Q.For oxidation of iron, $4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)$, entropy change is –549.4 JK$^{-1}$ mol$^{-1}$ at 298 K. In spite of the negative entropy change of this reaction, why is the reaction spontaneous? ($\Delta_r H^\ominus$ for this reaction is $-1648 \times 10^3$ J mol$^{-1}$)

Showing the 12 most recent of 27 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.What is $\Delta_r G^{\ominus}$ for the following reaction at 298 K? $2\,X(g) + Y(g) \rightarrow 2\,Z(g)$ ($\Delta U^{\ominus} = -10.5 \; \mathrm{kJ}$; $\Delta S^{\ominus} = -44 \; \mathrm{J\,K^{-1}}$; $R = 8.3 \; \mathrm{J\,K^{-1}mol^{-1}}$) (A) 12973 J (B) 13112 J (C) 139.0 J (D) 0.1390 J

   ›Reveal solutionSolution

   Chain two thermodynamic relations — $\Delta H = \Delta U + \Delta n_g RT$ then $\Delta G = \Delta H - T\Delta S$ — to get $\Delta_r G^{\ominus} \approx 139.0\ \mathrm{J}$.

   Concept and Intuition

   $\Delta U$ measures energy change at constant volume, but a real gas-phase reaction generally happens at constant pressure, where the system also does (or has done to it) $P\Delta V$ work as the mole number of gas changes. That correction, $\Delta n_g RT$, converts $\Delta U \to \Delta H$. Once you have $\Delta H$, Gibbs' definition $\Delta G = \Delta H - T\Delta S$ folds in the entropy cost/benefit of the reaction to tell you the true thermodynamic driving force.

   Step-by-Step Solution

   1. Reaction: $2X(g) + Y(g) \rightarrow 2Z(g)$. Moles of gas: reactants $=2+1=3$, products $=2$, so $\Delta n_g = 2-3 = -1$.
   2. $\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT = -10500\ \mathrm{J} + (-1)(8.3\ \mathrm{J\,K^{-1}mol^{-1}})(298\ \mathrm{K})$.
   3. $(8.3)(298) = 2473.4\ \mathrm{J}$, so $\Delta H^{\ominus} = -10500 - 2473.4 = -12973.4\ \mathrm{J}$.
   4. $T\Delta S^{\ominus} = 298 \times (-44) = -13112\ \mathrm{J}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.At $17°C$, the enthalpy change and entropy change of the reaction given below are respectively $-12.55\ kJ\,mol^{-1}$ and $5.0\ JK^{-1}$ $Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)$ What is $\Delta G$ (in kJ $mol^{-1}$) for this reaction? (A) -14 (B) +14 (C) -28 (D) +56

   ›Reveal solutionSolution

   A direct Gibbs–Helmholtz calculation; $\Delta G=-14$ kJ/mol, confirming the reaction is spontaneous.

   Concept and Intuition

   Gibbs free energy change combines the enthalpy driving force with the entropy (disorder) contribution, weighted by temperature: $\Delta G=\Delta H-T\Delta S$. A negative $\Delta G$ signals spontaneity. Unit consistency (J vs kJ) is the main thing to watch — $\Delta S$ is usually given in J/K while $\Delta H$ is in kJ.

   Step-by-Step Solution

   1. Convert temperature: $17°C = 17+273 = 290$ K.
   2. Convert $\Delta S$ to kJ: $5.0\ J\,K^{-1} = 5.0\times10^{-3}\ kJ\,K^{-1}$.
   3. Apply $\Delta G=\Delta H-T\Delta S = -12.55 - (290)(5.0\times10^{-3})$. …

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3. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A certain reaction is at equilibrium at 82°C. If the enthalpy change for the reaction is 21.3 kJ, the value of $\Delta S$ (in $J\ K^{-1}$) for the reaction is (A) 120 (B) 80 (C) 60 (D) 75

   ›Reveal solutionSolution

   At equilibrium $\Delta G=0$, so $\Delta S=\Delta H/T = 60$ J K$^{-1}$.

   Concept and Intuition

   A system "at equilibrium" (for a reaction, not phase transition specifically) has zero net driving force, i.e. $\Delta G = 0$. Since $\Delta G=\Delta H - T\Delta S$, setting it to zero directly gives $\Delta S$ in terms of $\Delta H$ and the absolute temperature.

   Step-by-Step Solution

   1. Condition of equilibrium: $\Delta G = \Delta H - T\Delta S = 0$.
   2. Convert temperature to Kelvin: $T = 82 + 273 = 355$ K. …

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4. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Consider the following reaction $A(g) + B(g) \rightleftharpoons C(g) + D(g)$ Which one of the following is zero for this reaction at 298 K? (A) $\Delta_r H^{\ominus}$ (B) $\Delta_r S^{\ominus}$ (C) $\Delta_r G$ (D) $\Delta_r G^{\ominus}$

   ›Reveal solutionSolution

   At equilibrium, the true (non-standard) Gibbs free energy change of a reaction is always zero — this is the defining condition of chemical equilibrium. The answer is $\Delta_rG$.

   Concept and Intuition

   Standard thermodynamic quantities like $\Delta_rH^{\ominus}$, $\Delta_rS^{\ominus}$ and $\Delta_rG^{\ominus}$ are fixed properties of a reaction at a given temperature (298 K here) — they don't depend on how far the reaction has actually proceeded. But the actual Gibbs energy change, $\Delta_rG$, depends on the current composition (reaction quotient $Q$) via $\Delta_rG = \Delta_rG^{\ominus} + RT\ln Q$. Equilibrium is precisely the state where the reaction has no further thermodynamic tendency to proceed in either direction, i.e. $\Delta_rG = 0$.

   Step-by-Step Solution

   1. Recall the relation $\Delta_rG = \Delta_rG^{\ominus} + RT\ln Q$.
   2. At equilibrium, by definition $Q = K$, so $\Delta_rG^{\ominus} = -RT\ln K$.
   3. Substitute back: $\Delta_rG = -RT\ln K + RT\ln K = 0$.
   4. This holds regardless of the actual values of $\Delta_rH^{\ominus}$ or $\Delta_rS^{\ominus}$, which are generally non-zero for a reaction like this one (unless coincidentally zero, which isn't implied here). …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Consider the following reaction $\mathrm{A(g) + 3B(g) \longrightarrow 2C(g)}$; $\Delta H^{\ominus} = -24$ kJ. At $25°C$ if $\Delta G^{\ominus}$ of the reaction is $-9$ kJ, the standard entropy change (in $\mathrm{JK^{-1}}$) of the same reaction at same temperature is (A) $-5.33$ (B) $-50.33$ (C) $-500.33$ (D) $-0.533$

   ›Reveal solutionSolution

   This tests the Gibbs-Helmholtz relation $\Delta G^\ominus=\Delta H^\ominus-T\Delta S^\ominus$ to extract entropy change, giving $\Delta S^\ominus\approx -50.33$ J K$^{-1}$.

   Concept and Intuition

   The spontaneity of a reaction is governed by the interplay of enthalpy and entropy through $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$; rearranging isolates $\Delta S^\ominus$ once $\Delta H^\ominus$, $\Delta G^\ominus$ and $T$ are known, provided units are made consistent (kJ vs J).

   Step-by-Step Solution

   1. Convert to consistent units: $\Delta H^\ominus = -24\text{ kJ} = -24000$ J; $\Delta G^\ominus = -9\text{ kJ} = -9000$ J; $T = 25°C = 298$ K.
   2. Rearrange: $\Delta S^\ominus = \dfrac{\Delta H^\ominus - \Delta G^\ominus}{T}$. …

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6. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.At 298 K, $\Delta_r U^{\ominus}$ and $\Delta_r S^{\ominus}$ for the following reaction are $-10.5$ kJ and $+44.1$ J K$^{-1}$; $\mathrm{2X(g) + Y(g) \longrightarrow 2Z(g)}$. What is $\Delta_r G^{\ominus}$ (in kJ) for this reaction ? ($R = 8.314$ J K$^{-1}$ mol$^{-1}$) (A) $+0.164$ (B) $-26.119$ (C) $-2.6119$ (D) $-0.082$

   ›Reveal solutionSolution

   This tests chaining $\Delta U \to \Delta H \to \Delta G$ using $\Delta n_g RT$ and the Gibbs equation; the answer is $-26.119$ kJ, option (B).

   Concept and Intuition

   Gibbs free energy change is $\Delta G = \Delta H - T\Delta S$, but we're given $\Delta U$, not $\Delta H$ — so we first need $\Delta H = \Delta U + \Delta n_g RT$, using the change in gas moles for the reaction as written.

   Step-by-Step Solution

   1. Reaction: $2X(g) + Y(g) \rightleftharpoons 2Z(g)$. Gas moles: reactants $=2+1=3$, products $=2$. So $\Delta n_g = 2-3 = -1$.
   2. Compute $\Delta n_g RT = (-1)(8.314)(298) = -2477.6$ J $= -2.4776$ kJ.
   3. $\Delta H = \Delta U + \Delta n_g RT = -10.5 + (-2.4776) = -12.9776$ kJ.
   4. Compute $T\Delta S = 298 \times 44.1 = 13141.8$ J $= 13.1418$ kJ. …

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7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $\Delta_r H^{\ominus}$ and $\Delta_r S^{\ominus}$ are standard enthalpy change and standard entropy change respectively for a reaction, the incorrect option is (A) $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = positive; spontaneous at all temperatures (B) $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = negative; non-spontaneous at low temperatures (C) $\Delta_r H^{\ominus}$ = positive; $\Delta_r S^{\ominus}$ = positive; non-spontaneous at low temperatures (D) $\Delta_r H^{\ominus}$ = negative; $\Delta_r S^{\ominus}$ = negative; spontaneous at low temperatures

   ›Reveal solutionSolution

   This tests how the signs of $\Delta H$ and $\Delta S$ combine, via $\Delta G=\Delta H-T\Delta S$, to fix the temperature range of spontaneity. The incorrect statement is (B).

   Concept and Intuition

   Spontaneity is governed by the sign of $\Delta_r G^{\ominus} = \Delta_r H^{\ominus} - T\Delta_r S^{\ominus}$, not by $\Delta H$ or $\Delta S$ alone. Since $T>0$ always, the term $-T\Delta S$ can flip sign relative to $\Delta H$ as $T$ changes, and that tug-of-war is what makes some reactions spontaneous only in a certain temperature window.

   Step-by-Step Solution

   1. Case $\Delta H<0,\ \Delta S>0$: both terms in $\Delta G=\Delta H-T\Delta S$ are negative for all $T>0$ (since $-T\Delta S<0$ too). So $\Delta G<0$ always — spontaneous at every temperature. (A) is a correct statement.
   2. Case $\Delta H<0,\ \Delta S<0$: $\Delta G=\Delta H-T\Delta S = (\text{negative}) + T|\Delta S|$. At small $T$, the negative $\Delta H$ dominates and $\Delta G<0$ (spontaneous). As $T$ increases, the positive term $T|\Delta S|$ grows and eventually $\Delta G$ turns positive (non-spontaneous). So this combination is spontaneous at low $T$, non-spontaneous at high $T$.
   3. Compare to (B): it says "non-spontaneous at low temperatures" for this exact case — that is backwards from step 2, so (B) is the incorrect statement.
   4. Compare to (D): it says "spontaneous at low temperatures" for the same $\Delta H<0,\Delta S<0$ case — that matches step 2, so (D) is correct (and consistent with, not contradicting, B being wrong). …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The signs of $\Delta_rH^{\circ}$ and $\Delta_rS^{\circ}$ for a reaction to be spontaneous at all temperatures respectively are (A) positive, positive (B) positive, negative (C) negative, negative (D) negative, positive

   ›Reveal solutionSolution

   Using $\Delta G = \Delta H - T\Delta S$, spontaneity at every temperature requires $\Delta H < 0$ and $\Delta S > 0$ so that $\Delta G$ stays negative for all $T$.

   Concept and Intuition

   Gibbs free energy change determines spontaneity: a process is spontaneous when $\Delta G < 0$. Since $\Delta G = \Delta H - T\Delta S$, whether $\Delta G$ stays negative as $T$ varies depends on the signs of $\Delta H$ and $\Delta S$. There are four sign combinations, and only one guarantees spontaneity at all temperatures (from absolute zero to arbitrarily high $T$): when $\Delta H$ is negative (release of energy always helps make $\Delta G$ more negative) and $\Delta S$ is positive (the $-T\Delta S$ term is then always negative too, since $T>0$), both contributions independently push $\Delta G$ negative regardless of how large $T$ gets.

   Step-by-Step Solution

   1. Write $\Delta G = \Delta H - T\Delta S$.
   2. If $\Delta H < 0$ and $\Delta S > 0$: $\Delta G = (\text{negative}) - T(\text{positive}) = (\text{negative}) + (\text{negative}) = \text{always negative}$, for any $T>0$. This is the only combination spontaneous at all temperatures. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.Consider the cell reaction, at 300 K. $\mathrm{A(s) + B^{2+}(aq) \rightleftharpoons A^{2+}(aq) + B(s)}$ Its $E^{\ominus}$ is 1.0 V. The $\Delta_r H^{\ominus}$ of the reaction is $-163\ \mathrm{kJ\,mol^{-1}}$. What is $\Delta_r S^{\ominus}$ (in $\mathrm{J\,K^{-1}}$) of the reaction? ($F = 96500\ \mathrm{C\,mol^{-1}}$) (A) 10 (B) 100 (C) 1000 (D) 10000

   ›Reveal solutionSolution

   Combine $\Delta G^\ominus=-nFE^\ominus$ with $\Delta G^\ominus=\Delta H^\ominus - T\Delta S^\ominus$ to solve for $\Delta S^\ominus = 100\ \mathrm{J\,K^{-1}}$ — option (B).

   Concept and Intuition

   For an electrochemical cell, the maximum non-expansion work available (the Gibbs free energy change) is directly tied to the cell EMF via $\Delta G^\ominus = -nFE^\ominus$, where $n$ is the number of moles of electrons transferred in the balanced cell reaction. Once we have $\Delta G^\ominus$, the standard thermodynamic relation $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ lets us extract the entropy change.

   Step-by-Step Solution

   1. From the cell reaction $A(s) + B^{2+}(aq) \rightarrow A^{2+}(aq) + B(s)$, each metal atom transfers 2 electrons ($A \to A^{2+} + 2e^-$ and $B^{2+}+2e^- \to B$), so $n = 2$.
   2. $\Delta G^\ominus = -nFE^\ominus = -(2)(96500\ \mathrm{C\,mol^{-1}})(1.0\ \mathrm{V}) = -193000\ \mathrm{J\,mol^{-1}} = -193\ \mathrm{kJ\,mol^{-1}}$.
   3. Use $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$: $-193 = -163 - (300)\Delta S^\ominus$ (kJ, with $\Delta S^\ominus$ in $\mathrm{kJ\,K^{-1}}$). …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.At 300 K, the $E^{\ominus}_{cell}$ of $\mathrm{A(s)} + \mathrm{B^{2+}(aq)} \rightleftharpoons \mathrm{A^{2+}(aq)} + \mathrm{B(s)}$ is 1.0 V. If $\Delta_r S^{\ominus}$ of this reaction is $100\ \mathrm{J\ K^{-1}}$, what is $\Delta_r H^{\ominus}$ (in $\mathrm{kJ\ mol^{-1}}$) of this reaction? ($F = 96500\ \mathrm{C\ mol^{-1}}$) (A) -163 (B) -223 (C) -193 (D) -163000

    ›Reveal solutionSolution

    Combining $\Delta_rG^\ominus=-nFE^\ominus$ with $\Delta_rG^\ominus=\Delta_rH^\ominus-T\Delta_rS^\ominus$ gives $\Delta_rH^\ominus = -163\ \mathrm{kJ\,mol^{-1}}$.

    Concept and Intuition

    Electrochemical cell thermodynamics links the cell EMF to the Gibbs free energy change of the underlying reaction via $\Delta_rG^\ominus = -nFE^\ominus_{cell}$. Once $\Delta_rG^\ominus$ is known, the standard Gibbs-Helmholtz relation $\Delta_rG^\ominus = \Delta_rH^\ominus - T\Delta_rS^\ominus$ lets us back out the enthalpy change if entropy change and temperature are given.

    Step-by-Step Solution

    1. The reaction $\mathrm{A(s)}+\mathrm{B^{2+}(aq)}\rightleftharpoons\mathrm{A^{2+}(aq)}+\mathrm{B(s)}$ involves transfer of 2 electrons (both A and B change between the 0 and +2 oxidation states), so $n = 2$.
    2. $\Delta_rG^\ominus = -nFE^\ominus_{cell} = -2 \times 96500 \times 1.0 = -193000\ \mathrm{J\,mol^{-1}} = -193\ \mathrm{kJ\,mol^{-1}}$.
    3. Given $\Delta_rS^\ominus = 100\ \mathrm{J\,K^{-1}}$ and $T = 300\ \mathrm{K}$: $T\Delta_rS^\ominus = 300 \times 100 = 30000\ \mathrm{J} = 30\ \mathrm{kJ}$. …

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11. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
    Q.$\Delta G^\circ$ (in kJ mol$^{-1}$) for the cell reaction is $\mathrm{Cu^{2+}(aq) + Fe(s) \rightarrow Fe^{2+}(aq) + Cu(s)}$ [Given $E^\circ_{Cu^{2+}/Cu} = 0.34$V, $E^\circ_{Fe^{2+}/Fe} = -0.44$V and F = 96,500 C mol$^{-1}$] (A) +150.5 (B) -150.5 (C) +140.2 (D) -140.2

    ›Reveal solutionSolution

    Using $\Delta G^\circ=-nFE^\circ_{cell}$ with $E^\circ_{cell}=0.78$ V and $n=2$ gives $\Delta G^\circ \approx -150.5$ kJ/mol.

    Concept and Intuition

    The standard Gibbs energy change of a redox reaction is directly linked to the cell's standard EMF via $\Delta G^\circ = -nFE^\circ_{cell}$ — a spontaneous reaction (negative $\Delta G^\circ$) corresponds to a positive cell EMF.

    Step-by-Step Solution

    1. Identify the half-reactions: $Cu^{2+}+2e^-\rightarrow Cu$ (reduction, cathode) and $Fe\rightarrow Fe^{2+}+2e^-$ (oxidation, anode).
    2. Compute $E^\circ_{cell} = E^\circ_{cathode}-E^\circ_{anode} = 0.34\ V - (-0.44\ V) = 0.78$ V.
    3. Number of electrons transferred, $n=2$ (both half-reactions involve 2 electrons). …

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12. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.If for an ideal gas reaction $2A(g) + B(g) \rightarrow 2C(g)$ the standard internal energy and entropy changes $\Delta U^\theta$, $\Delta S^\theta$ are $-8.80\ \mathrm{kJ\ mol^{-1}}$, $-50\ \mathrm{J\ K^{-1}\ mol^{-1}}$ respectively at 300 K, which of the following statements is correct? (A) $\Delta G^\theta = -3.71\ \mathrm{kJ\ mol^{-1}}$, the reaction is non-spontaneous (B) $\Delta G^\theta = 3.71\ \mathrm{kJ\ mol^{-1}}$, the reaction is spontaneous (C) $\Delta G^\theta = 3.71\ \mathrm{kJ\ mol^{-1}}$, the reaction is non-spontaneous (D) $\Delta G^\theta = -26.29\ \mathrm{kJ\ mol^{-1}}$, the reaction is spontaneous

    ›Reveal solutionSolution

    This tests converting $\Delta U$ to $\Delta H$ via $\Delta n_g RT$, then applying $\Delta G = \Delta H - T\Delta S$ to judge spontaneity; the answer is $\Delta G^\theta = +3.71\ \mathrm{kJ\,mol^{-1}}$, non-spontaneous.

    Concept and Intuition

    For a gas-phase reaction, $\Delta H = \Delta U + \Delta n_g RT$ accounts for the $PV$ work difference between reactants and products when the number of gas moles changes. Once $\Delta H$ and $\Delta S$ are known, the Gibbs free energy change $\Delta G = \Delta H - T\Delta S$ tells us spontaneity: $\Delta G < 0$ spontaneous, $\Delta G > 0$ non-spontaneous.

    Step-by-Step Solution

    1. Reaction: $2A(g) + B(g) \rightarrow 2C(g)$. Moles of gas: reactants = 3, products = 2, so $\Delta n_g = 2-3 = -1$.
    2. $\Delta H^\theta = \Delta U^\theta + \Delta n_g RT = -8.80\ \mathrm{kJ\,mol^{-1}} + (-1)(8.314\times10^{-3}\ \mathrm{kJ\,mol^{-1}K^{-1}})(300\ \mathrm{K})$.
    3. $\Delta n_g RT = -2.494\ \mathrm{kJ\,mol^{-1}}$, so $\Delta H^\theta = -8.80 - 2.494 = -11.294\ \mathrm{kJ\,mol^{-1}}$.
    4. $\Delta G^\theta = \Delta H^\theta - T\Delta S^\theta = -11.294\ \mathrm{kJ\,mol^{-1}} - (300\ \mathrm{K})(-0.050\ \mathrm{kJ\,K^{-1}mol^{-1}})$. …

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