| Table of contents |
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| Section A: Fill In The Blanks |
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| Section B. Multiple Choice Questions |
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| Section C: Assertion and Reason |
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| Section D: Answer the following Questions |
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Q1. Enthalpy of sublimation is the heat required to change a mole of substance from _______ state to its gaseous state at STP.
View Answer 
Ans: solid
Explanation: Enthalpy of sublimation is the heat required to change a substance from a solid to its gaseous state at STP.
Q2. Enthalpy of ________________ refers to the transition from one phase to another, which releases or absorbs a particular standard enthalpy.
View Answer Ans: phase transition
Explanation: Enthalpy of phase transition involves a phase change where heat is either released or absorbed depending on the phase change direction.
Q3. In thermodynamics, the system is a notable part of the universe which is kept under _______.
View Answer Ans: observation
Explanation: A system is the part of the universe that is kept under observation.
Q4. The surrounding refers to the remaining part of the universe except the system, which isn’t kept under _______.
View Answer Ans: observation
Explanation: The surrounding refers to everything else in the universe except the part of the system kept under observation.
Q5. The universe can be defined as the sum of the system and the _______.
View Answer Ans: surrounding
Explanation: The universe is defined as the sum of both the system and the surrounding.
Q1. For the reaction C(s) + O2(g) → CO2(g)
(a) △H >△U
(b) △H < △U
(c) △H = △U
(d) None of these
View Answer Ans: (c) △H = △U
Solution: Here △ng RT = 0 , because reactant and product contain same number of gaseous molecules. So that △H = △U + △ng RT ⇒ △H = △U
Q2. For an ideal gas, CV and CP are related as :
(a) CV – CP = R
(b) CV + CP = R
(c) CP – Cv = RT
(d) CP – Cv = R
View Answer Ans: (d) CP – Cv = R
Solution: For an ideal gas, CV and CP are related as CP – Cv = R
Q3. The least random state of the water system is:
(a) ice
(b) liquid water
(c) steam
(d) randomness is same
View Answer Ans: (a)
Solution: The least random state of the water system is ice.
Q5. Considering entropy(S) thermodynamic parameters the criteria for the spontaneity of any process is:
(a) △S system + △S surroundings > 0
(b) △S system – △S surroundings < 0
(c) △S system > 0
(d) △S surroundings > 0
View Answer Ans: (a)
Solution: The criteria for the spontaneity of any process is
△S system + △S surroundings > 0
Q5. The enthalpy change in a reaction does not depend upon
(a) the state of reactions and products
(b) the nature of the reactants and products
(c) different intermediate steps in the reaction
(d) initial and final enthalpy of the reaction
View Answer Ans: (c)
Solution: The enthalpy change is a state function so it doesn’t depend on different intermediate steps in the reaction.
Q.1. Assertion : Enthalpy of formation of graphite is zero but of diamond it is not zero.
Reason : Enthalpy of formation of the most stable allotrope is taken as zero.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
View Answer Answer: (a)
Exaplanation: Graphite, being the most stable form, has zero enthalpy of formation, while diamond does not.
Q.2. Assertion : Heat of neutralisation for both H2SO4 and HCl with NaOH is 53.7 kJ mol –1.
Reason : Both HCl and H2SO4 are strong acids.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
View Answer Answer: (a)
Exaplanation: Both HCl and H₂SO₄ are strong acids, leading to the same heat of neutralisation.
Q.3. Assertion : Some salts are sparingly soluble in water at room temperature.
Reason : The entropy increases on dissolving the salts.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
View Answer Answer: (b)
Exaplanation: Sparingly soluble salts dissolve due to Gibbs free energy, not just entropy.
Q.4. Assertion : Spontaneous process is an irreversible process and may be reversed by some external agency.
Reason : Decrease in enthalpy is a contributory factor for spontaneity.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
View Answer Answer: (b)
Exaplanation: Spontaneity depends on Gibbs free energy, not just enthalpy.
Q.5. Assertion : The heat absorbed during the isothermal expansion of an ideal gas against vaccum is zero.
Reason : The volume occupied by the molecules of an ideal gas is zero.
(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.
(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion
(c) Assertion is correct, reason is incorrect
(d) Assertion is incorrect, reason is correct.
View Answer Answer: (b)
Exaplanation: Heat absorption in an isothermal expansion depends on work done, not the volume of gas molecules.
Q.1. 18.0 g for the water completely evaporates at 100°C as well as 1 bar pressure, as well as the enthalpy change for the process is 40.79 kJ mol-1. What would be the enthalpy change for vaporising the two moles of the water under the given conditions? What is the standard enthalpy for the vaporisation of the water?
View Answer Ans: Given condition,
The amount of the water is 18.0 g, as well as the pressure, is 1 bar.
Now, we know that 18.0 g H2O = 1 mole H2O
Enthalpy change of vaporising the 1 mole of H2O = 40.79 kJ mol-1
So, Enthalpy change of vaporising the 2 moles of H2O = 2 × 40.79 kJ = 81.358kJ
Standard enthalpy for evaporation at 100℃ and 1 bar pressure is given as,
Δvap H2O = + 40.79 k J mol-1.
Q.2. Standard molar enthalpy of formation, ΔfH° is just a special case of enthalpy of reaction, ΔrH°. Is the ΔrH° for the following reaction same as ΔfH°? Give reason for your answer.
CaO(s) + CO2(g) → CaCO3(s); ΔrH° = -178.3 kJ mol⁻¹
View Answer Ans:No, the ΔrH° for the given reaction is not the same as ΔfH°.
The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, ΔfH°.
CaO(s) + C(s) + 3/2 O2(g) → CaCO3(s)
This reaction is different from the given reaction. Hence, ΔrH° ≠ ΔfH°.
Q.3. The value of ΔfH° for NH3 is -91.8 kJ mol⁻¹. Calculate enthalpy change for the following reaction:
2NH3 (g) → N2 (g) + 3H2 (g)
View Answer Ans: ΔfH° is given for formation. For the reverse reaction, ΔrH° changes sign because the opposite of an exothermic reaction is an endothermic reaction.
So, for one mole, ΔrH° for decomposition is – (-91.8) = 91.8.
However, two moles are decomposing here.
Therefore, ΔrH° = 2 × 91.8 = 183.6 kJ mol⁻¹.
Q.4. The enthalpy of atomisation for the reaction CH4(g) → C(g) + 4H(g) is 1665 kJ mol⁻¹. What is the bond energy of C–H bond?
View Answer Ans:Enthalpy of atomisation of 4 moles of C–H bonds = 1665 kJ mol⁻¹
∴ C–H bond energy, per mole = 1665 kJ mol⁻¹/4 = 416.2 kJ mol⁻¹.
Q.5. Use the following data to calculate Δlattice H° for NaBr.
Δsub H° for sodium metal = 108.4 kJ mol⁻¹
Ionization enthalpy of sodium = 496 kJ mol⁻¹
Electron gain enthalpy of bromine = -325 kJ mol⁻¹
Bond dissociation enthalpy of bromine = 192 kJ mol⁻¹
ΔfH° for NaBr (s) = -360.1 kJ mol⁻¹
View Answer Answer:
According to Hess’s Law,
ΔfH° = Δsub H° + ΔIEH° + ΔdissH° + ΔtgH° + U
Δsub H° for Na metal = 108.4 kJ/mol
I.E. of Na = 496 k/mol
ΔtgH° of Br = -325 kJ/mol
ΔdissH° of Br = 192 k/mol
ΔfH° for NaBr = -360.1 kJ/mol
ΔfH° = Δsub H° + I.E. of Na + ΔdissH° + ΔtgH° + U
-360.1 = 108.4 + 496 + 96 + (-325) - U
U = +735.5 kJ/mol
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