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Heat capacities and Hess’s law: 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.

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easy

1. For an ideal gas, the relation between molar heat capacities is:

  1. Cp + Cv = R
  2. Cp = Cv
  3. Cp/Cv = R
  4. Cp - Cv = R
Show answer and explanation

Answer: (D) Cp - Cv = R

For one mole of an ideal gas, ΔH = ΔU + RΔT, which gives Cp - Cv = R.

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medium

2. The heat needed to raise the temperature of m grams of a substance of specific heat c by ΔT is:

  1. q = cΔT/m
  2. q = mcΔT
  3. q = mc/ΔT
  4. q = m/(cΔT)
Show answer and explanation

Answer: (B) q = mcΔT

The heat absorbed is proportional to mass, specific heat capacity and temperature change: q = mcΔT.

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hard

3. Calculate the enthalpy of formation of anhydrous Al₂Cl₆(s) from the following data: (1) 2Al(s) + 6HCl(aq) → Al₂Cl₆(aq) + 3H₂(g); ΔH = -1004 kJ (2) H₂(g) + Cl₂(g) → 2HCl(g); ΔH = -184 kJ (3) HCl(g) + aq → HCl(aq); ΔH = -73 kJ (4) Al₂Cl₆(s) + aq → Al₂Cl₆(aq); ΔH = -643 kJ

  1. -708 kJ mol⁻¹
  2. -1994 kJ mol⁻¹
  3. -1351 kJ mol⁻¹
  4. -1522 kJ mol⁻¹
Show answer and explanation

Answer: (C) -1351 kJ mol⁻¹

Target: 2Al(s) + 3Cl₂(g) → Al₂Cl₆(s). Reaction = (1) + 3×(2) + 6×(3) - (4). ΔH = -1004 + 3(-184) + 6(-73) - (-643) = -1004 - 552 - 438 + 643 = -1994 + 643 = -1351 kJ mol⁻¹.

medium

4. The molar heat capacity of liquid mercury is 28.0 J K⁻¹ mol⁻¹. If 100.3 g of mercury absorbs 560 J of heat energy at constant pressure, what is the increase in temperature of the mercury? (Atomic mass of Hg = 200.6 g mol⁻¹)

  1. 40.0 K
  2. 20.0 K
  3. 10.0 K
  4. 80.0 K
Show answer and explanation

Answer: (A) 40.0 K

Number of moles of Hg, n = mass / molar mass = 100.3 g / 200.6 g mol⁻¹ = 0.50 mol. The heat absorbed at constant pressure is qp = n Cp ΔT. Substituting the values: 560 J = (0.50 mol) × (28.0 J K⁻¹ mol⁻¹) × ΔT, so ΔT = 560 / 14.0 = 40.0 K.

easy

5. Equal masses of three different liquids X, Y, and Z have specific heat capacities of 4.0 J g⁻¹ K⁻¹, 2.0 J g⁻¹ K⁻¹, and 1.0 J g⁻¹ K⁻¹ respectively. If identical amounts of heat are supplied to each liquid initially at 25 °C without phase change, which liquid will attain the highest final temperature?

  1. Liquid Z
  2. Liquid X
  3. Liquid Y
  4. All will attain the same final temperature
Show answer and explanation

Answer: (A) Liquid Z

From q = m × c × ΔT, the temperature rise is ΔT = q / (m × c). Since mass m and heat q are identical, ΔT is inversely proportional to specific heat capacity c. Liquid Z has the lowest specific heat capacity (1.0 J g⁻¹ K⁻¹), so it experiences the largest temperature rise and reaches the highest final temperature.

hard

6. For a gaseous reaction: A(g) + B(g) → C(g), the standard enthalpy change at 300 K is -40.0 kJ mol⁻¹. The molar heat capacities at constant pressure (Cp,m) for A(g), B(g), and C(g) are 30.0, 25.0, and 35.0 J K⁻¹ mol⁻¹ respectively, independent of temperature. What is the standard enthalpy of the reaction at 400 K?

  1. -40.2 kJ mol⁻¹
  2. -38.0 kJ mol⁻¹
  3. -42.0 kJ mol⁻¹
  4. -39.8 kJ mol⁻¹
Show answer and explanation

Answer: (C) -42.0 kJ mol⁻¹

According to Kirchhoff's equation, ΔH°(T₂) = ΔH°(T₁) + ΔCp(T₂ - T₁). Here, ΔCp = Cp,m(C) - [Cp,m(A) + Cp,m(B)] = 35.0 - (30.0 + 25.0) = -20.0 J K⁻¹ mol⁻¹ = -0.020 kJ K⁻¹ mol⁻¹. ΔT = 400 - 300 = 100 K. Therefore, ΔH°(400 K) = -40.0 + (-0.020 × 100) = -40.0 - 2.0 = -42.0 kJ mol⁻¹.

medium

7. Given the following thermochemical reactions: (i) 2NO(g) + O₂(g) → 2NO₂(g); ΔH₁° = -114.0 kJ (ii) 4NO₂(g) + O₂(g) → 2N₂O₅(g); ΔH₂° = -110.2 kJ What is the standard enthalpy change for the reaction: 2NO(g) + 3/2 O₂(g) → N₂O₅(g)?

  1. -224.2 kJ
  2. -57.1 kJ
  3. -112.1 kJ
  4. -169.1 kJ
Show answer and explanation

Answer: (D) -169.1 kJ

Desired equation: 2NO(g) + 3/2 O₂(g) → N₂O₅(g). This is obtained by: equation (i) + 1/2 × equation (ii). Therefore, ΔH° = ΔH₁° + 0.5 × ΔH₂° = -114.0 kJ + 0.5 × (-110.2 kJ) = -114.0 - 55.1 = -169.1 kJ.

medium

8. The heat capacity at constant pressure of a system is defined mathematically by which of the following expressions?

  1. (∂U / ∂T) at constant V
  2. (∂H / ∂P) at constant T
  3. (∂H / ∂T) at constant P
  4. (∂U / ∂P) at constant T
Show answer and explanation

Answer: (C) (∂H / ∂T) at constant P

By definition, the heat absorbed at constant pressure equals the enthalpy change, qP = dH. Therefore, heat capacity at constant pressure is given by Cp = (∂H / ∂T)P.

hard

9. Using the thermochemical data below: (1) P₄(white, s) + 5O₂(g) → P₄O₁₀(s); ΔH₁° = -2984 kJ mol⁻¹ (2) P(red, s) + 5/4 O₂(g) → 1/4 P₄O₁₀(s); ΔH₂° = -738.5 kJ mol⁻¹ Calculate the standard enthalpy of transition for: P₄(white, s) → 4P(red, s).

  1. -2245.5 kJ mol⁻¹
  2. +30 kJ mol⁻¹
  3. -30 kJ mol⁻¹
  4. +2245.5 kJ mol⁻¹
Show answer and explanation

Answer: (C) -30 kJ mol⁻¹

Multiplying equation (2) by 4 gives: 4P(red, s) + 5O₂(g) → P₄O₁₀(s) with ΔH° = 4 × (-738.5) = -2954 kJ mol⁻¹. Subtracting this modified equation from equation (1): P₄(white, s) → 4P(red, s). Thus, ΔH° = -2984 - (-2954) = -30 kJ mol⁻¹.

medium

10. The standard reaction enthalpy for the hydration of anhydrous copper(II) sulfate is determined indirectly via Hess's law: (1) CuSO₄(s) + 800 H₂O(l) → CuSO₄(aq, 800 H₂O); ΔH₁° = -66.5 kJ mol⁻¹ (2) CuSO₄·5H₂O(s) + 795 H₂O(l) → CuSO₄(aq, 800 H₂O); ΔH₂° = +11.7 kJ mol⁻¹ What is the enthalpy change for the hydration reaction: CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s)?

  1. -54.8 kJ mol⁻¹
  2. +78.2 kJ mol⁻¹
  3. -78.2 kJ mol⁻¹
  4. +54.8 kJ mol⁻¹
Show answer and explanation

Answer: (C) -78.2 kJ mol⁻¹

By Hess's law, subtract equation (2) from equation (1): CuSO₄(s) + 800 H₂O(l) - [CuSO₄·5H₂O(s) + 795 H₂O(l)] → 0. Rearranging gives CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s). Hence, ΔrH° = ΔH₁° - ΔH₂° = -66.5 kJ mol⁻¹ - (+11.7 kJ mol⁻¹) = -78.2 kJ mol⁻¹.

easy

11. Hess's law of constant heat summation is a direct consequence of:

  1. Second law of thermodynamics and entropy being a path function
  2. First law of thermodynamics and enthalpy being a state function
  3. Third law of thermodynamics and absolute zero attainment
  4. Zeroth law of thermodynamics and thermal equilibrium
Show answer and explanation

Answer: (B) First law of thermodynamics and enthalpy being a state function

Hess's law is an application of the law of conservation of energy (First law of thermodynamics) applied to chemical reactions, as enthalpy (H) is a state function whose change depends only on the initial and final states, not the pathway taken.

medium

12. For a reaction involving gases, ΔH and ΔU are related by:

  1. ΔH = ΔU × Δng
  2. ΔH = ΔU + PV only for solids
  3. ΔH = ΔU + Δng RT
  4. ΔH = ΔU - RT
Show answer and explanation

Answer: (C) ΔH = ΔU + Δng RT

ΔH = ΔU + PΔV, and for ideal gases PΔV = Δng RT, where Δng is the change in the number of moles of gas.

NCERT Class 11

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