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આ MCQ મોડ્યુલ આના પર આધારિત છે: Spontaneity Second Law
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Spontaneity Second Law
🎓 Class 11
Chemistry
CBSE
Theory
Ch 5 – Thermodynamics
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Spontaneity Second Law
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Spontaneity, Entropy and Gibbs Free Energy
5.15 Spontaneity — What Drives a Process?
A spontaneous process is one that has a natural tendency to occur without any external aid. Examples: water flowing downhill, iron rusting, ice melting at 25 °C, two gases mixing.
Caution: Spontaneous ≠ fast! Diamond → graphite is spontaneous but takes geological time. Spontaneity tells us the direction the process tends to go; it says nothing about rate.
Some spontaneous processes are exothermic (combustion of methane, ΔH = −890 kJ/mol) and some are endothermic (NH₄Cl dissolving in water, ΔH = +14 kJ/mol). So enthalpy alone cannot predict spontaneity. We need another driving force.
5.16 Entropy (S)
The missing ingredient is entropy (S) — a measure of the disorder or randomness of a system. The higher the disorder, the higher the entropy.
Entropy: A state function. The change in entropy for a reversible process is:
\[\Delta S = \frac{q_\text{rev}}{T}\]
Boltzmann's microscopic interpretation:
\[S = k_B \ln W\]
where W = number of accessible microstates and k_B = 1.38 × 10⁻²³ J/K.
5.16.1 Predicting the Sign of ΔS
| Process | ΔS | Reason |
|---|---|---|
| Solid → Liquid (melting) | + | More positions available |
| Liquid → Gas (vaporisation) | ++ (very large) | Huge increase in volume/disorder |
| Gas → Liquid (condensation) | − | Order increases |
| Reaction increases moles of gas | + | e.g., 2KClO₃ → 2KCl + 3O₂ |
| Reaction decreases moles of gas | − | e.g., N₂ + 3H₂ → 2NH₃ |
| Mixing of two gases | + | More ways to arrange |
| Increase in temperature | + | More kinetic states accessible |
5.17 The Second Law of Thermodynamics
Second Law (one statement): The entropy of the universe always increases in any spontaneous (real) process.
\[\Delta S_\text{universe} = \Delta S_\text{system} + \Delta S_\text{surroundings} > 0 \quad \text{(spontaneous)}\]
For a reversible process, ΔS_universe = 0. For a non-spontaneous process, ΔS_universe < 0.
Even an exothermic reaction with ΔS_system < 0 can be spontaneous if the surroundings gain enough entropy: ΔS_surr = −ΔH/T > 0, and the total can be positive.
5.17.1 Third Law of Thermodynamics
Third Law (Nernst): The entropy of a perfectly crystalline substance is zero at absolute zero (0 K). This gives an absolute reference point — unlike H or U, we can talk about absolute entropy values S° (not just ΔS).
5.18 Gibbs Free Energy (G)
For chemists working at constant T and P (typical lab/industrial conditions), it is inconvenient to compute ΔS_universe (need data on the surroundings). The American physicist J. W. Gibbs defined a new state function:
Gibbs Free Energy (G):
\[G = H - TS\]
At constant T and P:
\[\boxed{\Delta G = \Delta H - T\Delta S}\]
Why is this useful? Multiplying ΔS_universe by −T gives: \[-T\Delta S_\text{universe} = \Delta H_\text{system} - T\Delta S_\text{system} = \Delta G_\text{system}\] So ΔG measures the increase of disorder of the universe, but using only system properties!
5.18.1 Criteria for Spontaneity
| ΔG | Process |
|---|---|
| < 0 (negative) | Spontaneous in forward direction |
| = 0 | Equilibrium |
| > 0 (positive) | Non-spontaneous forward (reverse is spontaneous) |
5.18.2 Sign Combinations of ΔH and ΔS
| ΔH | ΔS | ΔG = ΔH − TΔS | Spontaneity |
|---|---|---|---|
| − (exo) | + (disorder ↑) | Always − | Spontaneous at all T (e.g., 2H₂O₂ → 2H₂O + O₂) |
| + (endo) | − (disorder ↓) | Always + | Never spontaneous (e.g., 3O₂ → 2O₃) |
| − (exo) | − (disorder ↓) | − at low T | Spontaneous only at LOW T (e.g., water freezing) |
| + (endo) | + (disorder ↑) | − at high T | Spontaneous only at HIGH T (e.g., NH₄Cl dissolving, decomposition reactions) |
5.19 Standard Free Energy & Equilibrium Constant
The standard Gibbs free energy change Δ_rG° is related to the equilibrium constant K by:
\[\Delta_r G^\circ = -RT \ln K = -2.303\,RT\,\log K\]| Δ_rG° | K | Direction at standard state |
|---|---|---|
| < 0 | K > 1 | Products favoured |
| = 0 | K = 1 | Equal amounts |
| > 0 | K < 1 | Reactants favoured |
🎯 Interactive: ΔG Spontaneity Predictor
Set ΔH and ΔS for a reaction. Slide T to see when the reaction becomes spontaneous (ΔG < 0).
ΔG = ΔH − TΔS = −10.20 kJ
SPONTANEOUS (forward) ✓
Switch-over T (T_eq = ΔH/ΔS) = 400 K
🧪 Activity 5.4 — When Does Ice Melt?
Setup: For ice → liquid water at 1 atm: ΔH_fus = +6.01 kJ/mol, ΔS_fus = +22.0 J K⁻¹ mol⁻¹.
Predict: Calculate ΔG at (a) 263 K (−10 °C), (b) 273 K (0 °C), (c) 283 K (+10 °C). At which T is melting spontaneous?
(a) T = 263 K: ΔG = 6010 − 263×22 = 6010 − 5786 = +224 J. Non-spontaneous — ice stays frozen.
(b) T = 273 K: ΔG = 6010 − 273×22 = 6010 − 6006 ≈ 0. Equilibrium — ice and water coexist (this is the melting point!).
(c) T = 283 K: ΔG = 6010 − 283×22 = 6010 − 6226 = −216 J. Spontaneous — ice melts.
Key insight: The melting point is precisely the temperature where ΔG = 0, i.e., T_m = ΔH/ΔS. Below it, freezing wins; above it, melting wins. ΔG = 0 is the natural definition of phase equilibrium.
Worked Example 5.7: Spontaneity at 298 K
For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.4 kJ and ΔS° = −198.3 J K⁻¹. Is the reaction spontaneous at 298 K? Above what temperature does it become non-spontaneous?
At 298 K: ΔG° = ΔH° − TΔS° = (−92.4 × 1000) − 298 × (−198.3) = −92400 + 59093 = −33307 J = −33.3 kJ → spontaneous ✓.
Switch-over temperature (ΔG° = 0): T = ΔH°/ΔS° = (−92400)/(−198.3) = 466 K (≈ 193 °C).
Above 466 K, ΔG° becomes positive — reaction becomes non-spontaneous. This is why industrial NH₃ synthesis uses high pressure to compensate for moderate temperatures (~700 K).
Switch-over temperature (ΔG° = 0): T = ΔH°/ΔS° = (−92400)/(−198.3) = 466 K (≈ 193 °C).
Above 466 K, ΔG° becomes positive — reaction becomes non-spontaneous. This is why industrial NH₃ synthesis uses high pressure to compensate for moderate temperatures (~700 K).
Worked Example 5.8: K from ΔG°
For a reaction at 298 K, Δ_rG° = −13.6 kJ/mol. Calculate the equilibrium constant K. (R = 8.314 J K⁻¹ mol⁻¹)
ΔG° = −2.303 RT log K
−13600 = −2.303 × 8.314 × 298 × log K
−13600 = −5705 × log K
log K = 2.383 → K = 10²·³⁸³ ≈ 242
A modest negative ΔG° (~13 kJ) gives K ~ 240, which is product-favoured but not overwhelmingly so.
−13600 = −2.303 × 8.314 × 298 × log K
−13600 = −5705 × log K
log K = 2.383 → K = 10²·³⁸³ ≈ 242
A modest negative ΔG° (~13 kJ) gives K ~ 240, which is product-favoured but not overwhelmingly so.
🎯 Competency-Based Questions
Q1. The criterion of spontaneity at constant T and P is: L1 Remember
Answer: (c) ΔG < 0. Neither ΔH nor ΔS_system alone is sufficient. ΔG combines them correctly.
Q2. Predict the sign of ΔS for: I₂(s) → I₂(g). L2 Understand
Answer: ΔS > 0 (large positive). Sublimation increases disorder dramatically — solid (highly ordered) → gas (highly disordered).
Q3. NH₄Cl dissolves in water with ΔH = +14 kJ/mol but the process is spontaneous at room temperature. Explain. L4 Analyse
Answer: Despite being endothermic (ΔH > 0), the dissolution scatters NH₄⁺ and Cl⁻ ions throughout the solution → ΔS > 0 (large). At room T, TΔS exceeds ΔH, making ΔG = ΔH − TΔS negative. The reaction proceeds because it increases disorder. The solution feels cold because heat is drawn from surroundings.
Q4. The standard ΔG° for a reaction at 300 K is +5.706 kJ. Calculate K. L3 Apply
Answer: ΔG° = −2.303 RT log K → 5706 = −2.303 × 8.314 × 300 × log K → 5706 = −5744 × log K → log K = −0.993 → K = 10⁻⁰·⁹⁹³ ≈ 0.10. Reaction is reactant-favoured (K << 1).
Q5. HOT (Evaluate): Life involves building highly ordered structures (proteins, DNA) — apparently violating the second law! How is this reconciled? L5 Evaluate
Answer: Life does NOT violate the second law because living systems are not isolated. They take in low-entropy molecules (e.g., glucose) and release high-entropy waste (CO₂, H₂O, heat). The local DECREASE in entropy of the cell is more than compensated by an INCREASE in entropy of the surroundings. The TOTAL entropy of the universe still increases. Biological order is built by exporting disorder.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Spontaneous processes have negative ΔG.
R: Spontaneous processes always release heat.
Answer: (C). A is TRUE. R is FALSE — many endothermic processes (NH₄Cl dissolving) are spontaneous because of large positive ΔS.
A: The entropy of a perfectly crystalline solid is zero at 0 K.
R: At 0 K, only one microstate is accessible (all atoms locked in place).
Answer: (A). Both true; R explains A. By Boltzmann's S = k ln W, when W = 1, S = k ln 1 = 0. This is the third law.
A: A reaction with ΔH > 0 and ΔS > 0 becomes spontaneous at high temperature.
R: The TΔS term grows in magnitude with T, eventually overcoming ΔH.
Answer: (A). Both true; R explains A. T_switch = ΔH/ΔS; above this temperature, ΔG becomes negative.
Frequently Asked Questions — Spontaneity, Entropy and Gibbs Free Energy
What is a spontaneous process?
A spontaneous process is one that occurs naturally under specified conditions without external intervention — though it may need a small initial push (activation energy). Examples include water flowing downhill, ice melting at 25°C, iron rusting in moist air and the dissolution of NaCl in water. Spontaneity does not imply speed: a spontaneous process can be very slow (rusting) or very fast (combustion). In NCERT Class 11 Chemistry Chapter 5, spontaneity is rigorously defined using the second law of thermodynamics in terms of entropy change of the universe (ΔS_total > 0) or Gibbs free energy (ΔG < 0).
What is entropy and how does it change?
Entropy (S) is a state function that measures the degree of disorder or randomness in a system. Higher entropy means greater disorder and more accessible microstates. Entropy increases when: a solid melts to liquid, liquid vaporises, a solute dissolves, temperature increases, the number of moles of gas increases in a reaction, or a gas expands. For an ideal gas isothermal expansion: ΔS = nR ln(V₂/V₁). NCERT Class 11 Chemistry Chapter 5 uses entropy alongside enthalpy to predict spontaneity via Gibbs free energy. ΔS is positive for natural processes in isolated systems.
What is the second law of thermodynamics?
The second law of thermodynamics states that the total entropy of an isolated system (or universe) always increases for any spontaneous process: ΔS_universe = ΔS_system + ΔS_surroundings ≥ 0. The equality holds for reversible processes and the inequality for irreversible (spontaneous) processes. NCERT Class 11 Chemistry Chapter 5 presents the second law as the criterion for spontaneity. It explains why heat flows from hot to cold spontaneously but not vice versa, and why all natural processes tend toward greater disorder. The second law cannot be derived — it is established by experiment.
What is Gibbs free energy?
Gibbs free energy (G) is a state function defined as G = H − TS. At constant temperature and pressure, the change in Gibbs free energy is ΔG = ΔH − TΔS. ΔG predicts spontaneity: ΔG < 0 means the process is spontaneous; ΔG = 0 means the system is at equilibrium; ΔG > 0 means the process is non-spontaneous (the reverse process is spontaneous). NCERT Class 11 Chemistry Chapter 5 uses ΔG to combine the favourable enthalpy and entropy criteria into a single quantity, and to connect thermodynamics to chemical equilibrium via ΔG° = −RT ln K.
How are ΔH, ΔS and ΔG related to spontaneity?
From ΔG = ΔH − TΔS, four cases arise: (1) ΔH < 0, ΔS > 0 → ΔG < 0 always, spontaneous at all temperatures; (2) ΔH > 0, ΔS < 0 → ΔG > 0 always, never spontaneous; (3) ΔH < 0, ΔS < 0 → ΔG < 0 only at low T; (4) ΔH > 0, ΔS > 0 → ΔG < 0 only at high T. NCERT Class 11 Chemistry Chapter 5 uses this analysis to predict why some reactions favour low temperature (e.g., NH₃ synthesis) and others favour high temperature (e.g., CaCO₃ decomposition). Crossover temperature is T = ΔH/ΔS.
What is the third law of thermodynamics?
The third law of thermodynamics states that the entropy of a perfectly crystalline substance at absolute zero (0 K) is zero. This provides an absolute reference point for measuring entropy, unlike enthalpy or internal energy which only have relative scales. From this, absolute entropies S° of substances at any temperature can be calculated using calorimetric data and tabulated as standard molar entropies. NCERT Class 11 Chemistry Chapter 5 introduces the third law to enable use of S° values for calculating ΔS° of reactions: ΔS°_reaction = Σ S°(products) − Σ S°(reactants).
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