Q.Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason (R): Decrease in enthalpy is a contributory factor for spontaneity.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.What is the enthalpy change (in kJ mol$^{-1}$) for the following reaction? CH$_4$(g) $\rightarrow$ C(g) + 4H(g) Given: ($\Delta_f H^\Theta$(CH$_4$) = $-74.8$ kJ mol$^{-1}$ ; H$_2$(g) $\rightarrow$ 2H(g) ; $\Delta_a H^\Theta$ = $435$ kJ mol$^{-1}$ ; C(s) $\rightarrow$ C(g) ; $\Delta_a H^\Theta$ = $716.7$ kJ mol$^{-1}$) (A) 396.67 (B) 1586.7 (C) 1661.5 (D) 415.37

   ›Reveal solutionSolution

   The enthalpy change for breaking CH₄ into gaseous atoms is the sum of the atomisation enthalpies of carbon and hydrogen, minus the formation enthalpy of methane. The result is 1661.5 kJ mol⁻¹, which corresponds to option (C).

   The key idea here is that we want the enthalpy for atomising methane — turning it into isolated gaseous atoms. We are given the standard enthalpy of formation of methane (from its elements in their standard states) and the atomisation enthalpies of carbon (solid → gas) and hydrogen (H₂ → 2H). By constructing a thermodynamic cycle (Hess’s law), we can find the desired value.

   Why this works:

   The formation of CH₄ from its elements in their standard states (C(s) and H₂(g)) releases energy. To go the other way — from CH₄ to atoms — we must first reverse that formation (costing +74.8 kJ mol⁻¹), then atomise the carbon solid (costing +716.7 kJ mol⁻¹), and finally atomise two H₂ molecules into four H atoms (costing 2 × 435 kJ mol⁻¹). Adding these steps gives the total enthalpy change.

   1. Write the target reaction:

   $\text{CH}_4(g) \rightarrow \text{C}(g) + 4\text{H}(g)$

   We need $\Delta H$ for this.

   2. List the given data:

      • $\Delta_f H^\Theta(\text{CH}_4) = -74.8\ \text{kJ mol}^{-1}$ (This is for: $\text{C}(s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)$)
      • $\text{H}_2(g) \rightarrow 2\text{H}(g)$; $\Delta_a H^\Theta = 435\ \text{kJ mol}^{-1}$
      • $\text{C}(s) \rightarrow \text{C}(g)$; $\Delta_a H^\Theta = 716.7\ \text{kJ mol}^{-1}$

   3. Reverse the formation of methane:

   $\text{CH}_4(g) \rightarrow \text{C}(s) + 2\text{H}_2(g) \quad \Delta H = +74.8\ \text{kJ mol}^{-1}$

   4. Atomise the carbon: $$$$

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2. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Identify the reaction that takes place in the Blast furnace at 1570 K - 2170 K (A) $\mathrm{FeO} + \mathrm{CO} \rightarrow \mathrm{Fe} + \mathrm{CO}_2$ (B) $\mathrm{Fe}_3\mathrm{O}_4 + 4\mathrm{CO} \rightarrow 3\mathrm{Fe} + 4\mathrm{CO}_2$ (C) $\mathrm{Fe}_2\mathrm{O}_3 + \mathrm{CO} \rightarrow 2\mathrm{FeO} + \mathrm{CO}_2$ (D) $\mathrm{FeO} + \mathrm{C} \rightarrow \mathrm{Fe} + \mathrm{CO}$

   ›Reveal solutionSolution

   In the hottest zone of the blast furnace (1570 K - 2170 K), the remaining iron(II) oxide ($\mathrm{FeO}$) is directly reduced to iron by carbon (coke). The correct reaction is $\boxed{\mathrm{FeO} + \mathrm{C} \rightarrow \mathrm{Fe} + \mathrm{CO}}$.

   Concept and Intuition

   The extraction of iron from its oxides in a blast furnace is a complex process involving multiple reactions occurring at different temperature zones. The fundamental principle is the reduction of iron oxides ($\mathrm{Fe_2O_3}$, $\mathrm{Fe_3O_4}$, $\mathrm{FeO}$) to metallic iron ($\mathrm{Fe}$) using reducing agents like carbon monoxide ($\mathrm{CO}$) and carbon ($\mathrm{C}$, from coke).

   The effectiveness of a reducing agent depends on temperature. This can be understood by considering the change in Gibbs free energy ($\Delta G$) for the reduction reactions. According to the Ellingham diagram, which plots $\Delta G^\circ$ versus temperature for the formation of oxides, the relative stability of oxides changes with temperature.

   • At lower temperatures (typically below 1073 K or 1123 K), carbon monoxide ($\mathrm{CO}$) is a more effective reducing agent for iron oxides. This is because the $\Delta G^\circ$ for the formation of $\mathrm{CO_2}$ from $\mathrm{CO}$ is more negative than for the formation of $\mathrm{FeO}$ from $\mathrm{Fe}$ in this range.
   • At higher temperatures (above 1073 K or 1123 K), carbon ($\mathrm{C}$) becomes a more powerful reducing agent. The $\Delta G^\circ$ for the formation of $\mathrm{CO}$ from $\mathrm{C}$ becomes more negative than for the formation of $\mathrm{FeO}$ from $\mathrm{Fe}$, making direct reduction by carbon thermodynamically favourable.

   The blast furnace is designed to exploit these thermodynamic principles by maintaining different temperature zones.

   Step-by-Step Analysis

   1. Understanding the Blast Furnace Temperature Zones:

      A blast furnace operates with a temperature gradient, with the coolest zone at the top and the hottest at the bottom.

      • Top Zone (500 K - 800 K): Here, iron(III) oxide ($\mathrm{Fe_2O_3}$) is gradually reduced by carbon monoxide ($\mathrm{CO}$) to iron(II,III) oxide ($\mathrm{Fe_3O_4}$) and then to iron(II) oxide ($\mathrm{FeO}$).
        • $\mathrm{3Fe_2O_3} + \mathrm{CO} \rightarrow \mathrm{2Fe_3O_4} + \mathrm{CO_2}$
        • $\mathrm{Fe_3O_4} + \mathrm{CO} \rightarrow \mathrm{3FeO} + \mathrm{CO_2}$
      • Middle Zone (900 K - 1200 K): In this zone, the iron(II) oxide ($\mathrm{FeO}$) formed in the upper regions is further reduced to metallic iron ($\mathrm{Fe}$) primarily by carbon monoxide.
        • $\mathrm{FeO} + \mathrm{CO} \rightarrow \mathrm{Fe} + \mathrm{CO_2}$
      • Lower Zone (1570 K - 2170 K): This is the hottest part of the furnace, near the tuyeres where hot air is blown in. Here, the remaining iron(II) oxide ($\mathrm{FeO}$) is reduced directly by carbon (coke). Also, limestone decomposes, and coke burns to produce heat and carbon monoxide.
        • $\mathrm{C} + \mathrm{O_2} \rightarrow \mathrm{CO_2}$ (combustion of coke)
        • $\mathrm{CO_2} + \mathrm{C} \rightarrow \mathrm{2CO}$ (Boudouard reaction)
        • $\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$ (decomposition of limestone)

   2. Identifying the Reaction at 1570 K - 2170 K:

      The question specifically asks for the reaction occurring in the temperature range of 1570 K - 2170 K. This corresponds to the lower, hottest zone of the blast furnace. In this zone, the direct reduction of iron(II) oxide by carbon becomes the dominant reaction for the final reduction step.

      Let's evaluate the given options:

      • (A) $\mathrm{FeO} + \mathrm{CO} \rightarrow \mathrm{Fe} + \mathrm{CO_2}$: This reaction is significant in the middle temperature range (900 K - 1200 K) where CO is an effective reducing agent for $\mathrm{FeO}$. While it might still occur to some extent at higher temperatures, it is not the characteristic or predominant reaction in the 1570 K - 2170 K range where carbon's reducing power is superior. …

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3. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Identify the correct statements from the following: I. The energy of an isolated system is constant II. In a closed system, exchange of both energy and matter with surroundings is possible III. Enthalpy is an extensive property (A) I, II, III (B) I, III only (C) II, III only (D) I, II only

   ›Reveal solutionSolution

   The question tests your grasp of the fundamental definitions of isolated, closed, and open systems, plus the distinction between extensive and intensive properties. The correct statements are I and III only, so the answer is option (B).

   Let’s unpack each statement one by one, because this is a classic concept-check in thermodynamics — and the trap is in statement II.

   1. Statement I: “The energy of an isolated system is constant.”

      An isolated system exchanges neither energy nor matter with its surroundings. By the first law of thermodynamics, if no heat or work crosses the boundary, the internal energy cannot change. So yes, the energy of an isolated system is constant. This is a direct consequence of the law of conservation of energy. Statement I is true.

   2. Statement II: “In a closed system, exchange of both energy and matter with surroundings is possible.”

      This is the common pitfall. A closed system allows exchange of energy (heat or work) with the surroundings, but not matter. The system that allows exchange of both energy and matter is called an open system. So statement II is false — it confuses closed with open. …

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4. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Match the following List - I (Type of drug) A Antacid B Antihistamine C Tranquilizer D Antibiotic List - II (Example) I Serotonin II Seldane III Ranitidine IV Chloramphenicol The correct answer is (A) A-III, B-I, C-IV, D-II (B) A-II, B-III, C-IV, D-I (C) A-III, B-II, C-I, D-IV (D) A-II, B-III, C-I, D-IV

   ›Reveal solutionSolution

   This question tests your knowledge of common drug categories and their specific examples. The correct matching is Antacid → Ranitidine, Antihistamine → Seldane, Tranquilizer → Serotonin, Antibiotic → Chloramphenicol, which corresponds to option (C).

   The key here is to understand what each type of drug does and which well-known example fits that function. Let’s break it down by the role of each drug class, not just by memorizing names.

   1. Antacid (A) – These neutralize stomach acid to relieve heartburn or indigestion. Ranitidine (III) is a classic example (it’s an H2 blocker that reduces acid production). So A matches III.

   2. Antihistamine (B) – These block histamine receptors to treat allergies (like hay fever). Seldane (II) is a well-known antihistamine (though now less common due to side effects). So B matches II. …

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5. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Observe the following two statements I. For an isolated system, $\Delta U = 0; \ q = 0$ II. For a closed system, $\Delta U \neq 0; \ q \neq 0$ The correct answer is (A) Both statements I and II are correct (B) Both statements I and II are not correct (C) Statement I is correct but statement II is not correct (D) Statement I is not correct but statement II is correct

   ›Reveal solutionSolution

   An isolated system exchanges neither matter nor energy, so $q = 0$ and $\Delta U = 0$; a closed system exchanges energy (though not matter), so $q$ and $\Delta U$ can be non-zero. Both statements are correct — option (A).

   The concept first

   Thermodynamics starts by drawing a boundary around the part of the universe we care about. Everything then follows from a single question: what is allowed to cross that boundary?

   System	Matter crosses?	Energy crosses?	Example
   Open	✓	✓	boiling water in an open beaker
   Closed	✗	✓	a sealed metal container being heated
   Isolated	✗	✗	(ideally) a thermos flask

   Now bring in the first law of thermodynamics, which is just the conservation of energy applied to that boundary:

   $\Delta U = q + w$

   where $q$ is the heat absorbed by the system and $w$ the work done on it. The law immediately tells you what each kind of system can do.

   Step-by-step

   1. Statement I — the isolated system. By definition its boundary permits no exchange of energy in any form.

   • No heat can cross ⇒ $q = 0$.
   • No energy can cross as work either; the boundary is rigid and impermeable ⇒ $w = 0$.
   • First law: $\Delta U = q + w = 0 + 0 = 0$. So $q = 0$ and $\Delta U = 0$. This is the sense in which the internal energy of an isolated system is constant — the whole universe, treated as one isolated system, has constant $U$. Statement I is correct.

   2. Statement II — the closed system. Its boundary blocks matter but permits energy. Heat may flow in or out, and work may be exchanged. Hence: …

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6. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.Observe the following properties I. Molar volume II. Mass III. Internal energy IV. Volume V. Enthalpy VI. Temperature VII. Density The intensive properties in the above list are (A) I, VI, VII only (B) I, IV, VI, VII only (C) I, III, IV, V only (D) II, III, V only

   ›Reveal solutionSolution

   Intensive properties are independent of the amount of substance; from the list, only temperature, density, and molar volume are intensive, so the correct set is I, VI, VII.

   Concept & Intuition

   The key distinction is between intensive and extensive properties. An intensive property does not change when you split or combine the system — it’s a local, sample-independent quantity (like temperature or density). An extensive property does scale with the size or amount of matter (like mass or volume).

   Molar volume is volume per mole, so it’s intensive even though volume itself is extensive. Internal energy and enthalpy are total energies, so they scale with amount — extensive. Temperature is famously intensive. Density is mass per volume, so it’s intensive.

   Thus we simply classify each item.

   Step-by-step reasoning

   1. Molar volume (I) – Volume per mole. If you take half the sample, the volume halves but the moles also halve, so the ratio stays the same. Intensive.

   2. Mass (II) – Doubling the sample doubles the mass. Extensive.

   3. Internal energy (III) – Total energy of all molecules; more matter means more energy. Extensive.

   4. Volume (IV) – Directly proportional to amount of substance. Extensive.

   5. Enthalpy (V) – Like internal energy, it’s a total quantity (H = U + PV). Extensive. …

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7. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
   Q.Out of the given options below, which one is an intensive property (A) $\dfrac{q_{\text{rev}}}{T}$ (B) $U + PV$ (C) $\dfrac{\Delta H}{\Delta T}$ (D) $\dfrac{RT}{P}$

   ›Reveal solutionSolution

   $\dfrac{RT}{P}$ is the molar volume $V_m$, which does not depend on the amount of substance — intensive. The other three are extensive.

   Check each:

   • (A) $\dfrac{q_{\text{rev}}}{T}=\Delta S$ (entropy) — extensive.
   • (B) $U+PV=H$ (enthalpy) — extensive.
   • (C) $\dfrac{\Delta H}{\Delta T}$ = heat capacity — extensive. …

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8. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Which of the following is/are "not correct" for CH$_3$OH + CH$_3$COOH mixture solution?

   a) $\Delta H_{mix} < 0$

   b) Does not obey Raoult's law

   c) $\Delta H_{mix} > 0$

   d) An example of ideal solution (A) d only (B) a, c only (C) a, b, c only (D) c, d only

   ›Reveal solutionSolution

   CH$_3$OH + CH$_3$COOH is a non-ideal solution with negative deviation ($\Delta H_{mix}<0$, disobeys Raoult's law), so the incorrect statements are (c) and (d).

   Concept — ideal vs non-ideal solutions. An ideal solution obeys Raoult's law at every composition, with $\Delta H_{mix}=0$ and $\Delta V_{mix}=0$; this requires solute–solvent interactions equal in strength to solute–solute and solvent–solvent interactions. When the new A–B interactions are stronger than the original A–A and B–B interactions, the solution shows negative deviation from Raoult's law and mixing is exothermic ($\Delta H_{mix}<0$).

   Step 1 — nature of the mixture. In a CH$_3$OH + CH$_3$COOH mixture, the alcoholic –OH and the carboxylic –COOH groups form strong intermolecular hydrogen bonds with each other, stronger than the interactions in the pure components. So the mixture is non-ideal with negative deviation.

   Step 2 — judge each statement. …

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9. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.The minimum temperature required for a non catalytic reaction between N$_2$ and O$_2$ is (A) 3000 K (B) 2000 K (C) 1000 K (D) 500 K

   ›Reveal solutionSolution

   The $\mathrm{N\equiv N}$ triple bond ($\approx 946\ \mathrm{kJ\,mol^{-1}}$) makes dinitrogen very inert, and $\mathrm{N_2} + \mathrm{O_2} \to 2\mathrm{NO}$ is endothermic. Without a catalyst the reaction needs a temperature of about 2000 K — option (B).

   The concept first

   Why is the air we breathe not a slowly exploding mixture? Nitrogen and oxygen sit side by side in every breath, and thermodynamically NO can form — yet nothing happens. The reason is the $\mathrm{N\equiv N}$ bond:

   $\Delta H_{\mathrm{N\equiv N}} \approx 946\ \mathrm{kJ\,mol^{-1}}$

   one of the strongest bonds in chemistry. Any reaction of $\mathrm{N_2}$ must first pay a huge activation cost to loosen that bond, so the rate at room temperature is effectively zero.

   There are only two ways round it:

   • Lower the barrier with a catalyst — this is what the Haber process does ($\mathrm{N_2 + 3H_2 \to 2NH_3}$ over iron at $\sim 773\ \mathrm{K}$), and what nitrogenase enzymes in root nodules do at ambient temperature.
   • Brute force with temperature — the route this question is about.

   Add to this that

   $\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\,\mathrm{NO(g)}, \qquad \Delta H \approx +180\ \mathrm{kJ\,mol^{-1}}$

   is endothermic, so by Le Chatelier's principle a high temperature shifts the equilibrium to the right as well as speeding it up. Both kinetics and equilibrium therefore demand extreme heat.

   Step-by-step

   1. Recognise the inertness of $\mathrm{N_2}$. High bond enthalpy $\Rightarrow$ high activation energy $\Rightarrow$ no reaction at ordinary temperatures.

   2. Recall the reaction and its thermochemistry. $\mathrm{N_2 + O_2 \rightleftharpoons 2NO}$, $\Delta H > 0$: heat is a "reactant" in the Le Chatelier sense. …

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10. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.The enthalpy of vaporisation of a compound is $840~\mathrm{J/mole}$ and its boiling point is $170~\mathrm{K}$. Its entropy of vaporisation is (A) $4.94~\mathrm{J/mol/K}$ (B) $12~\mathrm{J/mol/K}$ (C) $200~\mathrm{J/mol/K}$ (D) $49~\mathrm{J/mol/K}$

    ›Reveal solutionSolution

    The entropy of vaporisation is found using $\Delta S = \frac{\Delta H}{T}$ at the boiling point. Substituting the given values gives $\Delta S = \frac{840}{170} \approx 4.94~\mathrm{J/mol/K}$, so the correct option is (A).

    The key concept here is the relationship between enthalpy, entropy, and temperature at a phase transition. At the boiling point, the liquid and vapour are in equilibrium, so the Gibbs free energy change $\Delta G = 0$. This directly gives $\Delta S = \frac{\Delta H}{T}$, where $T$ is the boiling point in Kelvin. No extra steps or assumptions are needed — just a straightforward division.

    1. Recall the fundamental relation at equilibrium: For a phase change at constant temperature and pressure (like boiling), $\Delta G = \Delta H - T\Delta S = 0$. Rearranging:

    $\Delta S = \frac{\Delta H}{T}$

    This is the entropy change when one mole of substance vaporises at its boiling point.

    2. Identify the given values:

       • Enthalpy of vaporisation, $\Delta H = 840~\mathrm{J/mol}$
       • Boiling point, $T = 170~\mathrm{K}$

    3. Plug into the formula:

    $\Delta S = \frac{840}{170}$

    4. Perform the division:

    $\frac{840}{170} = \frac{84}{17} \approx 4.9412$

    So $\Delta S \approx 4.94~\mathrm{J/mol/K}$. …

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11. TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
    Q.The enthalpy of vaporisation of a compound is $840~\mathrm{J/mole}$ and its boiling point is $170~\mathrm{K}$. Its entropy of vaporisation is (A) $4.94~\mathrm{J/mol/K}$ (B) $12~\mathrm{J/mol/K}$ (C) $200~\mathrm{J/mol/K}$ (D) $49~\mathrm{J/mol/K}$

    ›Reveal solutionSolution

    The entropy of vaporisation is found using $\Delta S = \frac{\Delta H}{T}$ at the boiling point. Substituting the given values gives $\Delta S = \frac{840}{170} \approx 4.94~\mathrm{J/mol/K}$, so the correct option is (A).

    The key idea here is that at the boiling point, a liquid and its vapour are in equilibrium. For any phase change at constant temperature and pressure, the entropy change is simply the enthalpy change divided by the temperature. This comes directly from the definition of entropy in a reversible process: $dS = \frac{dq_\text{rev}}{T}$. At the boiling point, vaporisation occurs reversibly, so we can use this relation.

    Let’s work through it step by step:

    1. Recall the fundamental relation For a reversible phase change at constant temperature, the entropy change is

    $\Delta S = \frac{\Delta H}{T}$

    Here, $\Delta H$ is the enthalpy of vaporisation and $T$ is the boiling point (in Kelvin).

    2. Identify the given values

       • $\Delta H_\text{vap} = 840~\mathrm{J/mol}$
       • $T_\text{bp} = 170~\mathrm{K}$

    3. Plug into the formula

    $\Delta S_\text{vap} = \frac{840~\mathrm{J/mol}}{170~\mathrm{K}} = \frac{840}{170}~\mathrm{J/(mol \cdot K)}$

    4. Perform the division

    $\frac{840}{170} = \frac{84}{17} \approx 4.9412$

    So $\Delta S_\text{vap} \approx 4.94~\mathrm{J/(mol \cdot K)}$. …

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12. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.Which of the following compounds has the highest standard molar entropy? (A) $\mathrm{SO_2(g)}$ (B) $\mathrm{SO_3(g)}$ (C) $\mathrm{CO_2(g)}$ (D) $\mathrm{CO(g)}$

    ›Reveal solutionSolution

    Standard molar entropy increases with molecular complexity and the number of atoms, but also depends on molecular symmetry and the mass of atoms. Among the given gases, $\mathrm{SO_3(g)}$ has the most atoms and the heaviest atoms, giving it the highest entropy — answer is (B).

    Entropy is a measure of disorder or the number of microstates available to a system. For gases at the same temperature and pressure, standard molar entropy ($S^\circ$) is influenced by three main factors: the number of atoms in the molecule (more atoms = more vibrational and rotational modes), the mass of the atoms (heavier atoms give more closely spaced energy levels, increasing accessible states), and molecular symmetry (higher symmetry reduces entropy slightly). Let’s compare the four options.

    1. Count the atoms per molecule

       • $\mathrm{CO(g)}$: 2 atoms
       • $\mathrm{CO_2(g)}$: 3 atoms
       • $\mathrm{SO_2(g)}$: 3 atoms
       • $\mathrm{SO_3(g)}$: 4 atoms More atoms mean more ways to store energy (vibrations, rotations), so $\mathrm{SO_3}$ already has an edge.

    2. Compare atomic masses

       Sulfur (atomic mass ~32) is much heavier than carbon (~12). Oxygen (~16) is the same in all. Heavier atoms lead to lower vibrational frequencies, which means more vibrational states are populated at a given temperature, increasing entropy. So sulfur-containing compounds have an advantage over carbon-containing ones.

    3. Compare within the same number of atoms

       $\mathrm{SO_2}$ and $\mathrm{CO_2}$ both have 3 atoms, but $\mathrm{SO_2}$ has a heavier central atom (S vs. C) and is bent (lower symmetry) while $\mathrm{CO_2}$ is linear and symmetric. Lower symmetry means more rotational states, so $\mathrm{SO_2}$ has higher entropy than $\mathrm{CO_2}$.

    4. Now compare $\mathrm{SO_3}$ with $\mathrm{SO_2}$ …

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