NEETAceToday's NEET Daily

Gibbs energy and equilibrium constant: NEET MCQs with answers

89 questions on this topic of Chemical Thermodynamics (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

Practise Chemical Thermodynamics freeAll Chemical Thermodynamics questions

Free with a Google sign-in; NEET Daily needs no sign-in at all.

easy

1. The relationship between the equilibrium constant (K) and the standard Gibbs free energy change (ΔᵣG°) is given by ΔᵣG° = -RT ln K. If a reaction has K = 1 at temperature T, the value of ΔᵣG° is:

  1. Greater than 0
  2. Equal to 0
  3. Less than 0
  4. Equal to -RT
Show answer and explanation

Answer: (B) Equal to 0

From the equation ΔᵣG° = -RT ln K, when K = 1, ln(1) = 0. Therefore, ΔᵣG° = -RT(0) = 0.

medium

2. For a reaction, ΔH = 90 kJ mol⁻¹ and ΔS = 180 J K⁻¹ mol⁻¹. At 250 K, ΔG is:

  1. 45 kJ mol⁻¹
  2. −44910 kJ mol⁻¹
  3. −45 kJ mol⁻¹
  4. 135 kJ mol⁻¹
Show answer and explanation

Answer: (A) 45 kJ mol⁻¹

ΔG = ΔH − TΔS = 90 − 250 × (180/1000) = 45 kJ mol⁻¹ (convert ΔS to kJ).

hard

3. A hypothetical reaction A(g) ⇌ 2B(g) has an equilibrium constant K_p = 0.5 at 300 K. What is the standard Gibbs energy change ΔᵣG° for the reverse reaction: 2B(g) ⇌ A(g) at 300 K? (Take R = 8.314 J K⁻¹ mol⁻¹, ln 2 = 0.693)

  1. +1.73 kJ mol⁻¹
  2. +3.46 kJ mol⁻¹
  3. -1.73 kJ mol⁻¹
  4. -3.46 kJ mol⁻¹
Show answer and explanation

Answer: (C) -1.73 kJ mol⁻¹

For the forward reaction, K_p = 0.5. For the reverse reaction, K_p' = 1/K_p = 1/0.5 = 2. The standard Gibbs free energy change for the reverse reaction is ΔᵣG° = -RT ln(K_p') = -8.314 × 300 × ln(2) = -8.314 × 300 × 0.693 J mol⁻¹ ≈ -1728.5 J mol⁻¹ ≈ -1.73 kJ mol⁻¹.

medium

4. For a reaction, ΔH = −30 kJ mol⁻¹ and ΔS = −180 J K⁻¹ mol⁻¹. At 250 K, ΔG is:

  1. 15 kJ mol⁻¹
  2. −15 kJ mol⁻¹
  3. −75 kJ mol⁻¹
  4. 44970 kJ mol⁻¹
Show answer and explanation

Answer: (A) 15 kJ mol⁻¹

ΔG = ΔH − TΔS = -30 − 250 × (-180/1000) = 15 kJ mol⁻¹ (convert ΔS to kJ).

easy

5. Which of the following expressions correctly relates the reaction quotient Q, standard Gibbs energy change ΔG°, and the non-standard Gibbs energy change ΔG?

  1. ΔG = ΔG° - RT ln Q
  2. ΔG° = ΔG + RT ln Q
  3. ΔG = ΔG° + RT ln Q
  4. Q = exp(-ΔG / RT)
Show answer and explanation

Answer: (C) ΔG = ΔG° + RT ln Q

The thermodynamic relationship between the actual Gibbs energy change (ΔG) and standard Gibbs energy change (ΔG°) is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient.

hard

6. In a biochemical process at 310 K, the standard Gibbs free energy change is ΔG° = +17.1 kJ mol⁻¹. To drive this process, it is coupled with ATP hydrolysis which has ΔG° = -30.5 kJ mol⁻¹. What is the net standard Gibbs energy change and the approximate equilibrium constant of the coupled reaction? [Given: 2.303 RT at 310 K = 5.93 kJ mol⁻¹]

  1. +47.6 kJ mol⁻¹ and 9.3 × 10⁻⁹
  2. +13.4 kJ mol⁻¹ and 5.5 × 10⁻³
  3. -47.6 kJ mol⁻¹ and 1.1 × 10⁸
  4. -13.4 kJ mol⁻¹ and 1.8 × 10²
Show answer and explanation

Answer: (D) -13.4 kJ mol⁻¹ and 1.8 × 10²

ΔG°_net = +17.1 + (-30.5) = -13.4 kJ mol⁻¹. The equilibrium constant is found using log₁₀ K = -ΔG°_net / (2.303 RT) = -(-13.4) / 5.93 ≈ 2.26. Thus K = 10²·²⁶ ≈ 1.8 × 10².

medium

7. For a reaction, ΔH = +90 kJ mol⁻¹ and ΔS = +275 J K⁻¹ mol⁻¹. ΔG for the reaction at 550 K is:

  1. 89.73 kJ mol⁻¹
  2. -61.25 kJ mol⁻¹
  3. 241 kJ mol⁻¹
  4. 61.25 kJ mol⁻¹
Show answer and explanation

Answer: (B) -61.25 kJ mol⁻¹

ΔG = ΔH − TΔS = 90 − 550 × (275/1000) = -61.25 kJ mol⁻¹. Convert ΔS to kJ before subtracting.

Quick tip: ΔS is usually in J; ΔH in kJ. Divide ΔS by 1000 first.

NCERT Class 11, Chapter 5

medium

8. For a reaction, ΔH = −90 kJ mol⁻¹ and ΔS = −120 J K⁻¹ mol⁻¹. At 250 K, ΔG is:

  1. −60 kJ mol⁻¹
  2. 60 kJ mol⁻¹
  3. −120 kJ mol⁻¹
  4. 29910 kJ mol⁻¹
Show answer and explanation

Answer: (A) −60 kJ mol⁻¹

ΔG = ΔH − TΔS = -90 − 250 × (-120/1000) = −60 kJ mol⁻¹ (convert ΔS to kJ).

medium

9. For a reversible reaction, the standard enthalpy change is positive (ΔᵣH° > 0). If the temperature of the reaction mixture is increased, what will be the effect on ΔᵣG° and the equilibrium constant K?

  1. ΔᵣG° increases and K decreases
  2. ΔᵣG° decreases and K increases
  3. Both ΔᵣG° and K decrease
  4. Both ΔᵣG° and K increase
Show answer and explanation

Answer: (B) ΔᵣG° decreases and K increases

For an endothermic reaction (ΔᵣH° > 0, typically ΔᵣS° > 0 for favorable high-T equilibria), increasing T increases K (van 't Hoff equation: d ln K/dT = ΔH°/RT² > 0). Since K increases, ln K becomes more positive, and ΔᵣG° = -RT ln K becomes more negative (decreases).

medium

10. For a reaction, ΔH = −40 kJ mol⁻¹ and ΔS = −60 J K⁻¹ mol⁻¹. At 300 K, ΔG is:

  1. 17960 kJ mol⁻¹
  2. −22 kJ mol⁻¹
  3. 22 kJ mol⁻¹
  4. −58 kJ mol⁻¹
Show answer and explanation

Answer: (B) −22 kJ mol⁻¹

ΔG = ΔH − TΔS = -40 − 300 × (-60/1000) = −22 kJ mol⁻¹ (convert ΔS to kJ).

medium

11. A reaction has ΔH = +60 kJ mol⁻¹ and ΔS = +125 J K⁻¹ mol⁻¹. At 450 K its Gibbs energy change is:

  1. 116 kJ mol⁻¹
  2. -3.75 kJ mol⁻¹
  3. 59.88 kJ mol⁻¹
  4. 3.75 kJ mol⁻¹
Show answer and explanation

Answer: (D) 3.75 kJ mol⁻¹

ΔG = ΔH − TΔS = 60 − 450 × (125/1000) = 3.75 kJ mol⁻¹. Convert ΔS to kJ before subtracting.

Quick tip: ΔS is usually in J; ΔH in kJ. Divide ΔS by 1000 first.

NCERT Class 11, Chapter 5

medium

12. For a reaction, ΔH = 100 kJ mol⁻¹ and ΔS = 200 J K⁻¹ mol⁻¹. At 250 K, ΔG is:

  1. 50 kJ mol⁻¹
  2. −50 kJ mol⁻¹
  3. 150 kJ mol⁻¹
  4. −49900 kJ mol⁻¹
Show answer and explanation

Answer: (A) 50 kJ mol⁻¹

ΔG = ΔH − TΔS = 100 − 250 × (200/1000) = 50 kJ mol⁻¹ (convert ΔS to kJ).

Keep going with Chemical Thermodynamics. All 508 questions with instant answers, your mistakes saved for revision and your accuracy by topic.
Start practising

Other topics in Chemical Thermodynamics