Q.Predict in which of the following, entropy increases/decreases:

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Match the following List – I: A. $\Delta_rH°=+,\ \Delta_rS°=-,\ \Delta_rG°=+$ B. $\Delta_rH°=-,\ \Delta_rS°=-,\ \Delta_rG°=-$ C. $\Delta_rH°=+,\ \Delta_rS°=+,\ \Delta_rG°=+$ D. $\Delta_rH°=+,\ \Delta_rS°=+,\ \Delta_rG°=-$ List - II (Description about the reaction): I. Non spontaneous at low temperature II. Spontaneous at high temperature III. Non spontaneous at all temperatures IV. Spontaneous at low temperature (A) A-III, B-IV, C-II, D-I (B) A-III, B-IV, C-I, D-II (C) A-IV, B-III, C-II, D-I (D) A-II, B-IV, C-I, D-III

   ›Reveal solutionSolution

   Applying $\Delta G = \Delta H - T\Delta S$ to each sign-combination pins down its temperature-dependence: A→III, B→IV, C→I, D→II, matching option (B).

   Concept and Intuition

   Spontaneity is governed by the competition between $\Delta H$ (enthalpy term) and $-T\Delta S$ (entropy term scaled by temperature) inside $\Delta G = \Delta H - T\Delta S$. When both terms have the SAME sign effect on $\Delta G$ (both push it positive, or both push it negative), the spontaneity doesn't depend on temperature at all. When they OPPOSE each other, temperature decides which one wins — giving a spontaneity that flips at some crossover $T$.

   Step-by-Step Solution

   1. A: $\Delta H=+,\ \Delta S=-\ (\Rightarrow \Delta G=+)$. Here $-T\Delta S$ is positive too (since $\Delta S<0$), so BOTH terms make $\Delta G$ positive at every temperature — non-spontaneous no matter what $T$ is → matches III (non-spontaneous at all temperatures).
   2. B: $\Delta H=-,\ \Delta S=-\ (\Rightarrow \Delta G=-)$. $\Delta H$ is negative (favours spontaneity) but $-T\Delta S$ is positive (opposes it) and grows with $T$. At LOW $T$ the small $-T\Delta S$ term can't overcome the negative $\Delta H$, so $\Delta G$ stays negative — spontaneous at low $T$ → matches IV.
   3. C: $\Delta H=+,\ \Delta S=+\ (\Rightarrow \Delta G=+$ as given$)$. Both a positive $\Delta H$ (opposes spontaneity) and a negative $-T\Delta S$ (favours it, since $\Delta S>0$) are in play. At LOW $T$, $-T\Delta S$ is small, so $\Delta H$ dominates and $\Delta G$ stays positive — this is the non-spontaneous-at-low-$T$ regime → matches I.
   4. D: same combination $\Delta H=+,\ \Delta S=+$ but $\Delta G=-$ as given. This is the HIGH-$T$ regime of the same reaction, where $-T\Delta S$ has grown large enough to overcome $\Delta H$, making $\Delta G$ negative — spontaneous at high $T$ → matches II.
   5. Final match: A–III, B–IV, C–I, D–II.

   Common Mistakes

   • Treating C and D as unrelated reactions rather than recognising they're the SAME $\Delta H,\Delta S$ sign-combination at two different temperature regimes.
   • Forgetting the negative sign in front of $T\Delta S$ when judging how the entropy term affects $\Delta G$.

   ✓Final answer

   The correct option is (B) — A-III, B-IV, C-I, D-II.

   ANSWER: B

2. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.At T(K), a gas is adsorbed on the surface of a solid. The signs of $\Delta H$ and $\Delta S$ are respectively (A) negative, negative (B) positive, positive (C) positive, negative (D) negative, positive

   ›Reveal solutionSolution

   Adsorption is always exothermic ($\Delta H<0$) and always decreases entropy ($\Delta S<0$), since it converts free gas molecules into an ordered surface layer.

   Concept and Intuition

   When gas molecules adsorb onto a solid surface, they lose translational degrees of freedom and become confined to (approximately) two dimensions on the surface — this is always a decrease in disorder, so $\Delta S_{ads}$ is negative. For adsorption to be spontaneous ($\Delta G = \Delta H - T\Delta S < 0$) despite this entropy decrease, $\Delta H$ must be sufficiently negative (i.e. the process must release heat) to outweigh the unfavourable $-T\Delta S$ term. This is why adsorption is universally observed to be exothermic.

   Step-by-Step Solution

   1. Consider the process: gas (free, high entropy) $\rightarrow$ gas adsorbed on solid surface (confined, low entropy).
   2. The system becomes more ordered, so $\Delta S < 0$ (negative) always for adsorption.
   3. For the overall process to be thermodynamically favourable (as adsorption spontaneously is, under suitable conditions), $\Delta G=\Delta H-T\Delta S$ must be negative; with $\Delta S$ negative, $-T\Delta S$ is positive, so $\Delta H$ must be negative and large enough in magnitude to make $\Delta G$ negative overall.
   4. Hence adsorption is always exothermic: $\Delta H<0$.
   5. Both signs are negative: $\Delta H$ negative, $\Delta S$ negative.

   Common Mistakes

   • Confusing adsorption (gas sticking to a surface, entropy decreases) with desorption or with dissolution processes that can have different entropy signs.
   • Assuming spontaneity requires $\Delta S>0$ — forgetting that a sufficiently negative $\Delta H$ can make $\Delta G$ negative even when $\Delta S$ is negative.

   ✓Final answer

   The correct option is (A) — negative, negative.

   ANSWER: A

3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.For which of the following processes entropy change ($\Delta S$) is negative ? (only) I) Sublimation of dry ice II) Freezing of water III) Crystallisation of the dissolved substance IV) Burning of rocket fuel (A) I & II only (B) II & III only (C) III & IV only (D) I & IV only

   ›Reveal solutionSolution

   Tests recognizing entropy decrease vs increase from physical/chemical changes in state of order; freezing and crystallisation both decrease entropy, giving option (B).

   Concept and Intuition

   Entropy is a measure of disorder/randomness. Processes that increase randomness (solid → liquid → gas, dissolution, more gas moles produced) have positive $\Delta S$. Processes that increase order (gas/liquid → solid, particles coming together into an ordered lattice, decreasing number of gas moles) have negative $\Delta S$.

   Step-by-Step Solution

   1. I) Sublimation of dry ice: solid $CO_2$ directly becomes gaseous $CO_2$ — a large increase in disorder (more freedom of motion) → $\Delta S > 0$. Not negative.
   2. II) Freezing of water: liquid water (more disordered) becomes ice (ordered crystal lattice) → $\Delta S < 0$. Negative — qualifies.
   3. III) Crystallisation of a dissolved substance: solute molecules/ions moving freely in solution come together into an ordered crystalline solid → $\Delta S < 0$. Negative — qualifies.
   4. IV) Burning of rocket fuel: typically converts solid/liquid fuel and oxidizer into large volumes of hot gaseous products (e.g., $CO_2$, $H_2O$ vapor, etc.) — net increase in moles of gas and disorder → $\Delta S > 0$. Not negative.
   5. So only II and III show a negative entropy change.

   Common Mistakes

   • Assuming all combustion processes decrease entropy because they "release energy" — entropy change is about disorder, not energy release; exothermic reactions can still have positive $\Delta S_{system}$.
   • Confusing sublimation (ordered solid → disordered gas, entropy increases) with freezing (entropy decreases) due to not tracking the direction of the phase change carefully.

   ✓Final answer

   The correct option is (B) — II & III only.

   ANSWER: B

4. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.Given below are two statements Statement I: The entropy of pure crystalline solid approaches zero as the temperature approaches absolute zero value Statement II: For a reaction at equilibrium, $\Delta_rG$ is zero 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

   Statement I is the Third Law of Thermodynamics (true), and Statement II is the condition for chemical equilibrium (true). Therefore both statements are correct, and the answer is (A).

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   Concept and Intuition: Why This Approach Works

   This question tests two foundational ideas in thermodynamics that often appear together in exams.

   • Statement I is a direct recall of the Third Law of Thermodynamics — a cornerstone principle about the behavior of entropy at absolute zero.
   • Statement II is about the Gibbs free energy criterion for equilibrium — a central result from chemical thermodynamics.

   The trick is not to overthink: each statement stands on its own, and both are textbook-true. The only pitfall is confusing “entropy at absolute zero” with “entropy change” or misremembering the equilibrium condition.

   ---

   Step-by-Step Reasoning

   1. Evaluate Statement I: Entropy at absolute zero

      The Third Law of Thermodynamics states: The entropy of a perfect crystalline substance is zero at absolute zero (0 K).

      More precisely, as $T \to 0$, the entropy $S \to 0$ for a pure, perfectly ordered crystal.

      This is because at 0 K, the system is in its single, lowest-energy microstate, giving $S = k \ln W = k \ln 1 = 0$.

      ✓ Statement I is correct.

   2. Evaluate Statement II: Gibbs free energy at equilibrium

      For a reaction at constant temperature and pressure, the Gibbs free energy change is:

   $\Delta_r G = \Delta_r H - T \Delta_r S$

   At equilibrium, the forward and reverse rates are equal, and the system has no net tendency to change. The thermodynamic condition for equilibrium is:

   $\Delta_r G = 0$

   (If $\Delta_r G < 0$, the reaction proceeds forward spontaneously; if $\Delta_r G > 0$, it proceeds backward.)

   ✓ Statement II is correct.

   3. Combine the evaluations Both statements are independently true. Therefore the correct choice is the one that says both are correct.

   Watch out

   A common mistake is to think Statement II is false because “at equilibrium, the reaction quotient $Q = K$ and $\Delta_r G^\circ = -RT \ln K$ is not zero.” But note: $\Delta_r G^\circ$ (standard state) is not the same as $\Delta_r G$ (actual conditions). At equilibrium, $\Delta_r G = 0$ always — that is the definition.

   Tip

   Remember the mnemonic:

   • Third Law: Perfect crystal at 0 K → entropy zero.
   • Equilibrium: $\Delta G = 0$ (like a ball at the bottom of a valley — no net force).

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.Observe the following processes I. $\mathrm{O_2(l) \rightarrow O_2(g)}$ II. $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ III. $\mathrm{C(s, graphite) \rightarrow C(s, diamond)}$ IV. $\mathrm{N_2(g, 1\ atm) \rightarrow N_2(g, 10\ atm)}$ V. $\mathrm{H_2(g) \rightarrow 2H(g)}$ VI. Temperature of a crystalline solid is raised from 0 K to 115 K for how many of the above processes, change in entropy is negative (A) 2 (B) 3 (C) 4 (D) 1

   ›Reveal solutionSolution

   Entropy decreases when disorder/moles-of-gas/volume decreases; here processes II, III and IV are the ones with negative $\Delta S$, so the count is 3.

   Concept and Intuition

   Entropy tracks the number of accessible microstates — more gas moles, more volume, more freedom of motion, and higher temperature all raise it, while compression, bond formation that reduces particle count, and forming a more rigid/ordered solid lower it.

   Step-by-Step Solution

   1. I. $\mathrm{O_2(l)\to O_2(g)}$: vaporization always increases entropy (gas has far more freedom than liquid) → positive.
   2. II. $\mathrm{N_2+3H_2\to2NH_3}$: 4 moles of gas become 2 moles of gas, fewer independent particles/less disorder → negative.
   3. III. $\mathrm{C(graphite)\to C(diamond)}$: graphite's loosely stacked sp² layers have higher entropy than diamond's rigid covalent sp³ network (standard molar entropies confirm $S^\circ_{graphite}>S^\circ_{diamond}$) → negative.
   4. IV. $\mathrm{N_2(g,1\ atm)\to N_2(g,10\ atm)}$: compressing a gas at constant T reduces the volume available, reducing entropy → negative.
   5. V. $\mathrm{H_2(g)\to2H(g)}$: one mole of gas becomes two, more disorder/more translational states → positive.
   6. VI. Heating a crystalline solid from 0 K raises molecular motion/vibrational disorder → positive (this is exactly how the third law is used to compute standard entropies).
   7. Negative ones: II, III, IV → total 3.

   Common Mistakes

   • Assuming graphite → diamond increases entropy just because diamond seems "more special"; entropy tracks disorder, not hardness or preciousness.
   • Forgetting that compressing a gas at constant temperature also lowers entropy, not just cooling it.

   ✓Final answer

   The correct option is (B) — 3.

   ANSWER: B

6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.At 300 K, for the reaction A $\rightarrow$ P, the $\Delta S_{sys}$ is 5 JK$^{-1}$ mol$^{-1}$. What is the heat absorbed (in kJ mol$^{-1}$) by the system? (A) 1.5 (B) 15 (C) 1500 (D) 0.6

   ›Reveal solutionSolution

   This tests the thermodynamic definition of entropy change for a reversible process, $\Delta S = q_{rev}/T$; the heat absorbed works out to 1.5 kJ/mol.

   Concept and Intuition

   Entropy is defined (for a reversible path) as the heat exchanged divided by the absolute temperature at which the exchange occurs. So if we are told $\Delta S_{sys}$ at a given T, we can directly back out the reversible heat absorbed by the system — no need for any other thermodynamic data.

   Step-by-Step Solution

   1. Reversible-process relation: $\Delta S = \dfrac{q_{rev}}{T}$.
   2. Rearranging: $q_{rev} = \Delta S \times T = 5\ \text{J K}^{-1}\text{mol}^{-1} \times 300\ \text{K} = 1500\ \text{J mol}^{-1}$.
   3. Converting to kJ: $1500\ \text{J mol}^{-1} = 1.5\ \text{kJ mol}^{-1}$.

   Common Mistakes

   • Forgetting to convert the final answer from J to kJ.
   • Confusing $\Delta S$ with $\Delta G$ or trying to invoke $\Delta G = \Delta H - T\Delta S$ when no $\Delta G$ or $\Delta H$ data is given — this problem only needs the direct reversible-heat relation.

   ✓Final answer

   The correct option is (A) — 1.5.

   ANSWER: A

7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Identify the incorrect statements from the following I. $\Delta S_{system} = (\Delta S_{total} + \Delta S_{surr})$ II. $A(l) \rightarrow A(s)$; for this process entropy change decreases III. Entropy units are J K mol$^{-1}$ (A) I, III only (B) I, II only (C) I, II, III (D) II, III only

   ›Reveal solutionSolution

   This checks three commonly-confused entropy facts; only I and III are actually wrong, so (A) I, III only is correct.

   Concept and Intuition

   Total entropy change of the universe is the sum of the system's and surroundings' entropy changes — this is the Second Law's bookkeeping identity, and getting the terms on the wrong side of the equals sign is a very common slip. Physically, entropy tracks disorder/microstates, so freezing a liquid (more ordered arrangement) must decrease entropy. And entropy's units must carry an inverse power of temperature, since $\Delta S = q/T$.

   Step-by-Step Solution

   1. Statement I: The Second Law bookkeeping identity is $\Delta S_{total} = \Delta S_{system} + \Delta S_{surr}$. As written, "$\Delta S_{system} = \Delta S_{total} + \Delta S_{surr}$" has the terms transposed — incorrect.
   2. Statement II: $A(l) \to A(s)$ is a liquid solidifying — molecules become more ordered, so entropy decreases. This statement is correct (true), so it is not one of the "incorrect" statements being asked for.
   3. Statement III: Entropy has units of energy/temperature, i.e. $\text{J K}^{-1}\text{mol}^{-1}$. Writing it as "J K mol$^{-1}$" (missing the inverse power on K) is dimensionally wrong — incorrect.
   4. So the incorrect statements are I and III only.

   Common Mistakes

   • Reading statement II as "wrong" just because it says entropy "decreases" — freezing genuinely does decrease entropy, so II is a true (correct) statement, not one to select.
   • Overlooking the missing negative exponent in statement III's units.

   ✓Final answer

   The correct option is (A) — I, III only.

   ANSWER: A

8. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The molar heats of fusion and vapourisation of benzene are 10.9 and 31.0 kJ mol$^{-1}$ respectively. The changes in entropy for the solid $\rightarrow$ liquid and liquid $\rightarrow$ vapour transitions for benzene are x and y, J K$^{-1}$ mol$^{-1}$, respectively. The value of (y-x) (in J K$^{-1}$ mol$^{-1}$) is (At 1 atm, benzene melts at 5.5 °C and boils at 80 °C) (A) 87.8 (B) 48.7 (C) 39.1 (D) 28.7

   ›Reveal solutionSolution

   Entropy of a phase transition at its transition temperature is $\Delta H/T$; computing both fusion and vaporisation entropies for benzene and subtracting gives 48.7 J/K/mol.

   Concept and Intuition

   At a reversible phase transition occurring at constant temperature and pressure, $\Delta S = \Delta H/T$ (with $T$ in kelvin, the transition temperature). This applies separately to melting (fusion) and boiling (vaporisation), each at its own characteristic temperature.

   Step-by-Step Solution

   1. Fusion: $T=5.5°C=278.65$ K, $\Delta H_{fus}=10.9$ kJ/mol $=10900$ J/mol. $x = 10900/278.65 \approx 39.1$ J K$^{-1}$ mol$^{-1}$.
   2. Vaporisation: $T=80°C=353.15$ K, $\Delta H_{vap}=31.0$ kJ/mol $=31000$ J/mol. $y = 31000/353.15 \approx 87.8$ J K$^{-1}$ mol$^{-1}$.
   3. $y - x \approx 87.8-39.1 = 48.7$ J K$^{-1}$ mol$^{-1}$.

   Common Mistakes

   • Forgetting to convert Celsius to Kelvin before dividing.
   • Using kJ instead of J for $\Delta H$, giving an answer off by a factor of 1000.

   ✓Final answer

   The correct option is (B) — 48.7.

   ANSWER: B

9. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.Identify the correct statements from the following I. $\Delta_r G$ is zero for A $\rightleftharpoons$ B reaction II. The entropy of pure crystalline solids approaches zero as the temperature approaches absolute zero III. $\Delta U$ of a reaction can be determined using bomb calorimeter (A) I, II only (B) I, III only (C) II, III only (D) I, II, III

   ›Reveal solutionSolution

   This tests three independent thermodynamics facts — the condition at equilibrium, the Third Law, and what a bomb calorimeter measures — all three are correct.

   Concept and Intuition

   Thermodynamics distinguishes between the standard free energy change $\Delta_rG^\circ$ (a fixed reference quantity) and the free energy change $\Delta_rG$ under the actual reaction conditions, which becomes zero exactly at equilibrium. The Third Law fixes the reference point for absolute entropy. Calorimetry technique determines whether the measured heat corresponds to $\Delta U$ (constant volume, bomb calorimeter) or $\Delta H$ (constant pressure, coffee-cup calorimeter).

   Step-by-Step Solution

   1. Statement I: For any reaction at equilibrium, $\Delta_rG = \Delta_rG^\circ + RT\ln Q$, and at equilibrium $Q = K$, so $\Delta_rG^\circ = -RT\ln K$ and thus $\Delta_rG = 0$. Since A ⇌ B denotes an equilibrium, $\Delta_rG = 0$ holds. True.
   2. Statement II: The Third Law of Thermodynamics states that the entropy of a perfectly ordered crystalline substance is zero at absolute zero, and approaches zero as $T \to 0$ K. True.
   3. Statement III: In a bomb calorimeter, the reaction occurs in a sealed, rigid container (constant volume), so no $PV$ work is done and all the heat released/absorbed equals the change in internal energy: $q_v = \Delta U$. True.
   4. All three statements I, II, III are correct.

   Common Mistakes

   • Confusing $\Delta_rG$ (zero at equilibrium) with $\Delta_rG^\circ$ (a fixed, generally non-zero standard value).
   • Mixing up bomb calorimeter (constant volume, gives $\Delta U$) with constant-pressure calorimetry (gives $\Delta H$).

   ✓Final answer

   The correct option is (D) — I, II, III.

   ANSWER: D

10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Observe the following reactions I. $\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$ II. $\mathrm{Cl_2(g)} \rightarrow 2\mathrm{Cl(g)}$ III. $\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(s)}$ Identify the reactions in which entropy increases (A) I, II, III (B) I, II Only (C) I, III Only (D) II, III Only

    ›Reveal solutionSolution

    Entropy tracks disorder; it rises when a gas is produced or a bond is broken into more particles, and falls on freezing. Only reactions I and II show an entropy increase.

    Concept and Intuition

    Entropy is a measure of the number of ways energy/matter can be arranged (disorder). Processes that increase the number of gas molecules, break a substance into more independent particles, or convert a more-ordered phase (solid) into a less-ordered one (liquid/gas) increase entropy. The reverse — condensation, freezing, or reducing the number of gas particles — decreases entropy.

    Step-by-Step Solution

    1. I. $\mathrm{CaCO_3(s)}\to\mathrm{CaO(s)}+\mathrm{CO_2(g)}$: a solid decomposes to give a solid plus a gas. Gas has far higher entropy than solid, so overall entropy increases.
    2. II. $\mathrm{Cl_2(g)}\to 2\mathrm{Cl(g)}$: one mole of gas becomes two moles of gas (bond breaking increases the number of independent gaseous particles) — entropy increases.
    3. III. $\mathrm{H_2O(l)}\to\mathrm{H_2O(s)}$: this is freezing — a liquid (more disordered) becomes a solid (more ordered) — entropy DECREASES, it does not increase.
    4. So the entropy-increasing reactions are I and II only.

    Common Mistakes

    • Assuming any phase change or any reaction with a physical-state label automatically increases entropy — freezing is a clear counter-example.
    • Forgetting that dissociation of a diatomic molecule into atoms increases particle count and hence entropy.

    ✓Final answer

    The correct option is (B) — I, II Only.

    ANSWER: B

11. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.Which of the following reactions are endothermic and spontaneous?

    (i) $\dfrac{1}{2}\mathrm{H_2(g)} + \dfrac{1}{2}\mathrm{Cl_2(g)} \rightarrow \mathrm{HCl(g)}$

    (ii) $\mathrm{C(Graphite)} + 2\mathrm{S(l)} \rightarrow \mathrm{CS_2(l)}$

    (iii) $\dfrac{1}{2}\mathrm{N_2(g)} + \dfrac{3}{2}\mathrm{H_2(g)} \rightarrow \mathrm{NH_3(g)}$

    (iv) $\mathrm{H_2(g)} + \dfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}$

    (v) $\dfrac{1}{2}\mathrm{N_2(g)} + \mathrm{O_2(g)} \rightarrow \mathrm{NO_2(g)}$ (A) (ii),

    (v) only (B) (iii),

    (v) only (C) (i),

    (ii) only (D) (iii),

    (iv) only

    ›Reveal solutionSolution

    Among the five formation reactions, only CS₂ and NO₂ have positive (endothermic) enthalpies of formation — HCl, NH₃, and H₂O are all classic exothermic formations — so (ii) and (v) are the pair being asked for.

    Concept and Intuition

    "Endothermic and spontaneous" reactions are a useful teaching example because they show that spontaneity is governed by $\Delta G=\Delta H-T\Delta S$, not by $\Delta H$ alone — a positive $\Delta H$ (heat absorbed) can still be overwhelmed by a sufficiently large, positive $T\Delta S$ term to give an overall negative $\Delta G$. The first filter, though, is simply checking which of the five reactions are even endothermic in the first place (positive $\Delta H_f^\theta$), since a reaction that's exothermic to begin with cannot be the "endothermic" answer regardless of its spontaneity.

    Step-by-Step Solution

    1. (i) $\tfrac12\mathrm{H_2}+\tfrac12\mathrm{Cl_2}\to\mathrm{HCl}$: strongly exothermic ($\Delta H_f^\theta\approx-92\ \mathrm{kJ/mol}$) — not endothermic, eliminate.
    2. (ii) $\mathrm C(\text{graphite})+2\mathrm S(l)\to\mathrm{CS_2}(l)$: endothermic ($\Delta H_f^\theta$ is positive for CS₂ formation from the elements) — keep as a candidate.
    3. (iii) $\tfrac12\mathrm{N_2}+\tfrac32\mathrm{H_2}\to\mathrm{NH_3}$: exothermic (Haber process, $\Delta H_f^\theta\approx-46\ \mathrm{kJ/mol}$) — not endothermic, eliminate.
    4. (iv) $\mathrm{H_2}+\tfrac12\mathrm{O_2}\to\mathrm{H_2O}(l)$: strongly exothermic ($\Delta H_f^\theta\approx-286\ \mathrm{kJ/mol}$) — not endothermic, eliminate.
    5. (v) $\tfrac12\mathrm{N_2}+\mathrm{O_2}\to\mathrm{NO_2}(g)$: endothermic ($\Delta H_f^\theta$ is positive, since nitrogen oxides require energy input to form from stable N₂ and O₂) — keep as a candidate.
    6. Only (ii) and (v) qualify as endothermic reactions among the five, matching option (A).

    Common Mistakes

    • Assuming all combination/formation reactions from elements are exothermic by default — several oxides of nitrogen are a well-known exception.
    • Trying to judge spontaneity from $\Delta H$ sign alone, rather than first simply screening for which reactions are endothermic at all.

    ✓Final answer

    The correct option is (A) — (ii), (v) only.

    ANSWER: A

12. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
    Q.In adsorption, the thermodynamic requirement is (A) $\Delta H = +ve$ ; $\Delta S = +ve$ (B) $\Delta H = -ve$ ; $\Delta S = -ve$ (C) $\Delta H = -ve$ ; $\Delta S = +ve$ (D) $\Delta H = +ve$ ; $\Delta S = -ve$

    ›Reveal solutionSolution

    This tests the standard thermodynamic signs for physical/chemical adsorption: both enthalpy and entropy change are negative, and spontaneity is maintained by the dominant exothermicity.

    Concept and Intuition

    When a gas is adsorbed onto a solid surface, the gas molecules go from a highly disordered, freely-moving state to a more ordered, surface-bound state — this is always accompanied by a decrease in entropy ($\Delta S<0$). For the process to remain spontaneous ($\Delta G = \Delta H - T\Delta S < 0$) despite the unfavourable (negative) entropy term, $\Delta H$ must be negative and large enough in magnitude — i.e., adsorption is exothermic.

    Step-by-Step Solution

    1. Write the Gibbs-Helmholtz relation: $\Delta G = \Delta H - T\Delta S$.
    2. Adsorption decreases the randomness of gas molecules (they become localized on the surface), so $\Delta S = -ve$.
    3. For $\Delta G$ to stay negative (spontaneous) with $-T\Delta S$ being positive (since $\Delta S<0$), $\Delta H$ must be sufficiently negative — adsorption is exothermic, $\Delta H = -ve$.
    4. This matches the well-known thermodynamic requirement taught for adsorption.

    Common Mistakes

    • Assuming adsorption increases entropy because surface interactions seem to "add order" incorrectly interpreted as increased randomness.
    • Picking $\Delta H=+ve$ by confusing adsorption (exothermic) with desorption.

    ✓Final answer

    The correct option is (B) — $\Delta H = -ve$; $\Delta S = -ve$.

    ANSWER: B