How can chemists predict whether a chemical reaction will explode violently or not react at all just by calculating a single value called Gibbs Free Energy ($\Delta G$)? Enthalpy and Entropy govern thermodynamic spontaneity.
यह अध्याय क्यों महत्वपूर्ण है
In Class 11 Chemistry, "Thermodynamics" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
अध्ययन से पूर्व (आवश्यक ज्ञान)
Exothermic and endothermic reactions from Class 10.
First Law of Thermodynamics from Physics.
Work done.
इस अध्याय के लक्ष्य
Distinguish between System (Open, Closed, Isolated) and Surroundings; State Functions vs Path Functions.
State First Law of Thermodynamics: $\Delta U = q + w$ and define Enthalpy ($H = U + PV, \Delta H = q_p$).
State Hess's Law of Constant Heat Summation.
Define Entropy ($S = \frac{q_{rev}}{T}$) and the Second Law of Thermodynamics ($\Delta S_{\text{total}} > 0$).
Apply Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$; evaluate Spontaneity ($\Delta G < 0$ for spontaneous process).
अध्याय रूपरेखा एवं प्रगति
11. Internal Energy & Enthalpy
22. Entropy & The Second Law
33. Gibbs Free Energy ($\Delta G$) &...
सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन
1. Internal Energy & Enthalpy
First Law: $\mathbf{\Delta U = q + w}$. Internal energy $U$ is a state function.
Enthalpy ($H$): Heat content at constant pressure: $$\mathbf{\Delta H = \Delta U + \Delta n_g R T} \quad (\Delta n_g = \text{gaseous product moles} - \text{reactant moles})$$ Exothermic: $\Delta H < 0$; Endothermic: $\Delta H > 0$.
Hess's Law: Total enthalpy change is identical whether a reaction occurs in one step or multiple steps!
2. Entropy & The Second Law
Entropy ($S$): A thermodynamic measure of atomic randomness or disorder: $$\mathbf{\Delta S = \frac{q_{\text{rev}}}{T}} \quad (\text{J/K}\cdot\text{mol})$$ Second Law of Thermodynamics: In any spontaneous process, the total entropy of the universe increases: $\mathbf{\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0}$.
3. Gibbs Free Energy ($\Delta G$) & Spontaneity
The master criterion of chemical feasibility: $$\mathbf{\Delta G = \Delta H - T\Delta S}$$ • $\Delta G < 0$ (Negative): Reaction is strictly Spontaneous. • $\Delta G = 0$: Reaction is at Dynamic Equilibrium. • $\Delta G > 0$ (Positive): Non-spontaneous in forward direction.
Thermodynamics - Key Conceptual Architecture & Molecular Model
अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष
मुख्य बिंदु 1
Enthalpy Relation: $\Delta H = \Delta U + \Delta n_g RT$ linking isobaric and isochoric heats.
मुख्य बिंदु 2
Hess's Law: Enthalpy change is strictly path-independent (state function).
मुख्य बिंदु 3
Entropy ($S$): Fundamental metric quantifying microscopic chaos and atomic disorder.
मुख्य बिंदु 4
Gibbs Free Energy: $\Delta G = \Delta H - T\Delta S$ establishing universal chemical spontaneity.
मुख्य बिंदु 5
Equilibrium Relation: $\Delta G^\circ = -2.303 RT \log_{10} K$ connecting free energy to equilibrium constant.
स्व-मूल्यांकन अभ्यास (Check Your Understanding)
मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।
1
State Hess's Law of Constant Heat Summation.
उत्तर एवं व्याख्या देखें
उत्तर: Hess's Law states that if a chemical reaction can take place in several steps, the overall enthalpy change for the reaction is the same regardless of the pathway taken, because enthalpy is a state function. Total enthalpy change is independent of pathway.
2
For a reaction, $\Delta H = -110\text{ kJ/mol}$ and $\Delta S = -40\text{ J/K}\cdot\text{mol}$. At what temperature range will the reaction be spontaneous?
उत्तर एवं व्याख्या देखें
उत्तर: For spontaneity: $\Delta G = \Delta H - T\Delta S < 0 \implies -110,000 - T(-40) < 0 \implies 40T < 110,000 \implies T < \frac{110,000}{40} = 2750\text{ K}$. The reaction is spontaneous below 2750 K. Spontaneous below 2750 K.
3
Differentiate between an Extensive property and an Intensive property with examples.
उत्तर एवं व्याख्या देखें
उत्तर: An Extensive property depends on the amount of matter in the system (e.g. Mass, Volume, Enthalpy, Heat Capacity); an Intensive property is independent of the amount of substance (e.g. Temperature, Density, Pressure, Molar Heat Capacity). Dependent on mass (Extensive) vs independent of mass (Intensive).
4
Calculate $\Delta U$ when a system absorbs $700\text{ J}$ of heat and performs $300\text{ J}$ of work on surroundings.
उत्तर एवं व्याख्या देखें
उत्तर: By First Law: $\Delta U = q + w$. Heat absorbed $q = +700\text{ J}$. Work done by system $w = -300\text{ J}$. $\Delta U = 700 + (-300) = +400\text{ Joules}$. ΔU = +400 J.
5
Why is the standard entropy of a pure crystalline solid zero at absolute zero ($0\text{ K}$)?
उत्तर एवं व्याख्या देखें
उत्तर: By the Third Law of Thermodynamics, at absolute zero ($0\text{ K}$), all thermal atomic vibrations cease completely and a perfectly pure crystalline substance attains perfect atomic order, so its entropy is zero ($S = 0$). Third Law: perfect crystalline order at absolute zero.
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