Chapter 5: Thermodynamics Assertion-Reason Questions | chemistry Class 11 CBSE

Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 1

Logic Check

Assertion (A): Internal energy U is a state function. Reason (R): Joule showed that the same amount of adiabatic work always produces the same temperature change regardless of the method used to do the work.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 2

Logic Check

Assertion (A): For an isolated system, ΔU = 0. Reason (R): In an isolated system, there is no exchange of heat or work with the surroundings.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 3

Logic Check

Assertion (A): Enthalpy change ΔH equals heat absorbed at constant volume. Reason (R): At constant volume, w = 0, so ΔU = qV, and ΔH = ΔU + pΔV.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 4

Logic Check

Assertion (A): For free expansion of an ideal gas, ΔU = 0. Reason (R): In free expansion, pex = 0, so w = 0, and Joule showed that q = 0 for ideal gases under isothermal conditions.

Reasoning Options

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Assertion-Reason

Hard difficulty • Check the statement and reason

Hard

Question 5

Logic Check

Assertion (A): ΔH > ΔU for the reaction N2(g) + 3H2(g) → 2NH3(g). Reason (R): In this reaction Δng = -2, so ΔH = ΔU + ΔngRT gives ΔH < ΔU.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 6

Logic Check

Assertion (A): Reversible work is greater in magnitude than irreversible work for the same expansion. Reason (R): In reversible expansion, external pressure at each step equals the internal pressure, extracting maximum work from the system.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 7

Logic Check

Assertion (A): The standard enthalpy of formation of O2(g) is zero. Reason (R): O2 is the most stable allotrope of oxygen and is its reference state at 298 K and 1 bar.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 8

Logic Check

Assertion (A): Hess's law allows calculation of enthalpy changes for reactions that cannot be measured directly. Reason (R): Enthalpy is a state function, so the total enthalpy change is independent of the pathway taken.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 9

Logic Check

Assertion (A): All enthalpy changes for melting (fusion) of solids are positive. Reason (R): Melting is an endothermic process — heat must be supplied to overcome the ordered crystal lattice and convert solid to a less ordered liquid state.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 10

Logic Check

Assertion (A): The combustion of glucose and burning of glucose in the body give the same net enthalpy change. Reason (R): Hess's law states that total enthalpy change depends only on initial and final states, not on the pathway, so the multi-step biochemical process gives the same ΔH as direct combustion.

Reasoning Options

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Assertion-Reason

Hard difficulty • Check the statement and reason

Hard

Question 11

Logic Check

Assertion (A): The mean bond enthalpy of C-H in methane is 416 kJ mol-1. Reason (R): The four successive bond dissociation enthalpies of C-H bonds in methane are identical because all four bonds are chemically equivalent.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 12

Logic Check

Assertion (A): Entropy is a state function. Reason (R): The entropy change ΔS = qrev/T is independent of path and depends only on the initial and final states of the system.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 13

Logic Check

Assertion (A): A liquid crystallizing into a solid involves a decrease in entropy. Reason (R): The crystalline solid state is the state of lowest entropy for a given substance because it is the most ordered state.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 14

Logic Check

Assertion (A): For an exothermic reaction, the entropy of the surroundings increases. Reason (R): Heat released by the exothermic reaction increases the thermal disorder (entropy) of the surroundings.

Reasoning Options

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Assertion-Reason

Hard difficulty • Check the statement and reason

Hard

Question 15

Logic Check

Assertion (A): ΔU cannot distinguish between reversible and irreversible processes for ideal gas expansion. Reason (R): For an ideal gas, ΔU depends only on temperature, and both reversible and irreversible isothermal expansions occur at constant T, giving ΔU = 0 in both cases.

Reasoning Options

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Assertion-Reason

Easy difficulty • Check the statement and reason

Easy

Question 16

Logic Check

Assertion (A): For the reaction H2(g) → 2H(g), entropy increases. Reason (R): One gaseous molecule splits into two gaseous atoms, increasing the number of particles and thus the degree of disorder.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 17

Logic Check

Assertion (A): Gibbs energy G is an extensive property. Reason (R): G = H - TS; since both H and TS are extensive (H is extensive and T is intensive, S is extensive), G is also extensive.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 18

Logic Check

Assertion (A): When ΔG = 0 for a reaction, the system is at equilibrium. Reason (R): At equilibrium, the free energy of the system is at its minimum and the reaction has no tendency to proceed in either direction.

Reasoning Options

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Assertion-Reason

Hard difficulty • Check the statement and reason

Hard

Question 19

Logic Check

Assertion (A): A reaction with ΔrH° < 0 and ΔrS° < 0 is spontaneous at all temperatures. Reason (R): For such a reaction, ΔrG° = ΔrH° - TΔrS° is always negative regardless of temperature.

Reasoning Options

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Assertion-Reason

Medium difficulty • Check the statement and reason

Medium

Question 20

Logic Check

Assertion (A): The standard Gibbs energy change ΔrG° = -RT ln K. Reason (R): At equilibrium, ΔrG = 0, and the relationship between ΔrG° and the reaction quotient Q gives 0 = ΔrG° + RT ln K when Q = K.

Reasoning Options

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Chapter sections

Chapter 5: Thermodynamics | Assertion-Reason

Assertion (A): Internal energy U is a state function. Reason (R): Joule showed that the same amount of adiabatic work always produces the same temperature change regardless of the method used to do the work.

Assertion (A): For an isolated system, ΔU = 0. Reason (R): In an isolated system, there is no exchange of heat or work with the surroundings.

Assertion (A): Enthalpy change ΔH equals heat absorbed at constant volume. Reason (R): At constant volume, w = 0, so ΔU = qV, and ΔH = ΔU + pΔV.

Assertion (A): For free expansion of an ideal gas, ΔU = 0. Reason (R): In free expansion, pex = 0, so w = 0, and Joule showed that q = 0 for ideal gases under isothermal conditions.

Assertion (A): ΔH > ΔU for the reaction N2(g) + 3H2(g) → 2NH3(g). Reason (R): In this reaction Δng = -2, so ΔH = ΔU + ΔngRT gives ΔH < ΔU.

Assertion (A): Reversible work is greater in magnitude than irreversible work for the same expansion. Reason (R): In reversible expansion, external pressure at each step equals the internal pressure, extracting maximum work from the system.

Assertion (A): The standard enthalpy of formation of O2(g) is zero. Reason (R): O2 is the most stable allotrope of oxygen and is its reference state at 298 K and 1 bar.

Assertion (A): Hess's law allows calculation of enthalpy changes for reactions that cannot be measured directly. Reason (R): Enthalpy is a state function, so the total enthalpy change is independent of the pathway taken.

Assertion (A): All enthalpy changes for melting (fusion) of solids are positive. Reason (R): Melting is an endothermic process — heat must be supplied to overcome the ordered crystal lattice and convert solid to a less ordered liquid state.

Assertion (A): The mean bond enthalpy of C-H in methane is 416 kJ mol-1. Reason (R): The four successive bond dissociation enthalpies of C-H bonds in methane are identical because all four bonds are chemically equivalent.

Assertion (A): Entropy is a state function. Reason (R): The entropy change ΔS = qrev/T is independent of path and depends only on the initial and final states of the system.

Assertion (A): A liquid crystallizing into a solid involves a decrease in entropy. Reason (R): The crystalline solid state is the state of lowest entropy for a given substance because it is the most ordered state.

Assertion (A): For an exothermic reaction, the entropy of the surroundings increases. Reason (R): Heat released by the exothermic reaction increases the thermal disorder (entropy) of the surroundings.

Assertion (A): ΔU cannot distinguish between reversible and irreversible processes for ideal gas expansion. Reason (R): For an ideal gas, ΔU depends only on temperature, and both reversible and irreversible isothermal expansions occur at constant T, giving ΔU = 0 in both cases.

Assertion (A): For the reaction H2(g) → 2H(g), entropy increases. Reason (R): One gaseous molecule splits into two gaseous atoms, increasing the number of particles and thus the degree of disorder.

Assertion (A): Gibbs energy G is an extensive property. Reason (R): G = H - TS; since both H and TS are extensive (H is extensive and T is intensive, S is extensive), G is also extensive.

Assertion (A): When ΔG = 0 for a reaction, the system is at equilibrium. Reason (R): At equilibrium, the free energy of the system is at its minimum and the reaction has no tendency to proceed in either direction.

Assertion (A): A reaction with ΔrH° < 0 and ΔrS° < 0 is spontaneous at all temperatures. Reason (R): For such a reaction, ΔrG° = ΔrH° - TΔrS° is always negative regardless of temperature.

Assertion (A): The standard Gibbs energy change ΔrG° = -RT ln K. Reason (R): At equilibrium, ΔrG = 0, and the relationship between ΔrG° and the reaction quotient Q gives 0 = ΔrG° + RT ln K when Q = K.

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