Chemical Thermodynamics — Part A

Introduction, Basic Concepts, Internal Energy (ΔU), Heat, Work, First Law, Enthalpy (ΔH). — For NEET & Bihar Board

1. Introduction (परिचय)

Chemical Thermodynamics studies energy changes, particularly heat and work, that accompany chemical reactions and physical transformations. It provides quantitative rules (laws) that predict whether a reaction will occur and how much energy is exchanged.

Key idea: Thermodynamics does not tell the speed of reaction — only whether a change is possible and how much energy is involved.

Importance for NEET and Bihar Board: Questions often test definitions (system, surroundings), numerical problems (ΔU, ΔH, calorimetry), Hess's law, and conceptual understanding of Gibbs energy and spontaneity.

2. System, Surroundings and Types of Systems (प्रणाली और परिवेश)

System (प्रणाली): The part of the universe under study — e.g., reactants in a beaker. Surroundings (परिवेश): Everything outside the system. Universe = system + surroundings.

Types of systems:

• Open system — exchange of both matter and energy with surroundings.
• Closed system — exchange of energy but not matter.
• Isolated system — no exchange of matter or energy (ideal).

Example: A beaker open to air is an open system (mass can leave via evaporation). A sealed, rigid vessel is a closed system.

3. State Functions and Path Functions (राज्य फलन और पथ फलन)

State functions (राज्य फलन) depend only on the current state of the system, not on how it got there. Examples: Internal energy (U), Enthalpy (H), Pressure (P), Temperature (T), Volume (V), Entropy (S).

Path functions depend on the process or path. Examples: heat (q) and work (w).

If A → B, Δ(state function) = state(B) − state(A); independent of path.

4. Internal Energy (आंतरिक ऊर्जा — ΔU)

Internal energy (U) of a system is the sum of kinetic and potential energies of molecules. It is a state function. The change in internal energy is written as ΔU.

ΔU = U_final − U_initial

In practice, we measure ΔU by tracking heat and work interactions with the surroundings via the First Law of Thermodynamics (below).

Work and Heat

Work (w) is energy transfer by a force acting through a distance. Heat (q) is energy transfer due to temperature difference.

Sign convention (chemistry): q > 0 when heat flows into the system; w > 0 when work is done on the system.

PV Work (Expansion/Compression)

The most common work in chemistry is pressure–volume work: w_PV = −P_ext ΔV (when the system expands against external pressure, it does work on surroundings, so system loses energy).

Example — Gas expansion: If a gas expands at constant external pressure from V₁ to V₂, w = −P_ext(V₂ − V₁). If V increases, w is negative (system does work).

5. First Law of Thermodynamics (प्रथम नियम)

The First Law is conservation of energy: Energy cannot be created or destroyed, only transferred.

ΔU = q + w

Here, ΔU is change in internal energy, q is heat exchanged (positive into system), and w is work done on the system (positive if done on system).

Applications — Constant Volume and Constant Pressure

Constant Volume (isochoric) — calorimetry

If V is constant, ΔV = 0 so w_PV = 0. Then ΔU = q_v (heat at constant volume).

Bomb calorimeter measures ΔU for combustion reactions because volume is fixed.

Constant Pressure (isobaric)

At constant pressure, the heat exchanged q_p is related to enthalpy change ΔH (see below): q_p = ΔH (for only PV work).

Calorimetry (simple): For a solution calorimeter (constant pressure), heat absorbed by solution = mass × specific heat × ΔT.

6. Enthalpy (उष्मागतिक अवस्था फलन — ΔH)

Enthalpy (H) is defined as H = U + PV. It is a state function.

ΔH = ΔU + Δ(PV)

For most chemical reactions at constant pressure, with only PV work, the heat exchanged with the surroundings equals the change in enthalpy:
q_p = ΔH

Interpretation

If ΔH < 0 — reaction is exothermic (releases heat).
If ΔH > 0 — reaction is endothermic (absorbs heat).

Standard Enthalpy Changes

Standard enthalpy change ΔH° is defined at standard state (1 bar, pure substances in standard state, usually 298.15 K unless specified).

Hess's Law (brief intro)

Enthalpy is a state function → total ΔH for a reaction is independent of the path. This gives Hess's law: if a reaction can be expressed as sum of steps, ΔH = sum of ΔH of steps.

Simple Hess example: Suppose A → B has ΔH1 and B → C has ΔH2, then A → C has ΔH = ΔH1 + ΔH2.

7. Worked numerical problems (Solved examples)

Problem 1 — Constant volume calorimetry

A bomb calorimeter has a heat capacity of 10.0 kJ K⁻¹. Combustion of 1.00 g of a compound raises the calorimeter temperature by 2.50 K. What is the heat released by the combustion (ΔU)?

Solution: q_v = C × ΔT = 10.0 kJ K⁻¹ × 2.50 K = 25.0 kJ (released). Since heat is released, ΔU = −25.0 kJ for the system (the sample).

Problem 2 — PV work

2.00 mol of ideal gas expands from 5.00 L to 25.0 L against constant external pressure of 1.00 atm. Calculate w (in kJ).

Solution: w = −P_ext ΔV = −(1.00 atm) × (20.0 L) = −20.0 L·atm. Convert: 1 L·atm = 101.325 J → w = −20.0 × 101.325 J = −2026.5 J = −2.03 kJ (approx).

8. Quick summary & Key formulas

• ΔU = q + w
• w_PV = −P_ext ΔV
• H = U + PV
• At constant pressure (only PV work): q_p = ΔH
• Enthalpy is a state function → Hess's law applies

Part A covers: Introduction, system types, state & path functions, internal energy, heat & work, first law, enthalpy and solved examples.

Next: Part B will cover Entropy (S), Second Law, spontaneity, Gibbs free energy (ΔG), and temperature dependence of ΔG.

Chemical Thermodynamics — Part B

Entropy (S), Second Law, Spontaneity, Gibbs Free Energy (ΔG), and temperature dependence — For NEET & Bihar Board

1. Entropy (S) — Concept and Definitions

Entropy (S) is a state function that quantifies the degree of disorder or the number of microscopic ways a system's energy can be arranged. For reversible processes, the infinitesimal change in entropy is defined as:

dS = \frac{\delta q_{rev}}{T}

Where \(\delta q_{rev}\) is the reversible heat exchanged and T is the absolute temperature (in Kelvin).

Key idea: Entropy measures dispersal of energy — higher entropy means energy is more spread out among microstates.

Statistical interpretation (Boltzmann)

S = k_B \ln W

Here, \(k_B\) is the Boltzmann constant (1.380649 × 10⁻²³ J K⁻¹) and W is the number of microstates consistent with the macrostate.

Example: If a macrostate has 10 microstates, S = k ln(10). More microstates → larger S.

Entropy change for processes

ΔS = \int_{i}^{f} \frac{\delta q_{rev}}{T}

For constant temperature (isothermal) reversible processes: ΔS = q_rev / T.

2. Second Law of Thermodynamics (द्वितीय नियम)

The Second Law introduces direction to natural processes. There are many equivalent statements; two common ones are:

• Clausius statement: Heat cannot spontaneously flow from a colder body to a hotter body.
• Entropy statement: For any spontaneous process in an isolated system, the entropy increases: ΔS_univ > 0. For reversible processes ΔS_univ = 0; for non-spontaneous ΔS_univ < 0.

ΔS_univ = ΔS_system + ΔS_surroundings > 0 for spontaneous processes.

Entropy change of surroundings

For processes at constant pressure where the only heat exchange is q_p, the entropy change of surroundings is:

ΔS_surroundings = -\frac{q_{system}}{T} = -\frac{ΔH_{system}}{T} (for constant pressure)

Example — Freezing of water at 273.15 K: Water freezing releases latent heat (ΔH < 0). For system (water), ΔS_sys = ΔH/T (negative). Surroundings gain entropy because heat flows out. Overall ΔS_univ = 0 at equilibrium (melting/freezing point).

3. Gibbs Free Energy (ΔG) and Spontaneity

Gibbs Free Energy combines enthalpy and entropy to predict spontaneity at constant temperature and pressure. Defined as:

G = H - TS

ΔG = ΔH - TΔS

Sign of ΔG indicates spontaneity (at constant T and P):

• ΔG < 0 → process is spontaneous (thermodynamically favorable).
• ΔG = 0 → process is at equilibrium.
• ΔG > 0 → process is non-spontaneous (requires input of work).

Note: Spontaneity is not the same as speed. A reaction can be spontaneous (ΔG < 0) yet very slow (kinetic barrier).

Relation with entropy of universe

At constant T and P: ΔG_sys = -T ΔS_univ. Therefore if ΔS_univ > 0, ΔG_sys < 0: spontaneous.

Standard Gibbs free energy (ΔG°)

ΔG° is calculated at standard states (1 bar, specified T). For reactions, ΔG° is related to the equilibrium constant K:

ΔG° = -RT ln K

And for non-standard conditions: ΔG = ΔG° + RT ln Q, where Q is reaction quotient.

Example — Simple reaction: For A ⇌ B, if ΔG° = −5.7 kJ mol⁻¹ at 298 K, then K = e^(−ΔG°/RT) ≈ e^(−(−5700)/(8.314×298)) ≈ e^(2.30) ≈ 10.0. So equilibrium favors products.

4. Temperature Dependence of ΔG — Van 't Hoff and Gibbs-Helmholtz

The Gibbs-Helmholtz equation relates temperature dependence of ΔG to enthalpy:

\left(\frac{\partial (ΔG/T)}{\partial T}\right)_P = -\frac{ΔH}{T^2}

Gibbs-Helmholtz useful for estimating ΔG at different temperatures if ΔH is (approximately) constant.

Van 't Hoff equation

The Van 't Hoff equation relates temperature dependence of equilibrium constant K to enthalpy change:

\frac{d\ln K}{dT} = \frac{ΔH°}{RT^2}

Integrated form (when ΔH° is constant): ln(K_2/K_1) = -ΔH°/R (1/T_2 - 1/T_1).

Example: For an exothermic reaction (ΔH° < 0), increasing temperature decreases K (equilibrium shifts to reactants).

5. Calculations — ΔG°, ΔG, and Equilibrium

For a general reaction: aA + bB ⇌ cC + dD,

ΔG = ΔG° + RT ln Q = ΔG° + RT ln \frac{[C]^c [D]^d}{[A]^a [B]^b}

At equilibrium Q = K and ΔG = 0 → ΔG° = −RT ln K.

Worked Problem — ΔG from ΔH and ΔS

A reaction has ΔH° = −120 kJ mol⁻¹ and ΔS° = −200 J K⁻¹ mol⁻¹. Calculate ΔG° at 298 K and state spontaneity.

Solution: Convert ΔS to kJ: ΔS = −0.200 kJ K⁻¹ mol⁻¹. Then ΔG° = ΔH° − TΔS° = (−120) − (298)(−0.200) = −120 + 59.6 = −60.4 kJ mol⁻¹. Since ΔG° < 0, reaction is spontaneous at 298 K.

6. Criteria for Spontaneity (at constant T and P)

Use ΔG = ΔH − TΔS to determine spontaneity:

• If ΔH < 0 and ΔS > 0 → ΔG < 0 at all T (always spontaneous).
• If ΔH > 0 and ΔS < 0 → ΔG > 0 at all T (never spontaneous).
• If ΔH < 0 and ΔS < 0 → spontaneous at low T (because −TΔS positive magnitude small at low T).
• If ΔH > 0 and ΔS > 0 → spontaneous at high T.

Example — Melting of ice: ΔH > 0 (endothermic), ΔS > 0 (increased disorder). Melting spontaneous above 273.15 K.

7. Standard Thermodynamic Tables & Using Data

Textbook and NEET problems often provide standard values: ΔH°f (standard enthalpy of formation), S° (standard molar entropy), and ΔG°f (standard Gibbs free energy of formation) at 298 K. Use these to compute reaction values: ΔX°_reaction = Σ ν_products X°_f − Σ ν_reactants X°_f (where X = H, S or G).

Example: ΔH°_reaction = Σ ΔH°f(products) − Σ ΔH°f(reactants).

8. Special topics (brief)

Non-PV work and ΔG

The Gibbs free energy change equals the maximum non-expansion work obtainable from a process at constant T and P (reversible): w_{non-PV,max} = −ΔG.

Coupled reactions

In biochemical or industrial processes, a non-spontaneous reaction (ΔG > 0) can be driven by coupling it with a spontaneous reaction (ΔG < 0) such that overall ΔG_total < 0.

9. Practice Questions (NEET/Bihar style)

1. State the Second Law of Thermodynamics in terms of entropy. (1 mark)
2. Calculate ΔS for melting 2.00 mol of a solid with enthalpy of fusion 6.00 kJ mol⁻¹ at its melting point 350 K. (3 marks)
3. For the reaction N₂ + 3H₂ ⇌ 2NH₃, given ΔH° = −92.4 kJ and ΔS° = −198.6 J K⁻¹, determine spontaneity at 298 K. (3 marks)
4. Explain relationship between ΔG° and equilibrium constant K. (2 marks)

10. Key Formulas & Quick Revision

• dS = δq_rev / T
• ΔS_univ = ΔS_sys + ΔS_surr > 0 for spontaneous processes
• G = H − TS
• ΔG = ΔH − TΔS
• ΔG° = −RT ln K
• ΔG = ΔG° + RT ln Q
• Van 't Hoff: d ln K / dT = ΔH° / RT^2

Part B covers entropy, the second law, Gibbs energy and related calculations. Next: Part C will include Hess's law in depth, advanced calorimetry, reaction energy profiles, and common NEET numerical problems.

रासायनिक ऊष्मागतिकी (Chemical Thermodynamics) - Part C

भाग – C : हेस का नियम, कैलोरीमेट्री और अनुप्रयोग (Hess's Law, Calorimetry & Applications)

1. हेस का ऊष्मा योग नियम (Hess’s Law of Constant Heat Summation)

यह नियम कहता है कि किसी रासायनिक अभिक्रिया में यदि प्रारंभिक और अंतिम अवस्था एक ही है, तो अभिक्रिया की कुल एन्थैल्पी परिवर्तन (ΔH) अभिक्रिया के मार्ग पर निर्भर नहीं करती।

Hess's Law: ΔH (कुल अभिक्रिया) = ΔH₁ + ΔH₂ + ΔH₃ + ...

Example:
मान लीजिए हमारे पास निम्न अभिक्रियाएँ हैं —
(1) C(s) + O₂(g) → CO₂(g)  ΔH = -393.5 kJ/mol
(2) C(s) + 1/2O₂(g) → CO(g)  ΔH = -110.5 kJ/mol
(3) CO(g) + 1/2O₂(g) → CO₂(g)  ΔH = -283.0 kJ/mol

स्पष्ट है कि (2) + (3) = (1) और कुल ΔH = -393.5 kJ/mol = (-110.5) + (-283.0)

2. कैलोरीमेट्री (Calorimetry)

कैलोरीमेट्री एक प्रयोगात्मक तकनीक है जिसके द्वारा ऊष्मा परिवर्तन (heat change) को मापा जाता है।

मूल सिद्धांत: जब दो वस्तुएँ संपर्क में आती हैं, तो गर्म वस्तु ऊष्मा खोती है और ठंडी वस्तु ऊष्मा प्राप्त करती है।
अतः,

Heat lost = Heat gained

कैलोरीमीटर समीकरण:

m₁ c₁ (T₁ - T_f) = m₂ c₂ (T_f - T₂)

जहाँ —

• m = पदार्थ का द्रव्यमान
• c = विशिष्ट ऊष्मा (specific heat)
• T = तापमान
• T_f = अंतिम तापमान

3. ऊष्मा धारिता (Heat Capacity)

किसी पदार्थ का तापमान 1°C (या 1K) बढ़ाने हेतु आवश्यक ऊष्मा की मात्रा को ऊष्मा धारिता कहते हैं।

C = q / ΔT

जहाँ C = ऊष्मा धारिता, q = ऊष्मा, और ΔT = तापमान परिवर्तन।

विशिष्ट ऊष्मा धारिता (Specific Heat Capacity):

1 ग्राम पदार्थ का तापमान 1°C बढ़ाने हेतु आवश्यक ऊष्मा की मात्रा।

4. निर्माण ऊष्मा (Heat of Formation)

जब 1 मोल यौगिक उसके तत्त्वों से उनके मानक अवस्थाओं में बनता है, तब जो ऊष्मा उत्पन्न या अवशोषित होती है, उसे निर्माण ऊष्मा कहते हैं।

C(s) + 1/2 O₂(g) → CO(g) ΔH_f = -110.5 kJ/mol

5. अभिक्रिया की ऊष्मा (Heat of Reaction)

किसी रासायनिक अभिक्रिया में नियत दाब या नियत आयतन पर ऊष्मा परिवर्तन को अभिक्रिया की ऊष्मा कहते हैं।

यह अभिक्रिया के तापमान, दाब और अवस्था पर निर्भर करता है।

6. अभिक्रिया प्रोफ़ाइल (Reaction Energy Profile)

यह एक ग्राफ होता है जो अभिक्रिया के दौरान ऊर्जा परिवर्तन को दर्शाता है।

7. ऊष्मागतिकीय समीकरणों का अनुप्रयोग (Applications of Thermodynamics)

• रासायनिक अभिक्रियाओं की दिशा निर्धारण में।
• ऊर्जा रूपांतरण के अध्ययन में।
• औद्योगिक प्रक्रियाओं जैसे ईंधन दहन, बैटरी, और रॉकेट ईंधन में।
• भौतिक परिवर्तन (गलन, वाष्पीकरण आदि) की ऊष्मा गणना में।

8. अभ्यास प्रश्न (Practice Questions)

1. Hess के नियम को उदाहरण सहित समझाइए।
2. कैलोरीमीटर का सिद्धांत क्या है?
3. Heat of formation और heat of reaction में क्या अंतर है?
4. Heat capacity और specific heat में अंतर स्पष्ट कीजिए।
5. ऊष्मागतिकी के व्यावहारिक उपयोग क्या हैं?

9. महत्वपूर्ण सूत्र (Key Formulas)

• ΔH = q_p
• ΔU = q_v + W
• ΔG = ΔH - TΔS
• C = q / ΔT
• Heat lost = Heat gained

10. निष्कर्ष (Conclusion)

ऊष्मागतिकी रासायनिक विज्ञान की नींव है। यह हमें ऊर्जा परिवर्तन, अभिक्रियाओं की दिशा, तथा उनकी स्वस्फूर्तता (spontaneity) की गहरी समझ प्रदान करती है। Hess का नियम और कैलोरीमेट्री जैसी तकनीकें न केवल प्रयोगशाला में बल्कि उद्योगों और इंजीनियरिंग के क्षेत्र में भी अत्यंत उपयोगी हैं।

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© Willer Academy | Designed for NEET + Bihar Board Students

Chemical Thermodynamics — Part D (Revision & Advanced Applications)

Third Law, Advanced Numericals, Industrial Applications, Energy Diagrams, PYQ-style problems, and a concise formula sheet — For NEET & Bihar Board

1. Third Law of Thermodynamics

Third Law: The entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero (0 K).

S → 0 as T → 0 K (for a perfect crystal)

This provides an absolute reference for entropy and allows calculation of absolute entropies (S°) by integrating heat capacities from 0 K to the temperature of interest and adding contributions from phase changes.

Application: Standard molar entropies (S°) tabulated for many substances are based on the Third Law and are used in reaction entropy and ΔG calculations.

2. Advanced Numerical Problems (with detailed solutions)

Problem 1 — Bomb calorimeter & ΔU

A 0.500 g sample of a hydrocarbon is burned in a bomb calorimeter. The calorimeter constant is 8.50 kJ K⁻¹. If the temperature rises from 298.00 K to 302.75 K, calculate ΔU (in kJ mol⁻¹) for the combustion. Molar mass of the hydrocarbon = 86.0 g mol⁻¹.

Solution:

1. Heat released by combustion = C × ΔT = 8.50 kJ K⁻¹ × (302.75 − 298.00) K = 8.50 × 4.75 = 40.375 kJ (released).
2. For the sample (0.500 g), q = −40.375 kJ (negative because system released heat).
3. Moles of sample = 0.500 g / 86.0 g mol⁻¹ = 0.005814 mol.
4. ΔU per mole = q / n = (−40.375 kJ) / 0.005814 mol = −6943 kJ mol⁻¹ (approx −6.94 × 10³ kJ mol⁻¹).

Problem 2 — ΔG at non-standard conditions

For reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔG° = −33.0 kJ mol⁻¹ at 298 K. Calculate ΔG when pressures are P(N₂) = 0.80 bar, P(H₂) = 0.50 bar, P(NH₃) = 0.10 bar.

Solution:

1. Reaction quotient Q = (P_NH3)^2 / (P_N2 × P_H2^3) = (0.10)^2 / (0.80 × (0.50)^3) = 0.01 / (0.80 × 0.125) = 0.01 / 0.10 = 0.10.
2. ΔG = ΔG° + RT ln Q = −33.0 kJ + (8.314 × 10⁻³ kJ K⁻¹ mol⁻¹) × 298 K × ln(0.10).
   ln(0.10) = −2.3026 → RT ln Q = 8.314×10⁻³ × 298 × (−2.3026) = −5.70 kJ (approx).
3. ΔG = −33.0 + (−5.70) = −38.7 kJ mol⁻¹. Reaction remains spontaneous under given conditions.

Problem 3 — ΔS calculation from calorimetry

1.50 mol of a solid melts at its melting point 450 K with enthalpy of fusion 18.0 kJ mol⁻¹. Calculate ΔS for the melting process.

Solution: ΔS = ΔH / T = (18.0 kJ mol⁻¹) / 450 K = 0.040 kJ K⁻¹ mol⁻¹ = 40.0 J K⁻¹ mol⁻¹. For 1.50 mol, total ΔS = 1.50 × 40.0 = 60.0 J K⁻¹.

3. Thermochemistry in Daily Life & Industry

• Combustion engines and fuels: Enthalpy of combustion determines fuel efficiency and heat released per mole.
• Batteries & fuel cells: ΔG gives maximum electrical work obtainable from electrochemical cells.
• Chemical manufacturing: Reaction conditions (T, P) are chosen using ΔG and Le Chatelier's principle to maximize yield.
• Refrigeration & heat pumps: Based on thermodynamic cycles (Carnot limit), efficiency is related to temperature differences.
• Material processing: Melting, sintering, and phase changes use enthalpy and entropy data to set process parameters.

Example: Haber process — equilibrium and ΔG determine optimal T and P. Although high pressure favors NH₃ formation (Le Chatelier), very low temperature would be thermodynamically favorable but kinetically slow; industry uses compromise (≈450°C and 150–300 atm) with catalysts to improve rate and yield.

4. Energy Diagrams & Reaction Coordinate

Typical energy profile shows reactants to products with an activation energy (E_a) peak. Exothermic reactions have products at lower energy than reactants; endothermic reactions have products at higher energy.

Activation energy affects kinetics (reaction rate). Thermodynamics (ΔH, ΔG) predicts direction and energy change, not the rate.

5. Important NEET & Bihar Board Type Questions (Past Year Style)

1. Define entropy. For a reversible isothermal expansion of an ideal gas, derive the expression for ΔS. (5 marks)
2. State Hess’s law and use it to determine ΔH for the reaction: C₂H₄(g) + 3 O₂(g) → 2 CO₂(g) + 2 H₂O(l) given formation enthalpies. (6 marks)
3. Calculate ΔG° from given ΔH° and ΔS° and comment on spontaneity at two different temperatures. (4 marks)
4. Explain the third law and how absolute entropy values are obtained. (3 marks)

6. Quick Revision — Formula & Table

• ΔU = q + w
• w_PV = −P_ext ΔV
• H = U + PV
• ΔH = q_p; ΔU = q_v
• dS = δq_rev/T
• ΔS = Σ S°(products) − Σ S°(reactants)
• G = H − TS; ΔG = ΔH − TΔS
• ΔG° = −RT ln K; ΔG = ΔG° + RT ln Q
• Gibbs-Helmholtz: (∂(ΔG/T)/∂T)_P = −ΔH/T²

7. Conclusion & Study Tips

Part D completes the four-part chapter set. Revision tips:

• Memorize key formulas and understand sign conventions for q, w, ΔH, ΔU and ΔG.
• Practice Hess law by manipulating reaction equations (reverse, multiply) and adding their ΔH values.
• Solve calorimetry problems frequently — track sign conventions carefully.
• Always write units and convert ΔS in consistent units with ΔH when using ΔG = ΔH − TΔS.

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