Thermodynamics — Class 11 Chapter 5 · Complete NCERT Exercise Solutions
Every question fully stated · Every solution fully explained · IUPAC sign convention · Board + JEE + NEET aligned
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.
Topics Covered in This Exercise
Internal energy, work done on/by system, heat exchange — IUPAC sign convention applied and explained in every numerical.
ΔH calculations from formation, combustion & neutralisation enthalpies; multi-step thermochemical Hess's cycles explained from first principles.
ΔS, spontaneity criteria, ΔG = ΔH − TΔS, and ΔG° = −RT ln K — the critical link between thermodynamics and equilibrium.
Bonds broken vs bonds formed — a top-scoring topic in NEET & JEE. Full atomisation cycle shown with Hess's Law.
Sign convention errors, J vs kJ unit mistakes, isothermal vs adiabatic confusion — flagged and explained wherever they arise.
Each solution cross-referenced with Board, JEE Main & NEET question trends. Updated March 2026.
Maximising Your Marks
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
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
Choose the correct answer. A thermodynamic state function is a quantity
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:
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.
Evaluating Statements (i) through (iv)
Choose the correct answer. For the process to occur under adiabatic conditions, the correct condition is:
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:
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:
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.
Evaluating Each Condition
Consequence of the Adiabatic Condition
When \(q = 0\) is substituted into the First Law:
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.
Choose the correct answer. The enthalpies of all elements in their standard states are:
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.
Some examples of this convention:
Evaluating Each Option
\(\Delta U^\Theta\) of combustion of methane is \(-X\ \mathrm{kJ\ mol^{-1}}\). The value of \(\Delta H^\Theta\) is:
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 \(H = U + pV\):
For ideal gases, \(pV = nRT\), so \(\Delta(pV) = \Delta n_g \cdot RT\), giving:
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 1: Write the Balanced Combustion Equation for Methane
Step 2: Calculate Δn_g
Count only gaseous species on each side:
Step 3: Apply the ΔH–ΔU Relationship
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 \(\Delta U\):
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 \(\Delta H^\Theta < \Delta U^\Theta\).
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:
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:
The three combustion reactions we are given are:
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
Step 4: Add the Three Equations
When we add these three equations, species appearing on both sides cancel:
The net equation is:
Step 5: Sum the Enthalpy Changes
A reaction, \(\mathrm{A + B \rightarrow C + D + q}\) is found to have a positive entropy change. The reaction will be:
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).
The spontaneity rules are:
The four possible combinations of \(\Delta H\) and \(\Delta S\) signs, and their spontaneity consequences, are summarised below — this table is essential for MCQs:
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.
The question also states explicitly that the entropy change is positive:
Step 2: Apply the Gibbs Equation
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.
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 \(\Delta G < 0\).
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?
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:
The sign convention used in modern chemistry (IUPAC convention) defines everything from the system's perspective:
Step 1: Assign Signs to q and w
Step 2: Apply the First Law
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 \(701 - 394 = 307\ \mathrm{J}\).
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.
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 \(H = U + pV\):
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.
Step 1: Identify Gaseous Reactants and Products
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
Step 3: Identify All Given Values
Step 4: Calculate the Correction Term Δn_g RT
Step 5: Calculate ΔH
The result is slightly more negative than \(\Delta U\), consistent with \(\Delta n_g < 0\).
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 \(24\ \mathrm{J\ mol^{-1}\ K^{-1}}\).
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:
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).
Step 1: Calculate the Number of Moles of Aluminium
Molar mass of aluminium: Al = 27 g mol⁻¹
Step 2: Calculate the Temperature Change
Step 3: Apply the Heat Equation
Step 4: Convert to kJ
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.
\(C_p[\mathrm{H_2O(l)}] = 75.3\ \mathrm{J\ mol^{-1}\ K^{-1}}\)
\(C_p[\mathrm{H_2O(s)}] = 36.8\ \mathrm{J\ mol^{-1}\ K^{-1}}\)
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:
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:
Step 1: Cool Liquid Water from 10°C to 0°C
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:
Step 3: Cool Ice from 0°C to −10°C
Step 4: Sum All Contributions
All three contributions are negative, which makes physical sense — the entire process involves releasing heat (cooling and freezing).
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.
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:
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 1: Calculate Moles of CO₂ Formed
Molar mass of CO₂: \(12 + 2 \times 16 = 44\ \mathrm{g\ mol^{-1}}\)
Step 2: Scale the Enthalpy Change
Since 1 mol of CO₂ formation releases 393.5 kJ, the formation of 0.8 mol releases:
The negative sign confirms heat is released (exothermic). The magnitude of heat released is 314.8 kJ.
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:
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:
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 1: Identify and List All ΔfH° Values
Step 2: Calculate ΣΔfH° of Products
Products: 1 mol N₂O and 3 mol CO₂
Step 3: Calculate ΣΔfH° of Reactants
Reactants: 1 mol N₂O₄ and 3 mol CO
Step 4: Calculate Δ_r H°
Given:
What is the standard enthalpy of formation of \(\mathrm{NH_3}\) gas?
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 \(\Delta_f H^\Theta = 0\):
Step 1: Write the Formation Equation (for 2 mol NH₃)
Step 2: Apply the Enthalpy of Formation Formula
Since N₂(g) and H₂(g) are elements in their standard states:
Step 3: Substitute and Solve
Calculate the standard enthalpy of formation of \(\mathrm{CH_3OH(l)}\) from the following data:
Deriving ΔfH° from Combustion Data Using Hess's Law
The standard enthalpy of formation of methanol is the enthalpy change for:
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.
Step 1: Apply the Hess's Law Formula to the Combustion of Methanol
For the combustion of methanol:
Since \(\Delta_f H^\Theta(\mathrm{O_2}) = 0\) (element in standard state):
Step 2: Evaluate the Bracket
Step 3: Solve for ΔfH°(CH₃OH)
Calculate the enthalpy change for the process
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)
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:
For CCl₄, there are four equivalent C–Cl bonds. When all four are broken:
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:
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: \(\mathrm{C(graphite) \rightarrow C(g)}\)
For chlorine: The atomisation enthalpy of Cl₂ = 242 kJ mol⁻¹ refers to:
For 4 Cl atoms, we need 2 mol Cl₂:
Total enthalpy to form all atoms from standard-state elements:
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:
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):
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):
C–Cl bond enthalpy = 326 kJ mol⁻¹
For an isolated system, \(\Delta U = 0\), what will be \(\Delta S\)?
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:
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:
This inequality has two cases:
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.
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:
In either case, \(\Delta S \geq 0\). It is never negative in an isolated system.
For the reaction at 298 K,
At what temperature will the reaction become spontaneous considering \(\Delta H\) and \(\Delta S\) to be constant over the temperature range?
Finding the Temperature of Spontaneity
The spontaneity of a reaction is determined by the sign of the Gibbs free energy change:
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 \(\Delta G = 0\):
For \(T > T_\text{critical}\): \(\Delta G < 0\) → spontaneous.
For \(T < T_\text{critical}\): \(\Delta G > 0\) → non-spontaneous.
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.
Step 2: Set ΔG = 0 and Solve for T
Step 3: Substitute Values
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 \(\Delta H\).
For the reaction,
what are the signs of \(\Delta H\) and \(\Delta S\)?
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.
A decrease in the number of gas particles means a decrease in the randomness/disorder of the system, so entropy decreases. Therefore \(\Delta S < 0\).
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:
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:
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.
For the reaction
Calculate \(\Delta G^\Theta\) for the reaction, and predict whether the reaction may occur spontaneously at 298 K.
Complete Workflow: ΔU → ΔH → ΔG
This problem requires a two-step conversion:
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 (\(U\)).
Step 1: Calculate Δn_g
Step 2: Convert ΔU to ΔH
Step 3: Convert ΔS to kJ K⁻¹
Step 4: Calculate ΔG° Using the Gibbs Equation
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.
The equilibrium constant for a reaction is 10. What will be the value of \(\Delta G^\Theta\)?
\(R = 8.314\ \mathrm{J\ K^{-1}\ mol^{-1}},\ T = 300\ \mathrm{K}\)
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:
or equivalently, using base-10 logarithms:
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 \(K = 1\), \(\Delta G^\Theta = 0\).
Step 1: Write Down All Given Values
Step 2: Choose the Appropriate Formula
Since \(K = 10\) is a convenient power of 10, use the \(\log_{10}\) form to simplify calculation:
Step 3: Evaluate log K
Step 4: Substitute All Values
Step 5: Convert to kJ
The negative value confirms that since \(K > 1\), the forward reaction is favoured under standard conditions.
Comment on the thermodynamic stability of NO(g), given
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:
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 1: Stability of NO with Respect to N₂ and O₂ (its constituent elements)
The formation of NO from its elements is given by:
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:
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:
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:
• 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).
Calculate the entropy change in surroundings when 1.00 mol of \(\mathrm{H_2O(l)}\) is formed under standard conditions.
\(\Delta_f H^\Theta = -286\ \mathrm{kJ\ mol^{-1}}\)
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:
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:
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 \(\Delta H_\text{sys} < 0\).
Step 1: Write the Reaction
The formation of liquid water from hydrogen and oxygen:
Step 2: Convert ΔfH° to Joules
Step 3: Identify Temperature
Standard conditions: \(T = 298\ \mathrm{K}\)
Step 4: Apply the Formula for ΔS_surr
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.