Q.Hydrolysis of sucrose gives, Sucrose + H 2O Glucose + Fructose Equilibrium constant Kc for the reaction is 2 × 10¹³ at 300K. Calculate ∆G ° at 300K.

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Given below are two statements Statement I: The correct relationship between $\Delta_rG^{\ominus}$ and $K_c$ is, $\Delta_rG^{\ominus} = -RT\ln K_c$ Statement II: At 298 K, the correct relationship between $E^{\ominus}_{cell}$ and $K_c$ is, $E^{\ominus}_{cell} = \dfrac{0.059}{n}\ln K_c$ The correct answer is (A) Both statement I and statement II are correct (B) Both statement I and statement 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

   $\Delta_rG^\ominus = -RT\ln K_c$ is the correct general relation (Statement I true). But the $0.059$ V constant in the Nernst-derived cell equation is specifically tied to $\log_{10}$, not $\ln$ — so writing $E^\ominus_{cell}=\frac{0.059}{n}\ln K_c$ (Statement II) is incorrect.

   Concept and Intuition

   The link between standard Gibbs energy change and the equilibrium constant is $\Delta_rG^\ominus = -RT\ln K_c$ This holds in natural-log form universally — that's Statement I, and it is correct.

   Separately, the standard cell potential connects to $\Delta_rG^\ominus$ via $\Delta_rG^\ominus = -nFE^\ominus_{cell}$. Combining the two: $-nFE^\ominus_{cell} = -RT\ln K_c \implies E^\ominus_{cell} = \frac{RT}{nF}\ln K_c = \frac{2.303RT}{nF}\log_{10}K_c$ At $T=298$ K, $\frac{2.303RT}{F} = 0.059$ V — but this 0.059 factor already absorbs the 2.303 conversion from $\ln$ to $\log_{10}$. So the correctly written formula is $E^\ominus_{cell} = \frac{0.059}{n}\log_{10}K_c$ NOT with $\ln$. Using $\ln$ alongside the 0.059 constant (as Statement II does) double-counts/misapplies the 2.303 factor and is numerically wrong.

   Step-by-Step Solution

   1. Statement I: $\Delta_rG^\ominus = -RT\ln K_c$ — matches the standard thermodynamic identity exactly. Correct.
   2. Statement II: claims $E^\ominus_{cell} = \frac{0.059}{n}\ln K_c$ at 298 K.
   3. Derive the true relation: $E^\ominus_{cell} = \frac{RT}{nF}\ln K_c = \frac{2.303RT}{nF}\log_{10}K_c = \frac{0.059}{n}\log_{10}K_c$ (using $\frac{2.303RT}{F}=0.059$ V at 298 K). …

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2. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.At 298 K for the reaction, $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, enthalpy and entropy changes are $-92.4\ kJ$ and $-200\ J\ K^{-1}$ respectively. The value of $\log K_c$ for the reaction is ($R = 8.3\ J\ K^{-1}\ mol^{-1}$) (A) 5.75 (B) 6.75 (C) 7.65 (D) 5.67

   ›Reveal solutionSolution

   This tests the Gibbs free energy relation connecting $\Delta H$, $\Delta S$ and the equilibrium constant.

   Concept and Intuition

   The spontaneity/equilibrium-position link is $\Delta G° = \Delta H° - T\Delta S°$, and $\Delta G°$ in turn fixes the equilibrium constant via $\Delta G° = -RT\ln K = -2.303RT\log K$. So once $\Delta G°$ is known, $\log K_c$ follows directly.

   Step-by-Step Solution

   1. Compute $\Delta G°$ at 298 K: $\Delta G° = \Delta H° - T\Delta S° = -92400\ J - (298\ K)(-200\ J/K) = -92400 + 59600 = -32800\ J$.
   2. Use $\Delta G° = -2.303\,RT\log K_c$, so $\log K_c = \dfrac{-\Delta G°}{2.303RT} = \dfrac{32800}{2.303 \times 8.3 \times 298}$.
   3. Denominator: $2.303 \times 8.3 = 19.11$; $19.11 \times 298 = 5696.3$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.At 298 K for the reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, $\Delta H = -92.4$ kJ, $\log K_C$ value is 5.75. What is the entropy change (in $J\ K^{-1}$) for this reaction at the same temperature? ($R = 8.3\ J\ K^{-1}mol^{-1}$) (A) $-300$ (B) $-400$ (C) $-200$ (D) $-100$

   ›Reveal solutionSolution

   Compute $\Delta G$ from $K_C$ via $\Delta G=-2.303RT\log K_C$, then solve $\Delta G=\Delta H-T\Delta S$ for $\Delta S \approx -200$ J/K.

   Concept and Intuition

   The equilibrium constant links directly to the standard Gibbs free energy change through $\Delta G° = -2.303RT\log K$. Once $\Delta G°$ is known alongside $\Delta H°$, the entropy change follows immediately from the defining relation $\Delta G = \Delta H - T\Delta S$ — this is exactly how thermodynamic entropy changes are extracted from measurable equilibrium data.

   Step-by-Step Solution

   1. $\Delta G° = -2.303RT\log K_C = -2.303 \times 8.3\text{ J/Kmol} \times 298\text{ K} \times 5.75$.
   2. $2.303\times8.3 = 19.11$; $19.11\times298 = 5696.3$; $5696.3\times5.75 = 32753.7$ J. So $\Delta G° \approx -32753.7$ J $= -32.75$ kJ.
   3. Given $\Delta H = -92.4$ kJ $= -92400$ J. Using $\Delta G = \Delta H - T\Delta S$: …

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4. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Given below are two statements Statement I: Physical equilibrium takes place only in closed systems at a given temperature Statement II: For an equilibrium reaction, $\Delta_r G$ is zero The correct answer is (A) Both statement I and statement II are correct (B) Both statement I and statement 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

   Physical equilibrium genuinely requires a closed system at constant temperature, and at any true equilibrium (physical or chemical) the free-energy change of the process, $\Delta_r G$, is zero — so both statements are correct.

   Concept and Intuition

   Equilibrium, whether physical (e.g., water ⇌ water vapour in a sealed container) or chemical (a reversible reaction), is a dynamic balance: the forward and reverse processes occur at equal rates, so there is no net observable change. This balance can only be sustained if nothing is allowed to escape or enter — hence the requirement of a closed system. It is also always defined at a given (constant) temperature, because equilibrium constants and vapour pressures shift with temperature.

   Thermodynamically, a system moves toward equilibrium in the direction that decreases its Gibbs free energy $G$, and it stops changing once $G$ reaches a minimum for the given conditions — i.e., once $\Delta_r G = 0$. This is a general result true for both physical and chemical equilibria; it is distinct from $\Delta_r G^\ominus$ (the standard free-energy change), which is a fixed reference quantity related to the equilibrium constant $K$ by $\Delta_r G^\ominus = -RT\ln K$ and is generally nonzero.

   Step-by-Step Solution

   1. Statement I: "Physical equilibrium takes place only in closed systems at a given temperature." — Correct: e.g., liquid-vapour equilibrium requires a sealed container (closed system) so vapour cannot escape, at a specified constant temperature, since vapour pressure is temperature-dependent. …

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5. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.At 298 K, $\Delta_r G^{\ominus}$ for the following reaction is $165.469\ \mathrm{kJ\,mol^{-1}}$. What is the equilibrium constant for this reaction? ($R = 8.3\ \mathrm{J\,mol^{-1}K^{-1}}$) $\frac{3}{2}\mathrm{O_2(g)} \longrightarrow \mathrm{O_3(g)}$ (A) $10^{29}$ (B) $10^{-29}$ (C) $5\times10^{-27}$ (D) $5\times10^{+27}$

   ›Reveal solutionSolution

   This tests the relation between standard Gibbs energy change and the equilibrium constant; a large positive $\Delta_rG^{\ominus}$ means an extremely small $K$.

   Concept and Intuition

   The thermodynamic link between free energy change and equilibrium constant is $\Delta_rG^{\ominus} = -RT\ln K$. A large positive $\Delta_rG^{\ominus}$ means the reaction is highly non-spontaneous in the forward direction under standard conditions, so at equilibrium the products are present in a vanishingly small amount — $K \ll 1$.

   Step-by-Step Solution

   1. Write the relation: $\Delta_rG^{\ominus} = -RT\ln K$.
   2. Substitute values: $165469 = -(8.3)(298)\ln K$.
   3. $RT = 8.3\times298 = 2473.4\ \mathrm{J\,mol^{-1}}$.
   4. $\ln K = -\dfrac{165469}{2473.4} = -66.90$.
   5. Convert to base-10: $\log K = \dfrac{-66.90}{2.303} = -29.05$. …

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6. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.In a reaction $\mathrm{X(g) \longrightarrow Y(g)}$ at equilibrium, the partial pressure of "Y" is one third of partial pressure of "X". The standard Gibbs energy change ($\Delta G^{\ominus}$) of the reaction is (A) $-RT\ln 3$ (B) $RT\ln 3$ (C) $RT\log 3$ (D) $-RT\log 3$

   ›Reveal solutionSolution

   The equilibrium partial pressure ratio directly gives Kp, and applying the standard Gibbs energy relation gives $\Delta G^\ominus=RT\ln3$.

   Concept and Intuition

   The standard Gibbs energy change of a reaction is related to its equilibrium constant via $\Delta G^\ominus=-RT\ln K$. A K less than 1 (products less favored) corresponds to a positive $\Delta G^\ominus$, consistent with the reaction not proceeding very far forward.

   Step-by-Step Solution

   1. Given $P_Y=\dfrac13 P_X$ at equilibrium.
   2. $K_p=\dfrac{P_Y}{P_X}=\dfrac13$.
   3. $\Delta G^\ominus=-RT\ln K_p=-RT\ln\left(\dfrac13\right)=-RT(-\ln3)=RT\ln3$.

   Common Mistakes …

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7. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Consider the following cell reaction $2\mathrm{Fe}^{3+}(aq) + 2\mathrm{I}^-(aq) \rightleftharpoons 2\mathrm{Fe}^{2+}(aq) + \mathrm{I_2}(s)$ At 298 K, the cell emf is 0.237 V. The equilibrium constant for the reaction is $10^x$. The value of x is (F = 96500 C mol$^{-1}$; R = 8.3 J K$^{-1}$ mol$^{-1}$). (A) 8 (B) 7 (C) 6 (D) 9

   ›Reveal solutionSolution

   This tests the Nernst-equation link between standard cell emf and the equilibrium constant; $x = 8$.

   Concept and Intuition

   A cell's standard emf tells you how far the reaction is from equilibrium in its standard state. At equilibrium the cell can do no more useful work, so its emf is zero and the free-energy change equals $-RT\ln K$. Equating the two expressions for $\Delta G^\circ$ ($-nFE^\circ$ from electrochemistry and $-RT\ln K$ from thermodynamics) gives the bridge formula:

   $nFE^\circ = RT\ln K$

   The number of electrons $n$ must be read off the balanced equation — here $2\mathrm{Fe}^{3+}+2e^- \to 2\mathrm{Fe}^{2+}$ and $2\mathrm{I}^- \to \mathrm{I_2}+2e^-$, so $n=2$.

   Step-by-Step Solution

   1. Balanced reaction: $2\mathrm{Fe}^{3+}+2\mathrm{I}^-\rightleftharpoons 2\mathrm{Fe}^{2+}+\mathrm{I_2}$, so $n=2$.
   2. Bridge equation: $nFE^\circ = 2.303\,RT\log K$.
   3. Substitute: $2\times 96500\times 0.237 = 2.303\times 8.3\times 298 \times \log K$. …

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8. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.At 300 K, $\Delta_r G^{\ominus}$ for the reaction $\mathrm{2AX_3(g) + BO_2(g) \rightleftharpoons (AX_2)_2BO(aq) + X_2O(l)}$ is $-15\ \mathrm{kJ\,mol^{-1}}$ What is the value of $\log K$ for the reaction at the same temperature? ($R = 8.3\ \mathrm{J\,K^{-1}mol^{-1}}$) (A) 13.1 (B) 1.31 (C) 26.2 (D) 2.62

   ›Reveal solutionSolution

   Using $\Delta_rG^\ominus = -2.303RT\log K$ with the given numbers gives $\log K \approx 2.62$.

   Concept and Intuition

   The standard Gibbs energy change of a reaction is related to its equilibrium constant by $\Delta_rG^\ominus = -RT\ln K = -2.303RT\log K$. A negative $\Delta_rG^\ominus$ means $K>1$, i.e., $\log K$ is positive.

   Step-by-Step Solution

   1. Given $\Delta_rG^\ominus = -15\ \mathrm{kJ\,mol^{-1}} = -15000\ \mathrm{J\,mol^{-1}}$, $T = 300$ K, $R = 8.3\ \mathrm{J\,K^{-1}mol^{-1}}$.
   2. $\log K = \dfrac{-\Delta_rG^\ominus}{2.303RT} = \dfrac{15000}{2.303 \times 8.3 \times 300}$.
   3. Denominator: $2.303 \times 8.3 = 19.115$; $19.115 \times 300 = 5734.5$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.At 300 K, $\Delta_r G^{\ominus}$ for the reaction $\mathrm{A_2(g) \rightleftharpoons B_2(g)}$ is $-11.5\ \mathrm{kJ\,mol^{-1}}$. The equilibrium constant at 300 K is approximately ($R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$) (A) 10 (B) 100 (C) 1000 (D) 25

   ›Reveal solutionSolution

   Solving $\Delta G^\ominus = -RT\ln K$ with the given numbers gives $K \approx 100$.

   Concept and Intuition

   The standard Gibbs free energy change of a reaction is related to its equilibrium constant by $\Delta G^{\ominus} = -RT\ln K$. A negative $\Delta G^{\ominus}$ (as given) means $K>1$, i.e. the reaction is product-favoured at equilibrium.

   Step-by-Step Solution

   1. Convert $\Delta G^{\ominus}$ to joules: $-11.5\ \mathrm{kJ/mol} = -11500\ \mathrm{J/mol}$.
   2. Rearrange: $\ln K = -\dfrac{\Delta G^{\ominus}}{RT} = \dfrac{11500}{8.314 \times 300} = \dfrac{11500}{2494.2} \approx 4.61$. …

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10. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.Two statements are given below Statement I: The reaction $\mathrm{Cr_2O_3 + 2Al \rightarrow Al_2O_3 + 2Cr}$ ($\Delta G^{\ominus} = -421\ \mathrm{kJ}$) is thermodynamically feasible Statement II: The above reaction occurs at room temperature The correct answer is (A) Both the statements I & II are correct (B) Both the statements I & 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

    A negative $\Delta G^\ominus$ only guarantees thermodynamic feasibility, not that the reaction proceeds spontaneously at room temperature — the classic thermite reaction needs ignition, so Statement I is true and Statement II is false: option (C).

    Concept and Intuition

    $\Delta G^\ominus$ tells you about the thermodynamics (whether a reaction is favorable overall, comparing initial and final states) — it says nothing about the kinetics (the rate, or whether there's a large activation-energy barrier preventing the reaction from starting). Many thermodynamically favorable reactions simply do not occur at an appreciable rate without an initial energy input to overcome the activation barrier. The aluminothermic (thermite) reaction between $\mathrm{Al}$ and $\mathrm{Cr_2O_3}$ is a textbook example: it is hugely exothermic and has a very negative $\Delta G^\ominus$, yet a piece of aluminium and chromium(III) oxide sitting together at room temperature do not react — the reaction must be ignited (e.g., with a burning Mg ribbon or spark) to supply the activation energy, after which it becomes self-sustaining due to the heat released.

    Step-by-Step Solution

    1. Statement I says the reaction is "thermodynamically feasible" because $\Delta G^\ominus = -421\ \mathrm{kJ}$ (negative) — this is correct by definition (negative $\Delta G^\ominus$ = thermodynamically favorable). …

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11. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
    Q.The correct value for $\log_{10} K_c$ ($K_c$ = equilibrium constant) for the equilibrium reaction at 298 K $\mathrm{2Fe^{3+} + 3I^- \rightleftharpoons 2Fe^{2+} + I_3^-}$ is ($E^\circ$ of the cell is $\underline{x}$ V) (A) $\dfrac{0.059x}{3}$ (B) $\dfrac{0.029x}{3}$ (C) $\dfrac{2x}{0.059}$ (D) $\dfrac{3x}{0.059}$

    ›Reveal solutionSolution

    Applying the Nernst-derived relation $\log K_c = nE^\circ/0.059$ with $n=2$ gives $\log_{10}K_c = 2x/0.059$.

    Concept and Intuition

    The equilibrium constant of a redox reaction is connected to its standard cell potential through the combined relation $\Delta G^\circ = -nFE^\circ = -2.303RT\log K$, which rearranges to $\log K = \dfrac{nE^\circ}{0.0591}$ at 298 K (often approximated as 0.059).

    Step-by-Step Solution

    1. Write the reaction: $2Fe^{3+}+3I^-\rightleftharpoons2Fe^{2+}+I_3^-$.
    2. Determine $n$: the reduction half is $Fe^{3+}+e^-\rightarrow Fe^{2+}$ (×2 = 2 electrons gained); the oxidation half is $3I^-\rightarrow I_3^-+2e^-$ (2 electrons lost) — both consistent, so $n=2$.
    3. Use $\log_{10}K_c = \dfrac{nE^\circ_{cell}}{0.059}$ at 298 K. …

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12. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.If $\Delta G$ and $\Delta S$ for the reaction $A(g) \rightarrow B(g) + 2C(g)$ at 2000 K are -40 $\mathrm{kJ\ mol^{-1}}$ and 0.22 $\mathrm{kJ\ K^{-1}\ mol^{-1}}$ respectively, the change in internal energy for the same reaction approximately (in $\mathrm{kJ\ mol^{-1}}$) is (A) 366.7 (B) -366.7 (C) 433.3 (D) -433.3

    ›Reveal solutionSolution

    First get ΔH from ΔG = ΔH − TΔS, then convert ΔH to ΔU using ΔH = ΔU + Δn(g)RT, giving ≈366.7 kJ/mol.

    Concept and Intuition

    Gibbs free energy, enthalpy and entropy are linked by $\Delta G = \Delta H - T\Delta S$. Separately, enthalpy and internal energy for a gas-phase reaction differ by the $\Delta n_g RT$ "PV work" term, since $\Delta H = \Delta U + \Delta n_g RT$ at constant pressure.

    Step-by-Step Solution

    1. Rearrange for enthalpy: $\Delta H = \Delta G + T\Delta S = -40 + (2000\times0.22) = -40+440 = 400$ kJ/mol.
    2. Determine $\Delta n_g$ for $A(g)\rightarrow B(g)+2C(g)$: moles of gas go from 1 to 3, so $\Delta n_g = 3-1=2$.
    3. Use $\Delta H = \Delta U + \Delta n_g RT \Rightarrow \Delta U = \Delta H - \Delta n_g RT$. …

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