CBSE Class 11 Chemistry Thermodynamics MCQs Set 07

Download CBSE MCQs for Class 11 Chemistry: Chapter 05 Thermodynamics

Review structured MCQ sets for Class 11 Chemistry Chapter 05 Thermodynamics. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Chapter 05 Thermodynamics

Access the complete set of multiple-choice questions for Chapter 05 Thermodynamics below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.

Multiple Choice Questions

Question 1: For the reaction C(s) + O2(g) → CO2(g), which relation holds?
(a) ΔH < ΔU
(b) ΔH > ΔU
(c) ΔH = ΔU
(d) ΔH = 0
Show Answer & Explanation

Answer: (c) ΔH = ΔU

Explanation:
1. Δng = 1 − 1 = 0.
2. ΔH = ΔU + ΔngRT = ΔU.

Teacher's Note:
1. Count only gaseous species.
2. Solids do not contribute to Δng.

Question 2: For which of the following reactions is ΔH greater than ΔU?
(a) N2(g) + 3H2(g) → 2NH3(g)
(b) PCl5(g) → PCl3(g) + Cl2(g)
(c) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
(d) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Show Answer & Explanation

Answer: (b) PCl5(g) → PCl3(g) + Cl2(g)

Explanation:
1. ΔH − ΔU = ΔngRT, so ΔH > ΔU only when Δng > 0.
2. PCl5 → PCl3 + Cl2 has Δng = 2 − 1 = +1.
3. The first two have Δng = −2, and the last has no gases.

Teacher's Note:
1. Find Δng for each option.
2. More gas moles in the products means ΔH > ΔU.

Question 3: For the reaction PCl5(g) → PCl3(g) + Cl2(g), the correct relation is:
(a) ΔH = ΔU
(b) ΔH < ΔU
(c) ΔH > ΔU
(d) ΔH = 0
Show Answer & Explanation

Answer: (c) ΔH > ΔU

Explanation:
1. Δng = (1 + 1) − 1 = +1.
2. ΔH = ΔU + RT, so ΔH exceeds ΔU by RT.

Teacher's Note:
1. The positive Δng makes the pV term positive.

Question 4: The difference between ΔH and ΔU on a molar basis for the combustion of methane gas, CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), at 300 K is:
(a) zero
(b) −2RT
(c) −RT
(d) −3RT
Show Answer & Explanation

Answer: (b) −2RT

Explanation:
1. Gaseous reactants: 1 + 2 = 3 mol; gaseous products: 1 mol (water is liquid).
2. Δng = 1 − 3 = −2.
3. ΔH − ΔU = ΔngRT = −2RT.

Teacher's Note:
1. Liquid water is not counted in Δng.

Question 5: Consider the reaction H2(g) + Cl2(g) → 2HCl(g); ΔH° = −185 kJ at 300 K. If 2 mol of H2 react completely with 2 mol of Cl2, the value of ΔU° is:
(a) 0
(b) −185 kJ
(c) +370 kJ
(d) −370 kJ
Show Answer & Explanation

Answer: (d) −370 kJ

Explanation:
1. Δng = 2 − 2 = 0, so ΔU = ΔH.
2. For 2 mol of each reactant the reaction occurs twice: ΔH = 2 × (−185) = −370 kJ.
3. Hence ΔU = −370 kJ.

Teacher's Note:
1. Scale the enthalpy with the amount of reactants.
2. Δng = 0 makes ΔU and ΔH equal.

Question 6: The enthalpies of formation of N2O and NO at 298 K are +82 and +90 kJ mol⁻¹ respectively. The enthalpy change of N2O(g) + ½O2(g) → 2NO(g) is:
(a) −8 kJ
(b) +98 kJ
(c) −74 kJ
(d) +8 kJ
Show Answer & Explanation

Answer: (b) +98 kJ

Explanation:
1. ΔrH = Σ ΔfH(products) − Σ ΔfH(reactants).
2. = 2(+90) − (+82 + 0) = 180 − 82 = +98 kJ.

Teacher's Note:
1. O2 is an element in its standard state, so its ΔfH is zero.
2. Multiply by the coefficient 2 for NO.

Question 7: The enthalpies of solution of anhydrous CuSO4 and CuSO4·5H2O are −15.89 and +2.80 kcal mol⁻¹ respectively. The enthalpy of hydration of CuSO4 to CuSO4·5H2O is:
(a) −18.69 kcal
(b) +13.09 kcal
(c) −13.09 kcal
(d) +18.69 kcal
Show Answer & Explanation

Answer: (a) −18.69 kcal

Explanation:
1. CuSO4(s) + aq → CuSO4(aq); ΔH = −15.89 kcal.
2. CuSO4·5H2O(s) + aq → CuSO4(aq); ΔH = +2.80 kcal.
3. Subtract the second from the first: CuSO4(s) + 5H2O → CuSO4·5H2O; ΔH = −15.89 − 2.80 = −18.69 kcal.

Teacher's Note:
1. Write both solution equations and arrange them to give the target equation.
2. Hess's law lets you subtract equations.

Question 8: Given H2(g) + ½O2(g) → H2O(g) + q₁ and H2(g) + ½O2(g) → H2O(l) + q₂, the enthalpy of vaporisation of water is equal to:
(a) q₁ + q₂
(b) q₁ − q₂
(c) q₂ − q₁
(d) q₁q₂
Show Answer & Explanation

Answer: (c) q₂ − q₁

Explanation:
1. H2O(l) → H2O(g) is obtained by subtracting the second equation from the first.
2. Heat released in forming the liquid (q₂) is larger than for the vapour (q₁).
3. Heat absorbed in vaporisation = q₂ − q₁.

Teacher's Note:
1. Vaporisation always absorbs heat, so the answer must be positive.

Question 9: The enthalpies of all elements in their standard states are taken as zero. The enthalpy of formation of a compound:
(a) should always be negative
(b) should always be positive
(c) is equal to twice the energy of combination
(d) may be positive or negative
Show Answer & Explanation

Answer: (d) may be positive or negative

Explanation:
1. ΔfH° can be negative (CO2, H2O) or positive (NO, N2O, C6H6).
2. It depends on whether the formation from elements releases or absorbs heat.

Teacher's Note:
1. Positive values mean the compound is less stable than its elements in energy terms.

Question 10: Which of the following reactions represents the standard enthalpy of formation of AgCl?
(a) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
(b) Ag(s) + ½Cl2(g) → AgCl(s)
(c) AgCl(s) → Ag(s) + ½Cl2(g)
(d) Ag(s) + AuCl(s) → AgCl(s) + Au(s)
Show Answer & Explanation

Answer: (b) Ag(s) + ½Cl2(g) → AgCl(s)

Explanation:
1. Formation means 1 mol of the compound is made from its elements in their standard (reference) states.
2. Ag(s) and Cl2(g) are the elements, so Ag(s) + ½Cl2(g) → AgCl(s) fits.
3. The other reactions start from ions or compounds, or are decompositions.

Teacher's Note:
1. Reactants must be elements only.
2. The product must be exactly one mole of the compound.

Question 11: Which of the following equations represents the standard enthalpy of formation of methane?
(a) C(graphite) + 2H2(g) → CH4(g)
(b) C(diamond) + 2H2(g) → CH4(g)
(c) C(diamond) + 4H(g) → CH4(g)
(d) C(graphite) + 4H(g) → CH4(g)
Show Answer & Explanation

Answer: (a) C(graphite) + 2H2(g) → CH4(g)

Explanation:
1. The reference state of carbon is graphite, and hydrogen is H2(g).
2. Equations with atomic H(g) are not formation reactions, and diamond is not the reference state.

Teacher's Note:
1. Reference state of carbon: graphite.

Question 12: For the reaction 2H2O(l) → 2H2(g) + O2(g); ΔH = +571.6 kJ, the standard enthalpy of formation of liquid water is:
(a) +285.8 kJ mol⁻¹
(b) +1143.2 kJ mol⁻¹
(c) −1143.2 kJ mol⁻¹
(d) −285.8 kJ mol⁻¹
Show Answer & Explanation

Answer: (d) −285.8 kJ mol⁻¹

Explanation:
1. The given reaction is the reverse of forming 2 mol of water.
2. Forming 2 mol of H2O(l) has ΔH = −571.6 kJ.
3. Per mole: −571.6/2 = −285.8 kJ mol⁻¹.

Teacher's Note:
1. Reverse the equation: change the sign.
2. Divide by 2 to get one mole.

Question 13: SO2(g) + ½O2(g) → SO3(g); ΔH = −98.3 kJ. If the enthalpy of formation of SO3(g) is −395.4 kJ mol⁻¹, the enthalpy of formation of SO2(g) is:
(a) +493.7 kJ
(b) −493.7 kJ
(c) −297.1 kJ
(d) +297.1 kJ
Show Answer & Explanation

Answer: (c) −297.1 kJ

Explanation:
1. ΔrH = ΔfH(SO3) − ΔfH(SO2).
2. −98.3 = −395.4 − ΔfH(SO2), so ΔfH(SO2) = −395.4 + 98.3 = −297.1 kJ mol⁻¹.

Teacher's Note:
1. O2 contributes zero.
2. Rearrange the formula for the unknown.

Question 14: The standard enthalpy of formation at 298 K is arbitrarily taken as zero for:
(a) liquid bromine
(b) gaseous bromine atoms
(c) gaseous bromine molecules
(d) solid bromine
Show Answer & Explanation

Answer: (a) liquid bromine

Explanation:
1. The reference state of an element is its most stable form at 298 K and 1 bar.
2. Bromine is a liquid under these conditions, so Br2(l) has ΔfH° = 0.
3. Br2(g) and Br(g) require energy to form from the liquid, so they have non-zero values.

Teacher's Note:
1. Reference state: the most stable physical form at 298 K.

Question 15: The enthalpies of formation of two compounds A and B are −84 kJ and −156 kJ respectively. Which statement is correct?
(a) A and B are endothermic compounds
(b) A is less stable than B
(c) A is more stable than B
(d) Both A and B are unstable
Show Answer & Explanation

Answer: (b) A is less stable than B

Explanation:
1. Both ΔfH values are negative, so both compounds are exothermic.
2. B releases more heat on formation, so it is lower in energy.
3. Hence A is less stable than B.

Teacher's Note:
1. The more negative the enthalpy of formation, the more stable the compound (energetically).

Question 16: When 0.5 g of sulphur is burnt to SO2, 4.6 kJ of heat is liberated. The enthalpy of formation of sulphur dioxide is (S = 32 g mol⁻¹):
(a) +294.4 kJ mol⁻¹
(b) −147.2 kJ mol⁻¹
(c) −294.4 kJ mol⁻¹
(d) +147.2 kJ mol⁻¹
Show Answer & Explanation

Answer: (c) −294.4 kJ mol⁻¹

Explanation:
1. Heat per mole of S = 4.6 × (32/0.5) = 4.6 × 64 = 294.4 kJ.
2. The heat is released, so ΔfH(SO2) = −294.4 kJ mol⁻¹.

Teacher's Note:
1. Scale the heat to one mole of sulphur.
2. Heat released means a negative sign.

Question 17: The enthalpy of formation of Fe2O3(s) is −824.2 kJ mol⁻¹. The ΔH for the reaction 2Fe2O3(s) → 4Fe(s) + 3O2(g) is:
(a) −412.1 kJ
(b) −1648.4 kJ
(c) −3296.8 kJ
(d) +1648.4 kJ
Show Answer & Explanation

Answer: (d) +1648.4 kJ

Explanation:
1. The reaction is the reverse of forming 2 mol of Fe2O3 from its elements.
2. ΔH = −2 × (−824.2) = +1648.4 kJ.

Teacher's Note:
1. Reverse reaction: change sign.
2. Multiply by 2 for the coefficient.

Question 18: The ΔH° for 4S(s) + 6O2(g) → 4SO3(g) is −1583.2 kJ. The standard enthalpy of formation of sulphur trioxide is:
(a) −395.8 kJ
(b) −3166.4 kJ
(c) +3166.4 kJ
(d) +395.8 kJ
Show Answer & Explanation

Answer: (a) −395.8 kJ

Explanation:
1. The equation forms 4 mol of SO3 from elements.
2. Per mole: −1583.2/4 = −395.8 kJ mol⁻¹.

Teacher's Note:
1. Divide by the coefficient of the compound.

Question 19: The heat of formation of PCl5(s) from the data 2P(s) + 3Cl2(g) → 2PCl3(l), ΔH = −151.8 kcal, and PCl3(l) + Cl2(g) → PCl5(s), ΔH = −32.8 kcal, is:
(a) +108.7 kcal
(b) −184.6 kcal
(c) −108.7 kcal
(d) +184.6 kcal
Show Answer & Explanation

Answer: (c) −108.7 kcal

Explanation:
1. Divide the first equation by 2: P(s) + 1.5Cl2(g) → PCl3(l); ΔH = −75.9 kcal.
2. Add the second equation: P(s) + 2.5Cl2(g) → PCl5(s); ΔH = −75.9 − 32.8 = −108.7 kcal.

Teacher's Note:
1. Make the number of moles of P equal to 1 first.
2. Then add the two steps.

Question 20: Given S + O2 → SO2, ΔH = −298.2 kJ; SO2 + ½O2 → SO3, ΔH = −98.2 kJ; SO3 + H2O → H2SO4, ΔH = −130.2 kJ; H2 + ½O2 → H2O, ΔH = −287.3 kJ. The enthalpy of formation of H2SO4 at 298 K is:
(a) −433.7 kJ
(b) −650.3 kJ
(c) +320.5 kJ
(d) −813.9 kJ
Show Answer & Explanation

Answer: (d) −813.9 kJ

Explanation:
1. Target: H2 + S + 2O2 → H2SO4.
2. Adding all four equations, SO2, SO3 and H2O cancel and give exactly this reaction.
3. ΔH = −298.2 − 98.2 − 130.2 − 287.3 = −813.9 kJ.

Teacher's Note:
1. Check that intermediates cancel when equations are added.
2. Add the ΔH values.

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Multiple Choice Questions (MCQs) for Class 11 Chemistry Chapter 05 Thermodynamics

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