307\ \mathrm{J}\)
08

The reaction of cyanamide, \(\mathrm{NH_2CN(s)}\), with dioxygen was carried out in a bomb calorimeter, and \(\Delta U\) was found to be \(-742.7\ \mathrm{kJ\ mol^{-1}}\) at 298 K. Calculate enthalpy change for the reaction at 298 K.

\[\mathrm{NH_2CN(g) + \dfrac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)}\]
NumericalBomb CalorimeterΔH from ΔU
Background Concept

Bomb Calorimeter — Why ΔU ≠ ΔH for Gas-Phase Reactions

A bomb calorimeter is a sealed, rigid container — the volume is constant during the reaction. Since the volume cannot change, no PV expansion work is done, and the heat measured equals the change in internal energy (\(\Delta U\)) rather than enthalpy (\(\Delta H\)).

Most chemical reactions, however, occur in open flasks at constant atmospheric pressure, where any volume change due to gas production or consumption involves work done against (or by) the atmosphere. The heat measured at constant pressure is the enthalpy change \(\Delta H\).

To convert between the two, we use the relationship derived from \(H = U + pV\):

\[\Delta H = \Delta U + \Delta n_g RT\]

where \(\Delta n_g = n_g(\text{gaseous products}) - n_g(\text{gaseous reactants})\), \(R = 8.314 \times 10^{-3}\ \mathrm{kJ\ K^{-1}\ mol^{-1}}\), and \(T\) is in Kelvin.

Important: Water produced as liquid H₂O(l) is not counted in \(\Delta n_g\) because liquids do not contribute significantly to gas-phase \(pV\) changes.
Step-by-Step Solution

Step 1: Identify Gaseous Reactants and Products

\[\mathrm{NH_2CN(\mathbf{g}) + \tfrac{3}{2}O_2(\mathbf{g}) \rightarrow N_2(\mathbf{g}) + CO_2(\mathbf{g}) + H_2O(\mathbf{l})}\]

Gaseous reactants: NH₂CN (1 mol) + O₂ (3/2 mol) → total = \(1 + \tfrac{3}{2} = \tfrac{5}{2}\ \mathrm{mol}\)
Gaseous products: N₂ (1 mol) + CO₂ (1 mol) → total = \(1 + 1 = 2\ \mathrm{mol}\)
(H₂O is liquid — excluded from count)

Step 2: Calculate Δn_g

\[\begin{aligned} \Delta n_g &= n_g(\text{products}) - n_g(\text{reactants})\\ &= 2 - \tfrac{5}{2}\\ &= \tfrac{4}{2} - \tfrac{5}{2} = -\tfrac{1}{2} \end{aligned}\]

Step 3: Identify All Given Values

\[\begin{aligned} \Delta U &= -742.7\ \mathrm{kJ\ mol^{-1}}\\ \Delta n_g &= -\tfrac{1}{2}\\ R &= 8.314 \times 10^{-3}\ \mathrm{kJ\ K^{-1}\ mol^{-1}}\\ T &= 298\ \mathrm{K} \end{aligned}\]

Step 4: Calculate the Correction Term Δn_g RT

\[\begin{aligned} \Delta n_g RT &= \left(-\tfrac{1}{2}\right) \times (8.314 \times 10^{-3}) \times 298\\ &= -\tfrac{1}{2} \times 2.4776\\ &= -1.24\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]

Step 5: Calculate ΔH

\[\begin{aligned} \Delta H &= \Delta U + \Delta n_g RT\\ &= -742.7 + (-1.24)\\ &= -743.94\ \mathrm{kJ\ mol^{-1}}\\ &\approx -743.9\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]

The result is slightly more negative than \(\Delta U\), consistent with \(\Delta n_g < 0\).

Answer
\(\Delta H \approx -743.9\ \mathrm{kJ\ mol^{-1}}\)
09

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 \(24\ \mathrm{J\ mol^{-1}\ K^{-1}}\).

NumericalHeat CapacityTemperature Change
Background Concept

Molar Heat Capacity and Sensible Heat

When the temperature of a substance is raised without any phase change occurring, the heat absorbed is called sensible heat. The amount of heat required depends on three factors: the number of moles of substance, the temperature change, and the molar heat capacity of the substance.

The molar heat capacity \(C_m\) is defined as the amount of heat required to raise the temperature of one mole of a substance by one kelvin (or one degree Celsius, since degree sizes are equal). The formula relating heat, moles, heat capacity, and temperature change is:

\[q = n \cdot C_m \cdot \Delta T\]

where \(n\) = number of moles, \(C_m\) = molar heat capacity in \(\mathrm{J\ mol^{-1}\ K^{-1}}\), and \(\Delta T\) = temperature change in K (or °C — the size of one degree is identical).

Unit Note: The molar heat capacity given is in J mol⁻¹ K⁻¹. The final answer must be converted to kJ by dividing by 1000. This unit conversion step must be shown explicitly in Board examinations.
Step-by-Step Solution

Step 1: Calculate the Number of Moles of Aluminium

Molar mass of aluminium: Al = 27 g mol⁻¹

\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{60.0\ \mathrm{g}}{27\ \mathrm{g\ mol^{-1}}} = 2.222\ \mathrm{mol}\]

Step 2: Calculate the Temperature Change

\[\Delta T = T_{\text{final}} - T_{\text{initial}} = 55°\mathrm{C} - 35°\mathrm{C} = 20\ \mathrm{K}\]

Step 3: Apply the Heat Equation

\[\begin{aligned} q &= n \cdot C_m \cdot \Delta T\\ &= 2.222\ \mathrm{mol} \times 24\ \mathrm{J\ mol^{-1}\ K^{-1}} \times 20\ \mathrm{K}\\ &= 2.222 \times 480\\ &= 1066.7\ \mathrm{J} \end{aligned}\]

Step 4: Convert to kJ

\[q = \frac{1066.7}{1000} = 1.067\ \mathrm{kJ} \approx 1.07\ \mathrm{kJ}\]
Answer
\(q \approx 1.07\ \mathrm{kJ}\)
10

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at −10.0°C.

\(\Delta_\text{fus} H = 6.03\ \mathrm{kJ\ mol^{-1}}\) at 0°C.
\(C_p[\mathrm{H_2O(l)}] = 75.3\ \mathrm{J\ mol^{-1}\ K^{-1}}\)
\(C_p[\mathrm{H_2O(s)}] = 36.8\ \mathrm{J\ mol^{-1}\ K^{-1}}\)

NumericalPhase TransitionMulti-step Enthalpy
Background Concept

Enthalpy Changes Involving Temperature Change and Phase Change

When a substance undergoes a change that involves both a temperature change (within a single phase) and a phase change (e.g., freezing, melting), the total enthalpy change must be calculated in separate steps and then summed — this is an application of Hess's Law.

For a temperature change within one phase (sensible heat), the formula is:

\[q = n \cdot C_p \cdot \Delta T\]

For a phase change (latent heat — no temperature change), heat is absorbed or released at constant temperature. For melting (fusion), heat is absorbed: \(\Delta H = +\Delta_\text{fus}H\). For freezing (the reverse of melting), heat is released: \(\Delta H = -\Delta_\text{fus}H\).

In this problem, liquid water at 10°C is converted to ice at −10°C. This requires three distinct steps:

\[\begin{array}{ll} \text{Step 1:} & \text{Cool liquid water from } 10°\text{C to } 0°\text{C}\\ \text{Step 2:} & \text{Freeze water at } 0°\text{C (phase change)}\\ \text{Step 3:} & \text{Cool ice from } 0°\text{C to } -10°\text{C} \end{array}\]
Step-by-Step Solution

Step 1: Cool Liquid Water from 10°C to 0°C

\[\begin{aligned} q_1 &= n \cdot C_p[\mathrm{H_2O(l)}] \cdot \Delta T\\ &= 1.0\ \mathrm{mol} \times 75.3\ \mathrm{J\ mol^{-1}\ K^{-1}} \times (0 - 10)\ \mathrm{K}\\ &= 1.0 \times 75.3 \times (-10)\\ &= -753\ \mathrm{J} = -0.753\ \mathrm{kJ} \end{aligned}\]

The negative sign means heat is released as the water cools — correct.

Step 2: Freeze Water at 0°C (Phase Change)

Freezing is the reverse of melting (fusion). The enthalpy of fusion is +6.03 kJ mol⁻¹ (heat absorbed during melting). Therefore, during freezing, heat is released and the enthalpy change is negative:

\[\begin{aligned} q_2 &= -\Delta_\text{fus}H\\ &= -6.03\ \mathrm{kJ} \end{aligned}\]

Step 3: Cool Ice from 0°C to −10°C

\[\begin{aligned} q_3 &= n \cdot C_p[\mathrm{H_2O(s)}] \cdot \Delta T\\ &= 1.0\ \mathrm{mol} \times 36.8\ \mathrm{J\ mol^{-1}\ K^{-1}} \times (-10 - 0)\ \mathrm{K}\\ &= 1.0 \times 36.8 \times (-10)\\ &= -368\ \mathrm{J} = -0.368\ \mathrm{kJ} \end{aligned}\]

Step 4: Sum All Contributions

\[\begin{aligned} \Delta H_\text{total} &= q_1 + q_2 + q_3\\ &= (-0.753) + (-6.03) + (-0.368)\\ &= -7.151\ \mathrm{kJ}\\ &\approx -7.15\ \mathrm{kJ} \end{aligned}\]

All three contributions are negative, which makes physical sense — the entire process involves releasing heat (cooling and freezing).

Answer
\(\Delta H \approx -7.15\ \mathrm{kJ}\)
11

Enthalpy of combustion of carbon to \(\mathrm{CO_2}\) is \(-393.5\ \mathrm{kJ\ mol^{-1}}\). Calculate the heat released upon formation of 35.2 g of \(\mathrm{CO_2}\) from carbon and dioxygen gas.

NumericalEnthalpy of CombustionMole Calculation
Background Concept

Enthalpy of Combustion — Per Mole Basis

The enthalpy of combustion is always quoted on a per mole basis — it represents the heat released when exactly one mole of the substance is completely burned in excess oxygen. The balanced equation for the combustion of carbon is:

\[\mathrm{C(s) + O_2(g) \rightarrow CO_2(g)} \quad \Delta H = -393.5\ \mathrm{kJ\ mol^{-1}}\]

This tells us that the formation of 1 mole of CO₂ releases 393.5 kJ of heat. For any other quantity, we scale proportionally using the number of moles actually formed.

Step-by-Step Solution

Step 1: Calculate Moles of CO₂ Formed

Molar mass of CO₂: \(12 + 2 \times 16 = 44\ \mathrm{g\ mol^{-1}}\)

\[n(\mathrm{CO_2}) = \frac{\text{given mass}}{\text{molar mass}} = \frac{35.2\ \mathrm{g}}{44\ \mathrm{g\ mol^{-1}}} = 0.8\ \mathrm{mol}\]

Step 2: Scale the Enthalpy Change

Since 1 mol of CO₂ formation releases 393.5 kJ, the formation of 0.8 mol releases:

\[\begin{aligned} q &= n \times \Delta H_c\\ &= 0.8\ \mathrm{mol} \times (-393.5\ \mathrm{kJ\ mol^{-1}})\\ &= -314.8\ \mathrm{kJ} \end{aligned}\]

The negative sign confirms heat is released (exothermic). The magnitude of heat released is 314.8 kJ.

Answer
314.8 kJ of heat is released
12

Enthalpies of formation of \(\mathrm{CO(g)}\), \(\mathrm{CO_2(g)}\), \(\mathrm{N_2O(g)}\) and \(\mathrm{N_2O_4(g)}\) are \(-110\), \(-393\), \(81\) and \(9.7\ \mathrm{kJ\ mol^{-1}}\) respectively. Find the value of \(\Delta_r H\) for the reaction:

\[\mathrm{N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)}\]
NumericalEnthalpies of FormationHess's Law
Background Concept

Calculating Reaction Enthalpy from Enthalpies of Formation

The most general and powerful application of Hess's Law is the formula that expresses the standard enthalpy change of any reaction in terms of the standard enthalpies of formation of all reactants and products:

\[\Delta_r H^\Theta = \sum \nu_{\text{products}} \Delta_f H^\Theta(\text{products}) - \sum \nu_{\text{reactants}} \Delta_f H^\Theta(\text{reactants})\]

where \(\nu\) are the stoichiometric coefficients in the balanced equation. In words: sum of (coefficient × ΔfH°) for all products minus sum of (coefficient × ΔfH°) for all reactants.

The logic is: we "decompose" all reactants into elements (costing/releasing ΔfH° each), then "form" all products from those elements (releasing/costing ΔfH° each). The elements themselves all have ΔfH° = 0 and cancel out, leaving only the net reaction.

Step-by-Step Solution

Step 1: Identify and List All ΔfH° Values

\[\begin{aligned} \Delta_f H^\Theta[\mathrm{CO(g)}] &= -110\ \mathrm{kJ\ mol^{-1}}\\ \Delta_f H^\Theta[\mathrm{CO_2(g)}] &= -393\ \mathrm{kJ\ mol^{-1}}\\ \Delta_f H^\Theta[\mathrm{N_2O(g)}] &= +81\ \mathrm{kJ\ mol^{-1}}\\ \Delta_f H^\Theta[\mathrm{N_2O_4(g)}] &= +9.7\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]

Step 2: Calculate ΣΔfH° of Products

Products: 1 mol N₂O and 3 mol CO₂

\[\begin{aligned} \sum \Delta H_f^\Theta(\text{products}) &= 1 \times \Delta_f H^\Theta[\mathrm{N_2O}] + 3 \times \Delta_f H^\Theta[\mathrm{CO_2}]\\ &= 1(+81) + 3(-393)\\ &= +81 - 1179\\ &= -1098\ \mathrm{kJ} \end{aligned}\]

Step 3: Calculate ΣΔfH° of Reactants

Reactants: 1 mol N₂O₄ and 3 mol CO

\[\begin{aligned} \sum \Delta H_f^\Theta(\text{reactants}) &= 1 \times \Delta_f H^\Theta[\mathrm{N_2O_4}] + 3 \times \Delta_f H^\Theta[\mathrm{CO}]\\ &= 1(+9.7) + 3(-110)\\ &= +9.7 - 330\\ &= -320.3\ \mathrm{kJ} \end{aligned}\]

Step 4: Calculate Δ_r H°

\[\begin{aligned} \Delta_r H^\Theta &= \sum \Delta H_f^\Theta(\text{products}) - \sum \Delta H_f^\Theta(\text{reactants})\\ &= (-1098) - (-320.3)\\ &= -1098 + 320.3\\ &= -777.7\ \mathrm{kJ} \end{aligned}\]
Answer
\(\Delta_r H^\Theta = -777.7\ \mathrm{kJ}\)
13

Given:

\[\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g);\ \Delta_r H^\Theta = -92.4\ kJ\ mol^{-1}}\]

What is the standard enthalpy of formation of \(\mathrm{NH_3}\) gas?

NumericalStandard Enthalpy of FormationStoichiometry
Background Concept

Standard Enthalpy of Formation — The Per-Mole Concept

The standard enthalpy of formation \(\Delta_f H^\Theta\) of a compound is defined as the enthalpy change when exactly one mole of that compound is formed from its constituent elements in their standard states. This one-mole basis is crucial.

The given reaction \(\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}\) produces two moles of NH₃, not one. Therefore, the given \(\Delta_r H^\Theta = -92.4\ \mathrm{kJ\ mol^{-1}}\) corresponds to the formation of 2 mol of NH₃. The enthalpy of formation of one mole of NH₃ will be half of this value.

We can verify this using the general formula. Since N₂ and H₂ are elements in their standard states, their \(\Delta_f H^\Theta = 0\):

\[\Delta_r H^\Theta = \sum \nu \Delta_f H^\Theta(\text{products}) - \sum \nu \Delta_f H^\Theta(\text{reactants})\]
Step-by-Step Solution

Step 1: Write the Formation Equation (for 2 mol NH₃)

\[\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)} \quad \Delta_r H^\Theta = -92.4\ \mathrm{kJ}\]

Step 2: Apply the Enthalpy of Formation Formula

\[\begin{aligned} \Delta_r H^\Theta &= 2\,\Delta_f H^\Theta(\mathrm{NH_3}) - \bigl[\Delta_f H^\Theta(\mathrm{N_2}) + 3\,\Delta_f H^\Theta(\mathrm{H_2})\bigr] \end{aligned}\]

Since N₂(g) and H₂(g) are elements in their standard states:

\[\Delta_f H^\Theta(\mathrm{N_2}) = 0 \quad \text{and} \quad \Delta_f H^\Theta(\mathrm{H_2}) = 0\]

Step 3: Substitute and Solve

\[\begin{aligned} -92.4 &= 2\,\Delta_f H^\Theta(\mathrm{NH_3}) - [0 + 3(0)]\\ -92.4 &= 2\,\Delta_f H^\Theta(\mathrm{NH_3}) \end{aligned}\]
\[\Delta_f H^\Theta(\mathrm{NH_3}) = \frac{-92.4}{2} = -46.2\ \mathrm{kJ\ mol^{-1}}\]
Answer
\(\Delta_f H^\Theta(\mathrm{NH_3}) = -46.2\ \mathrm{kJ\ mol^{-1}}\)
14

Calculate the standard enthalpy of formation of \(\mathrm{CH_3OH(l)}\) from the following data:

\[\begin{aligned} &\mathrm{CH_3OH(l) + \tfrac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l);} &&\Delta_r H^\Theta = -726\ \mathrm{kJ\ mol^{-1}}\\ &\mathrm{C_{(graphite)} + O_2(g) \rightarrow CO_2(g);} &&\Delta_r H^\Theta = -393\ \mathrm{kJ\ mol^{-1}}\\ &\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l);} &&\Delta_r H^\Theta = -286\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]
NumericalHess's LawFormation from Combustion Data
Background Concept

Deriving ΔfH° from Combustion Data Using Hess's Law

The standard enthalpy of formation of methanol is the enthalpy change for:

\[\mathrm{C(graphite) + 2H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow CH_3OH(l)} \quad \Delta_f H^\Theta = \;?\]

This reaction cannot be carried out directly. However, using the combustion data provided and Hess's Law, we can combine the three given reactions algebraically to derive it. Alternatively, we can use the standard formula involving enthalpies of formation directly.

Method 1: Using the Reaction Enthalpy Formula

Step 1: Apply the Hess's Law Formula to the Combustion of Methanol

For the combustion of methanol:

\[\Delta_r H^\Theta = \bigl[\Delta_f H^\Theta(\mathrm{CO_2}) + 2\,\Delta_f H^\Theta(\mathrm{H_2O})\bigr] - \bigl[\Delta_f H^\Theta(\mathrm{CH_3OH}) + \tfrac{3}{2}\,\Delta_f H^\Theta(\mathrm{O_2})\bigr]\]

Since \(\Delta_f H^\Theta(\mathrm{O_2}) = 0\) (element in standard state):

\[-726 = [-393 + 2(-286)] - \Delta_f H^\Theta(\mathrm{CH_3OH})\]

Step 2: Evaluate the Bracket

\[\begin{aligned} -393 + 2(-286) &= -393 - 572 = -965\ \mathrm{kJ} \end{aligned}\]

Step 3: Solve for ΔfH°(CH₃OH)

\[\begin{aligned} -726 &= -965 - \Delta_f H^\Theta(\mathrm{CH_3OH})\\ \Delta_f H^\Theta(\mathrm{CH_3OH}) &= -965 + 726 \end{aligned}\]
\[\Delta_f H^\Theta(\mathrm{CH_3OH(l)}) = -239\ \mathrm{kJ\ mol^{-1}}\]
Answer
\(\Delta_f H^\Theta(\mathrm{CH_3OH(l)}) = -239\ \mathrm{kJ\ mol^{-1}}\)
15

Calculate the enthalpy change for the process

\[\mathrm{CCl_4(g) \rightarrow C(g) + 4\,Cl(g)}\]

and calculate bond enthalpy of C–Cl in \(\mathrm{CCl_4(g)}\).

\(\Delta_\text{vap} H^\Theta(\mathrm{CCl_4}) = 30.5\ \mathrm{kJ\ mol^{-1}}\)
\(\Delta_f H^\Theta(\mathrm{CCl_4}) = -135.5\ \mathrm{kJ\ mol^{-1}}\)
\(\Delta_\text{a} H^\Theta(\mathrm{C}) = 715.0\ \mathrm{kJ\ mol^{-1}}\)  (enthalpy of atomisation of carbon)
\(\Delta_\text{a} H^\Theta(\mathrm{Cl_2}) = 242\ \mathrm{kJ\ mol^{-1}}\)  (enthalpy of atomisation of Cl₂)

NumericalBond EnthalpyAtomisationHess's Law
Background Concept

Bond Enthalpy and the Atomisation Cycle

The bond enthalpy of a bond X–Y is the enthalpy required to break one mole of that bond in the gaseous phase to produce gaseous atoms:

\[\mathrm{X-Y(g) \rightarrow X(g) + Y(g)} \quad \Delta H = \text{bond enthalpy}\]

For CCl₄, there are four equivalent C–Cl bonds. When all four are broken:

\[\mathrm{CCl_4(g) \rightarrow C(g) + 4\,Cl(g)} \quad \Delta H = 4 \times \text{(C–Cl bond enthalpy)}\]

Since we cannot measure this directly, we use a Hess's Law cycle. The strategy is to construct two paths from the same elements in standard states to the same gaseous atoms:

\[\text{Path 1: Elements (standard)} \xrightarrow{+\,\Delta_\text{atomisation}} \text{Gaseous atoms}\] \[\text{Path 2: Elements (standard)} \xrightarrow{-\,\Delta_f(\mathrm{CCl_4,l})} \mathrm{CCl_4(l)} \xrightarrow{+\,\Delta_\text{vap}} \mathrm{CCl_4(g)} \xrightarrow{+\,\Delta H_\text{required}} \text{Gaseous atoms}\]
Step-by-Step Solution

Step 1: Find ΔH for Converting Elements in Standard States to Gaseous Atoms

We need 1 mol C(g) and 4 mol Cl(g). The standard states of these elements are C(graphite) and Cl₂(g).

For carbon: \(\mathrm{C(graphite) \rightarrow C(g)}\)

\[\Delta H = +715.0\ \mathrm{kJ\ mol^{-1}}\]

For chlorine: The atomisation enthalpy of Cl₂ = 242 kJ mol⁻¹ refers to:

\[\mathrm{Cl_2(g) \rightarrow 2\,Cl(g)} \quad \Delta H = +242\ \mathrm{kJ\ mol^{-1}}\]

For 4 Cl atoms, we need 2 mol Cl₂:

\[\mathrm{2\,Cl_2(g) \rightarrow 4\,Cl(g)} \quad \Delta H = 2 \times 242 = +484\ \mathrm{kJ}\]

Total enthalpy to form all atoms from standard-state elements:

\[\Delta H_1 = 715 + 484 = +1199\ \mathrm{kJ}\]

Step 2: Find ΔfH° of Gaseous CCl₄

The given formation enthalpy refers to liquid CCl₄. We must add the enthalpy of vaporisation to get the formation enthalpy of gaseous CCl₄:

\[\begin{aligned} \Delta_f H^\Theta[\mathrm{CCl_4(g)}] &= \Delta_f H^\Theta[\mathrm{CCl_4(l)}] + \Delta_\text{vap}H^\Theta\\ &= -135.5 + 30.5\\ &= -105.0\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]

Step 3: Apply Hess's Law to Find ΔH for the Atomisation of CCl₄(g)

By Hess's Law, the enthalpy for atomising CCl₄(g) is the difference between Path 1 (cost of making atoms from elements) and Path 2 (gain from making CCl₄ from elements):

\[\begin{aligned} \Delta H[\mathrm{CCl_4(g) \rightarrow C(g) + 4Cl(g)}] &= \Delta H_1 - \Delta_f H^\Theta[\mathrm{CCl_4(g)}]\\ &= +1199 - (-105)\\ &= +1199 + 105\\ &= +1304\ \mathrm{kJ\ mol^{-1}} \end{aligned}\]

Step 4: Calculate the C–Cl Bond Enthalpy

The 1304 kJ is the total energy required to break all four C–Cl bonds in one mole of CCl₄. Assuming all four bonds are equivalent (which is valid for a symmetric molecule like CCl₄):

\[\text{Bond enthalpy (C–Cl)} = \frac{1304}{4} = 326\ \mathrm{kJ\ mol^{-1}}\]
Answer
Enthalpy of atomisation of CCl₄(g) = +1304 kJ mol⁻¹
C–Cl bond enthalpy = 326 kJ mol⁻¹
16

For an isolated system, \(\Delta U = 0\), what will be \(\Delta S\)?

ConceptualSecond LawEntropyIsolated System
Background Concept

Isolated Systems and the Second Law of Thermodynamics

An isolated system is one that exchanges neither matter nor energy (neither heat nor work) with its surroundings. Because of this total isolation:

\[q = 0, \quad w = 0 \implies \Delta U = q + w = 0\]

The internal energy of an isolated system is therefore constant. The universe as a whole can be considered an isolated system.

However, the constancy of internal energy (\(\Delta U = 0\)) does not tell us whether a spontaneous process can occur or in which direction it proceeds. This is governed by the Second Law of Thermodynamics, which deals with entropy.

The Second Law states that for any spontaneous process occurring in an isolated system, the total entropy of the system increases. The entropy can never decrease spontaneously in an isolated system. In its most compact form:

\[\Delta S_{\text{isolated}} \geq 0\]

This inequality has two cases:

\[\begin{array}{ll} \Delta S > 0 & \text{Process is spontaneous (naturally occurring)}\\ \Delta S = 0 & \text{System is at equilibrium (no net change)}\\ \Delta S < 0 & \text{Impossible in an isolated system} \end{array}\]

Entropy \(S\) is a measure of the disorder, randomness, or dispersal of energy in a system. Natural processes — heat flowing from hot to cold, gases expanding to fill a container, crystal dissolving in a solvent — all involve an increase in entropy of the isolated system. Nature always moves spontaneously toward greater disorder.

Answer and Explanation

What ΔS Can Be for an Isolated System

Even though \(\Delta U = 0\) (energy is conserved), spontaneous processes can and do occur inside isolated systems. For example, a gas confined to one half of an isolated box will spontaneously expand to fill the whole box even though \(\Delta U = 0\). In this process, entropy increases significantly.

The question therefore has two possible answers depending on what the system is doing:

\[\begin{aligned} \text{For a spontaneous (irreversible) process:} \quad &\Delta S > 0\\ \text{For a system at equilibrium (reversible):} \quad &\Delta S = 0 \end{aligned}\]

In either case, \(\Delta S \geq 0\). It is never negative in an isolated system.

Answer
\(\Delta S \geq 0\) — The entropy change of an isolated system is either positive (for any spontaneous process) or zero (when the system is at equilibrium). It can never be negative.
17

For the reaction at 298 K,

\[\mathrm{2A + B \rightarrow C}\] \[\Delta H = 400\ \mathrm{kJ\ mol^{-1}} \quad \text{and} \quad \Delta S = 0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}\]

At what temperature will the reaction become spontaneous considering \(\Delta H\) and \(\Delta S\) to be constant over the temperature range?

NumericalGibbs EnergySpontaneity Temperature
Background Concept

Finding the Temperature of Spontaneity

The spontaneity of a reaction is determined by the sign of the Gibbs free energy change:

\[\Delta G = \Delta H - T\Delta S\]

At a given temperature, a reaction is: spontaneous if \(\Delta G < 0\), at equilibrium if \(\Delta G = 0\), and non-spontaneous if \(\Delta G > 0\).

In this question, \(\Delta H = +400\ \mathrm{kJ\ mol^{-1}}\) (positive — endothermic) and \(\Delta S = +0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}\) (positive). Looking at the spontaneity table, a reaction with \(\Delta H > 0\) and \(\Delta S > 0\) is spontaneous only at high temperatures, because the \(T\Delta S\) term must be large enough to overcome the unfavourable positive \(\Delta H\).

The critical temperature — the threshold above which the reaction becomes spontaneous — is found by setting \(\Delta G = 0\):

\[\Delta G = 0 \implies T_\text{critical} = \frac{\Delta H}{\Delta S}\]

For \(T > T_\text{critical}\): \(\Delta G < 0\) → spontaneous.
For \(T < T_\text{critical}\): \(\Delta G > 0\) → non-spontaneous.

Step-by-Step Solution

Step 1: Verify Units Are Consistent

\(\Delta H = 400\ \mathrm{kJ\ mol^{-1}}\) and \(\Delta S = 0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}\). Both are in kJ, so no conversion is needed.

Common Mistake: If \(\Delta S\) is given in \(\mathrm{J\ K^{-1}\ mol^{-1}}\) and \(\Delta H\) in \(\mathrm{kJ\ mol^{-1}}\), one must convert to the same unit before dividing. Here, \(\Delta S = 0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}} = 200\ \mathrm{J\ K^{-1}\ mol^{-1}}\).

Step 2: Set ΔG = 0 and Solve for T

\[\begin{aligned} \Delta G &= \Delta H - T\Delta S\\ 0 &= \Delta H - T_\text{critical}\Delta S\\ T_\text{critical}\Delta S &= \Delta H\\ T_\text{critical} &= \frac{\Delta H}{\Delta S} \end{aligned}\]

Step 3: Substitute Values

\[T_\text{critical} = \frac{400\ \mathrm{kJ\ mol^{-1}}}{0.2\ \mathrm{kJ\ K^{-1}\ mol^{-1}}} = 2000\ \mathrm{K}\]

Step 4: State the Conclusion Clearly

The reaction is non-spontaneous below 2000 K (at 298 K, \(\Delta G = 400 - 298 \times 0.2 = +340.4\ \mathrm{kJ}\) — highly positive). The reaction becomes spontaneous above 2000 K, where the entropy contribution \(T\Delta S\) exceeds \(\Delta H\).

Answer
The reaction becomes spontaneous above \(T = 2000\ \mathrm{K}\)
18

For the reaction,

\[\mathrm{2\,Cl(g) \rightarrow Cl_2(g)}\]

what are the signs of \(\Delta H\) and \(\Delta S\)?

ConceptualBond FormationEntropy Signs
Background Concept

Enthalpy and Entropy of Bond Formation

To answer this question, we need to reason about what happens at the molecular level when two chlorine atoms combine to form a chlorine molecule.

Enthalpy (\(\Delta H\)): When a chemical bond forms between two atoms, the resulting molecule is more stable (lower energy) than the two separated atoms. Energy must be released to the surroundings for the system to reach this lower-energy bonded state. Bond formation is therefore always exothermic (\(\Delta H < 0\)). (Conversely, bond breaking is always endothermic.)

Entropy (\(\Delta S\)): Entropy measures the degree of randomness, disorder, or dispersal in a system. The number of particles is a key contributor to entropy: more particles means more ways to arrange their positions and momenta, hence more disorder. In this reaction, two separate gaseous particles (2 Cl atoms) combine into one particle (Cl₂ molecule). The number of moles of gas decreases from 2 to 1.

\[\Delta n_g = 1 - 2 = -1 < 0\]

A decrease in the number of gas particles means a decrease in the randomness/disorder of the system, so entropy decreases. Therefore \(\Delta S < 0\).

Detailed Reasoning

Sign of ΔH

The Cl–Cl bond is formed in this reaction. Bond formation always releases energy because the bonded state is more stable. The bond enthalpy of Cl–Cl is +242 kJ mol⁻¹ (energy needed to break it). Therefore, forming the bond releases 242 kJ mol⁻¹:

\[\Delta H = -242\ \mathrm{kJ\ mol^{-1}} < 0 \quad \text{(exothermic)}\]

Sign of ΔS

Two moles of individual gaseous Cl atoms (which can move independently in many different directions and positions) combine into one mole of Cl₂ molecules (which are constrained to move together). This represents a significant loss of randomness:

\[\Delta S < 0 \quad \text{(disorder decreases: 2 gas particles → 1 gas particle)}\]

Spontaneity Implications

With \(\Delta H < 0\) and \(\Delta S < 0\), the Gibbs equation gives: \(\Delta G = \Delta H - T\Delta S = (-) - T(-) = (-) + T(+)\). This means the reaction is spontaneous only at low temperatures where the negative \(\Delta H\) outweighs the positive \(T|\Delta S|\) term. This is physically sensible — at very high temperatures, Cl₂ dissociates back into atoms.

Answer
\(\Delta H < 0\) (exothermic — bond formation releases energy)  and  \(\Delta S < 0\) (entropy decreases — two particles become one)
19

For the reaction

\[\mathrm{2A(g) + B(g) \rightarrow 2D(g)}\] \[\Delta U^\Theta = -10.5\ \mathrm{kJ} \quad \text{and} \quad \Delta S^\Theta = -44.1\ \mathrm{J\ K^{-1}}\]

Calculate \(\Delta G^\Theta\) for the reaction, and predict whether the reaction may occur spontaneously at 298 K.

NumericalGibbs Free EnergySpontaneityΔH from ΔU
Background Concept

Complete Workflow: ΔU → ΔH → ΔG

This problem requires a two-step conversion:

\[\text{Step 1: } \Delta U \xrightarrow{+ \Delta n_g RT} \Delta H\] \[\text{Step 2: } \Delta G = \Delta H - T\Delta S\]

The reason we cannot directly use ΔU in the Gibbs equation is that the Gibbs equation is defined at constant pressure, where the relevant energy quantity is enthalpy (\(H = U + pV\)), not internal energy (\(U\)).

Unit Alert: \(\Delta S^\Theta = -44.1\ \mathrm{J\ K^{-1}}\) is in Joules, but \(\Delta H\) will be in kJ after calculation. Convert \(\Delta S\) to kJ before applying the Gibbs equation: \(\Delta S = -44.1\ \mathrm{J\ K^{-1}} = -0.0441\ \mathrm{kJ\ K^{-1}}\).
Step-by-Step Solution

Step 1: Calculate Δn_g

\[\begin{aligned} n_g(\text{reactants}) &= 2 + 1 = 3\\ n_g(\text{products}) &= 2\\ \Delta n_g &= 2 - 3 = -1 \end{aligned}\]

Step 2: Convert ΔU to ΔH

\[\begin{aligned} \Delta H &= \Delta U + \Delta n_g RT\\ &= -10.5\ \mathrm{kJ} + (-1) \times (8.314 \times 10^{-3}\ \mathrm{kJ\ K^{-1}\ mol^{-1}}) \times 298\ \mathrm{K}\\ &= -10.5 + (-1)(2.4776)\\ &= -10.5 - 2.48\\ &= -12.98\ \mathrm{kJ} \end{aligned}\]

Step 3: Convert ΔS to kJ K⁻¹

\[\Delta S = -44.1\ \mathrm{J\ K^{-1}} = \frac{-44.1}{1000}\ \mathrm{kJ\ K^{-1}} = -0.0441\ \mathrm{kJ\ K^{-1}}\]

Step 4: Calculate ΔG° Using the Gibbs Equation

\[\begin{aligned} \Delta G &= \Delta H - T\Delta S\\ &= -12.98 - (298)(-0.0441)\\ &= -12.98 - (-13.14)\\ &= -12.98 + 13.14\\ &= +0.16\ \mathrm{kJ} \end{aligned}\]

Step 5: Interpret the Sign of ΔG

\(\Delta G^\Theta = +0.16\ \mathrm{kJ} > 0\). Since \(\Delta G\) is positive (even though very small), the reaction is not spontaneous at 298 K under standard conditions. The entropy decrease (\(\Delta S < 0\)) slightly outweighs the enthalpy advantage (\(\Delta H < 0\)) at this temperature.

Answer
\(\Delta G^\Theta = +0.16\ \mathrm{kJ}\) — The reaction is NOT spontaneous at 298 K.
20

The equilibrium constant for a reaction is 10. What will be the value of \(\Delta G^\Theta\)?

\(R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}},\ T = 300\ \mathrm{K}\)

NumericalΔG° and Equilibrium ConstantThermodynamics–Equilibrium Link
Background Concept

The Fundamental Link Between ΔG° and the Equilibrium Constant K

One of the most profound results in physical chemistry is the relationship connecting the standard Gibbs free energy change (\(\Delta G^\Theta\)) to the equilibrium constant (\(K\)) of a reaction. This equation bridges thermodynamics and kinetics/equilibrium:

\[\Delta G^\Theta = -RT\ln K\]

or equivalently, using base-10 logarithms:

\[\Delta G^\Theta = -2.303\,RT\log K\]

The physical interpretation is: if \(K > 1\) (products favoured at equilibrium), then \(\ln K > 0\), so \(\Delta G^\Theta < 0\) — the reaction is thermodynamically downhill in the forward direction under standard conditions. If \(K < 1\) (reactants favoured), \(\Delta G^\Theta > 0\). If \(K = 1\), \(\Delta G^\Theta = 0\).

Important distinction: \(\Delta G^\Theta\) is the free energy change under standard conditions (all species at unit activity). The actual free energy change \(\Delta G\) at non-standard conditions is \(\Delta G = \Delta G^\Theta + RT\ln Q\), where \(Q\) is the reaction quotient.
Step-by-Step Solution

Step 1: Write Down All Given Values

\[K = 10, \quad R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}}, \quad T = 300\ \mathrm{K}\]

Step 2: Choose the Appropriate Formula

Since \(K = 10\) is a convenient power of 10, use the \(\log_{10}\) form to simplify calculation:

\[\Delta G^\Theta = -2.303\,RT\log K\]

Step 3: Evaluate log K

\[\log(10) = 1\]

Step 4: Substitute All Values

\[\begin{aligned} \Delta G^\Theta &= -2.303 \times 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}} \times 300\ \mathrm{K} \times 1\\ &= -2.303 \times 2494.2\ \mathrm{J\ mol^{-1}}\\ &= -5742.6\ \mathrm{J\ mol^{-1}}\\ &\approx -5743\ \mathrm{J\ mol^{-1}} \end{aligned}\]

Step 5: Convert to kJ

\[\Delta G^\Theta = \frac{-5743}{1000}\ \mathrm{kJ\ mol^{-1}} = -5.74\ \mathrm{kJ\ mol^{-1}}\]

The negative value confirms that since \(K > 1\), the forward reaction is favoured under standard conditions.

Answer
\(\Delta G^\Theta = -5743\ \mathrm{J\ mol^{-1}} = -5.74\ \mathrm{kJ\ mol^{-1}}\)
21

Comment on the thermodynamic stability of NO(g), given

\[\mathrm{\tfrac{1}{2}N_2(g) + \tfrac{1}{2}O_2(g) \rightarrow NO(g);\ \Delta_r H^\Theta = +90\ kJ\ mol^{-1}}\] \[\mathrm{NO(g) + \tfrac{1}{2}O_2(g) \rightarrow NO_2(g);\ \Delta_r H^\Theta = -74\ kJ\ mol^{-1}}\]
ConceptualThermodynamic StabilityEnthalpy Analysis
Background Concept

Thermodynamic Stability — What It Means

The thermodynamic stability of a compound refers to its stability with respect to its constituent elements (or other possible compounds) as judged by energy considerations. A compound is said to be thermodynamically stable if it has lower energy than its constituent elements — i.e., it forms spontaneously from those elements. Conversely, if forming the compound requires absorption of energy (positive \(\Delta_f H^\Theta\)), the compound has higher energy than the elements and is thermodynamically unstable.

We assess stability by examining:

\[\begin{array}{ll} \Delta_f H^\Theta < 0 & \text{Compound is more stable than its elements (stable)}\\ \Delta_f H^\Theta > 0 & \text{Compound is less stable than its elements (unstable)} \end{array}\]

Additionally, if the compound readily reacts with surrounding molecules (like O₂ in air) to form even more stable products (with negative \(\Delta_r H^\Theta\)), the compound is described as thermodynamically unstable with respect to those products too.

Analysis

Analysis 1: Stability of NO with Respect to N₂ and O₂ (its constituent elements)

The formation of NO from its elements is given by:

\[\mathrm{\tfrac{1}{2}N_2(g) + \tfrac{1}{2}O_2(g) \rightarrow NO(g)} \quad \Delta_r H^\Theta = +90\ \mathrm{kJ\ mol^{-1}}\]

The positive \(\Delta_r H^\Theta = +90\ \mathrm{kJ\ mol^{-1}}\) means this reaction is endothermic. Energy must be absorbed to form NO. This means that NO has a higher energy content than the elements N₂ and O₂. Therefore:

\[\text{NO(g) is \textbf{thermodynamically unstable} with respect to N}_2\text{(g) and O}_2\text{(g)}\]

In other words, if there were no kinetic barrier (activation energy), NO would spontaneously decompose back into N₂ and O₂ by the reverse reaction, which would be exothermic (\(-90\ \mathrm{kJ}\)). The fact that NO persists in the atmosphere is due to a high kinetic barrier, not thermodynamic stability.

Analysis 2: Stability of NO with Respect to NO₂

The reaction of NO with oxygen to form NO₂ is:

\[\mathrm{NO(g) + \tfrac{1}{2}O_2(g) \rightarrow NO_2(g)} \quad \Delta_r H^\Theta = -74\ \mathrm{kJ\ mol^{-1}}\]

This reaction is exothermic (\(-74\ \mathrm{kJ\ mol^{-1}}\)), meaning NO₂ is more stable than NO. Nitric oxide therefore has a natural thermodynamic tendency to oxidise further to nitrogen dioxide.

Overall Conclusion

NO is thermodynamically unstable in two senses:

\[\begin{aligned} &\text{1. Unstable with respect to its elements } (\Delta_f H^\Theta = +90\ \mathrm{kJ\ mol^{-1}} > 0)\\ &\text{2. Unstable with respect to NO}_2 \text{ (readily oxidised, releasing } 74\ \mathrm{kJ\ mol^{-1}}\text{)} \end{aligned}\]
Answer
NO(g) is thermodynamically unstable:
• Its formation from N₂ and O₂ is endothermic (+90 kJ mol⁻¹) — it has more energy than its elements.
• It readily converts to the more stable NO₂ in an exothermic reaction (−74 kJ mol⁻¹).
22

Calculate the entropy change in surroundings when 1.00 mol of \(\mathrm{H_2O(l)}\) is formed under standard conditions.

\(\Delta_f H^\Theta = -286\ \mathrm{kJ\ mol^{-1}}\)

NumericalEntropy of SurroundingsHeat–Entropy Relationship
Background Concept

Entropy Change of the Surroundings

The total entropy change of the universe is the sum of the entropy change of the system and the entropy change of the surroundings:

\[\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surroundings}\]

For a process occurring at constant temperature \(T\) and constant pressure, the surroundings exchange heat with the system. If the system releases heat (exothermic, \(\Delta H_\text{sys} < 0\)), the surroundings absorb that heat and their entropy increases. If the system absorbs heat (endothermic), the surroundings lose heat and their entropy decreases.

The entropy change of the surroundings is given by:

\[\Delta S_\text{surr} = -\frac{\Delta H_\text{system}}{T}\]

The negative sign arises because when the system releases heat (\(\Delta H_\text{sys} < 0\)), the surroundings absorb it (\(q_\text{surr} = -\Delta H_\text{sys} > 0\)), increasing their entropy. The formula correctly gives a positive \(\Delta S_\text{surr}\) when \(\Delta H_\text{sys} < 0\).

Unit Conversion Required: \(\Delta H\) is given in kJ, but \(\Delta S\) is typically expressed in J K⁻¹. Convert \(\Delta H\) to J by multiplying by 1000 before dividing by T.
Step-by-Step Solution

Step 1: Write the Reaction

The formation of liquid water from hydrogen and oxygen:

\[\mathrm{H_2(g) + \tfrac{1}{2}O_2(g) \rightarrow H_2O(l)} \quad \Delta_f H^\Theta = -286\ \mathrm{kJ\ mol^{-1}}\]

Step 2: Convert ΔfH° to Joules

\[\Delta H_\text{sys} = -286\ \mathrm{kJ\ mol^{-1}} = -286{,}000\ \mathrm{J\ mol^{-1}}\]

Step 3: Identify Temperature

Standard conditions: \(T = 298\ \mathrm{K}\)

Step 4: Apply the Formula for ΔS_surr

\[\begin{aligned} \Delta S_\text{surr} &= -\frac{\Delta H_\text{sys}}{T}\\ &= -\frac{-286{,}000\ \mathrm{J\ mol^{-1}}}{298\ \mathrm{K}}\\ &= +\frac{286{,}000}{298}\ \mathrm{J\ K^{-1}\ mol^{-1}}\\ &= +959.7\ \mathrm{J\ K^{-1}\ mol^{-1}}\\ &\approx +960\ \mathrm{J\ K^{-1}\ mol^{-1}} \end{aligned}\]

Step 5: Physical Interpretation

The formation of water is highly exothermic — it releases 286 kJ of heat to the surroundings. This large influx of heat into the surroundings greatly increases their disorder (entropy), giving a large positive \(\Delta S_\text{surr} = +960\ \mathrm{J\ K^{-1}\ mol^{-1}}\). This entropy increase in the surroundings is one of the reasons why the formation of water is highly spontaneous.

Answer
\(\Delta S_\text{surr} \approx +960\ \mathrm{J\ K^{-1}\ mol^{-1}}\)

Academia Aeternum — Class 11 Chemistry | Thermodynamics Chapter 5

NCERT Exercise Solutions — All 22 Questions Fully Solved

Updated March 2026 · IUPAC Sign Convention · Board · JEE Main · NEET · For Academic Use Only

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    Frequently Asked Questions

    Thermodynamics is the branch of chemistry that deals with energy changes accompanying physical and chemical processes. It studies heat, work, temperature, and the laws governing energy transformations in macroscopic systems.

    A thermodynamic system is the part of the universe selected for study, separated from the surroundings by a boundary. The rest of the universe is called surroundings.

    Systems are of three types: (i) Open system – exchanges both matter and energy, (ii) Closed system – exchanges energy but not matter, (iii) Isolated system – exchanges neither matter nor energy.

    Surroundings refer to everything outside the system that can exchange energy and/or matter with it.

    An open system can exchange both mass and energy with surroundings, while a closed system can exchange only energy, not matter.

    The state of a system is described by measurable properties such as pressure (P), volume (V), temperature (T), and composition at a given moment.

    State functions are properties that depend only on the initial and final states of a system, not on the path taken. Examples: internal energy (U), enthalpy (H), entropy (S), and Gibbs free energy (G).

    Path functions depend on the specific path followed during a process. Heat (q) and work (w) are path functions.

    Internal energy is the total energy contained within a system, including kinetic and potential energies of molecules.

    The first law states: “Energy can neither be created nor destroyed, only transformed.” Mathematically, \( \Delta U = q + w \).

    Heat absorbed by the system is positive (+q), heat released is negative (-q). Work done on the system is positive (+w), work done by the system is negative (-w).

    Expansion work is the work done when a system expands against external pressure: \( w = -P_{ext}\Delta V \).

    For a reversible process, work is maximum and given by \( w = -\int P_{ext} dV \). In ideal reversible expansion: \( w = -nRT \ln \frac{V_2}{V_1} \).

    Enthalpy is a thermodynamic quantity defined as \( H = U + PV \). It represents total heat content at constant pressure.

    At constant pressure, \( \Delta H = q_p \), meaning enthalpy change equals heat exchanged at constant pressure.

    THERMODYNAMICS – Learning Resources

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