PARIKSHA SUTRA

Home › Practice › Class 11 › Chemistry

Chemical Thermodynamics MCQs

Class 11 Chemistry — questions with answers and worked explanations.

20 free questions Class 11 Chemistry Answers + explanations No sign-up

These questions are drawn from the Pariksha Sutra question bank for Chemical Thermodynamics, part of the Class 11 Chemistry syllabus. Each one shows the correct answer and, where a method helps, the working behind it.

Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.

Questions

Question 1
A system that can exchange both matter and energy with its surroundings is called:
  1. Open system
  2. Adiabatic system
  3. Isolated system
  4. Closed system
Answer: Open system
  1. Matter can leave or enter the system, so the system is not isolated or closed.
  2. Energy can also be transferred as heat or work across the boundary.
  3. Therefore a system that exchanges both matter and energy with surroundings is an Open system.
Question 2
For expansion of a gas into vacuum (free expansion), the work done is:
  1. Zero
  2. Negative and large
  3. Equal to -p*deltaV with p large
  4. Positive and large
Answer: Zero
  1. Work done by a gas is w = ∫ p_ext dV, where p_ext is the pressure exerted by the surroundings.
  2. In a free expansion the gas expands against a vacuum, so the external pressure is zero throughout the process.
  3. Therefore the integral gives zero work, so the work done is Zero.
Question 3
At constant volume, the heat absorbed by a system equals its:
  1. deltaG
  2. deltaU
  3. deltaH
  4. deltaS
Answer: deltaU
  1. First law: ΔU = q + w.
  2. At constant volume, the system does no expansion work, so w = 0.
  3. Hence the heat absorbed q equals ΔU, the change in internal energy. deltaU
Question 4
The quantity of heat required to raise the temperature of 1 g of water by 1 degree C is about:
  1. 4.18 J
  2. 41.8 J
  3. 4.18 kJ
  4. 1.00 J
Answer: 4.18 J
  1. Specific heat (c) is the heat needed to raise 1 g of a substance by 1 °C.
  2. For water, c = 4.18 J g
Question 5
Given deltaHf of CO2(g) = -393.5 kJ/mol, H2O(l) = -285.8 kJ/mol and CH4(g) = -74.8 kJ/mol, the enthalpy of combustion of CH4 (CH4 + 2O2 -> CO2 + 2H2O) is:
  1. -604.5 kJ/mol
  2. -890.3 kJ/mol
  3. -965.1 kJ/mol
  4. +890.3 kJ/mol
Answer: -890.3 kJ/mol
  1. ΔH° = [–393.5 + 2(–285.8)] – (–74.8) = –965.1 + 74.8 = **–890 kJ mol⁻¹** (combustion of CH₄).
Question 6
Hess's law is mainly used to calculate enthalpy changes that are:
  1. Independent of temperature
  2. Difficult to measure directly
  3. Always zero
  4. Only for gases
Answer: Difficult to measure directly
  1. Enthalpy change for a reaction depends only on initial and final states, not on the path taken.
  2. By constructing a hypothetical series of steps whose individual ΔH values are known, we can sum them to obtain the overall ΔH.
  3. This allows us to find ΔH for reactions that are hard to measure in a single experiment.
  4. Thus Hess’s law is mainly used to calculate enthalpy changes that are difficult to measure directly.
Question 7
A gas expands against a constant external pressure of 2 atm from 1 L to 5 L. The work done by the gas is (1 L atm = 101.3 J):
  1. -101.3 J
  2. +810.4 J
  3. -405.2 J
  4. -810.4 J
Answer: -810.4 J
  1. Work = Pₑₓₜ ΔV = 2 atm × (5 L – 1 L) = 8 L·atm
  2. 8 L·atm × 101.3 J / L·atm ≈ 8.1 × 10² J
  3. So the gas does about **8 L·atm (≈ 8.1 × 10² J)** of work.
Question 8
At constant pressure, the heat absorbed by a system equals its:
  1. Change in internal energy
  2. Change in entropy
  3. Change in enthalpy
  4. Work done
Answer: Change in enthalpy
  1. At constant pressure the heat transferred to a system is defined as the heat of reaction, qₚ.
  2. From the definition of enthalpy, H = U + pV, its differential at constant pressure gives dH = dU + p dV = δqₚ.
  3. Thus the heat absorbed equals the change in enthalpy, i.e., ΔH. Change in enthalpy.
Question 9
In bomb calorimetry, the heat of reaction measured is at constant:
  1. Temperature
  2. Enthalpy
  3. Pressure
  4. Volume
Answer: Volume
  1. In a bomb calorimeter the reacting mixture is sealed in a rigid container, so its volume does not change during the reaction.
  2. Because the volume is fixed, the heat measured corresponds to the change in internal energy (ΔU) at constant volume.
  3. Thus the heat of reaction is recorded under constant‑volume conditions. Volume
Question 10
For a reaction, deltaH(reaction) equals the sum of standard enthalpies of formation of:
  1. Products minus reactants
  2. Reactants only
  3. Products plus reactants
  4. Reactants minus products
Answer: Products minus reactants
  1. Enthalpy is a state function, meaning the total change in enthalpy depends solely on the initial state (reactants) and final state (products), independent of the reaction pathway (Hess's Law).
  2. To transform reactants into products:
  3. 1-Decomposing the reactants into their constituent standard elements requires the reverse of formation
  4. 2- Synthesizing the products from those standard elements releases or absorbs their enthalpy of formation
  5. Adding both paths together yields products minus reactants.
Question 11
Given deltaH for A -> B = +20 kJ. Then deltaH for B -> A is:
  1. 0 kJ
  2. +40 kJ
  3. -20 kJ
  4. +20 kJ
Answer: -20 kJ
  1. ΔH for a reaction is the heat change when reactants convert to products.
  2. Reversing the reaction changes the sign of the enthalpy change because the heat absorbed becomes heat released (and vice‑versa).
  3. Therefore ΔH for B → A = –20 kJ.
Question 12
Among solid, liquid and gas of the same substance, entropy is highest for the:
  1. Gas
  2. Solid
  3. Liquid
  4. All equal
Answer: Gas
  1. Entropy S = k ln W, where W is the number of accessible micro‑states.
  2. In a gas molecules move freely in translation, rotation and vibration, giving the largest W compared to liquids (restricted movement) and solids (vibrations about fixed positions).
  3. Hence the entropy is highest for the gas.
Question 13
A thermos flask (ideal) that exchanges neither matter nor energy with the surroundings is an example of a(n):
  1. Homogeneous system
  2. Closed system
  3. Isolated system
  4. Open system
Answer: Isolated system
  1. A system that can exchange matter but not energy with surroundings is a closed system; one that can exchange both is open.
  2. If neither matter nor energy can cross the boundary, the system is isolated.
  3. Hence the ideal thermos flask is an isolated system.
Question 14
Internal energy of a system is best described as its total:
  1. Kinetic plus potential energy of constituents
  2. Potential energy only
  3. Kinetic energy only
  4. Heat content at constant pressure
Answer: Kinetic plus potential energy of constituents
  1. Internal energy (U) of a system is defined as the total microscopic energy possessed by its particles.
  2. It includes the translational, rotational and vibrational kinetic energies of the molecules as well as the intermolecular potential energies.
  3. Therefore it is the kinetic plus potential energy of constituents.
Question 15
The relation between deltaH and deltaU for a reaction involving gases is:
  1. deltaH = deltaU - delta(ng)RT
  2. deltaH = deltaU + RT
  3. deltaH = deltaU
  4. deltaH = deltaU + delta(ng)RT
Answer: deltaH = deltaU + delta(ng)RT
  1. For a reaction at constant pressure, enthalpy change ΔH = ΔU + Δ(PV).
  2. For ideal gases PV = nRT, so the change in PV equals Δ(n_g)·R·T.
  3. Substituting gives ΔH = ΔU + Δ(n_g)RT.
  4. Hence ΔH = ΔU + Δ(ng)RT.
Question 16
How much heat is required to raise the temperature of 100 g of water by 10 degree C? (c = 4.18 J/g/degree C)
  1. 4180 J
  2. 418 J
  3. 41800 J
  4. 41.8 J
Answer: 4180 J
  1. q = m c ΔT
  2. Substitute m = 100 g, c = 4.18 J g⁻¹ °C⁻¹, ΔT = 10 °C → q = 100 × 4.18 × 10
  3. q = 4180 J, which matches the given answer. 4180 J
Question 17
The enthalpy of formation of water from its elements is -286 kJ/mol. The enthalpy of decomposition of one mole of water is:
  1. -143 kJ/mol
  2. +286 kJ/mol
  3. -286 kJ/mol
  4. +143 kJ/mol
Answer: +286 kJ/mol
  1. Enthalpy of formation is for H₂(g)+½O₂(g)→H₂O(l) and is –286 kJ per mole of water formed.
  2. The decomposition reaction is the exact reverse: H₂O(l)→H₂(g)+½O₂(g).
  3. Reversing a reaction changes the sign of ΔH while the magnitude stays the same.
  4. Therefore the enthalpy change for decomposition is +286 kJ/mol. +286 kJ/mol
Question 18
Entropy is a measure of the:
  1. Randomness or disorder of a system
  2. Work done by a system
  3. Total energy of a system
  4. Heat content at constant pressure
Answer: Randomness or disorder of a system
  1. Entropy (S) is defined by the statistical relation S = k ln W, where W is the number of possible microstates of the system.
  2. A larger number of microstates means the particles can be arranged in many ways, indicating higher randomness.
  3. Therefore entropy measures the randomness or disorder of a system. Randomness or disorder of a system
Question 19
Reactants sealed in a closed steel container exchange with the surroundings only:
  1. Neither matter nor energy
  2. Both matter and energy
  3. Energy but not matter
  4. Matter but not energy
Answer: Energy but not matter
  1. A sealed steel container is rigid, so no mass can pass through its walls.
  2. Heat can be transferred through the steel by conduction or radiation.
  3. Thus only energy (as heat) can be exchanged while matter remains confined.
  4. Energy but not matter.
Question 20
The mathematical statement of the first law of thermodynamics is:
  1. deltaU = w - q
  2. deltaH = q + w
  3. deltaU = q + w
  4. deltaU = q - w
Answer: deltaU = q + w
  1. First law states that the energy of an isolated system is constant, so any change in its internal energy (ΔU) must be accounted for by energy transferred as heat (q) and as work (w).
  2. Using the sign convention of chemistry, heat absorbed by the system is taken as positive (q > 0) and work done on the system is also positive (w > 0).
  3. Hence the deltaU = q + w

40 more questions on this chapter

The full chapter test is timed, gives an All-India rank and a subject-wise breakdown of where you lost marks.

See the test series →

Other chapters

Hydrocarbons Ray Optics and Optical Instruments Organic Chemistry – Some Basic Principles and Techniques Structure of Atom Equilibrium Some Basic Concepts of Chemistry Mental Maths practice