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Thermodynamics: questions with answers

Class 11 Chemistry, Chapter 5 (NCERT). 15 questions by concept. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Systems, surroundings, properties, states, and processes

Question 1

Two identical samples at the same temperature and pressure are combined without reaction. Which statement is correct?

  1. (a)Temperature and pressure double because intensive properties scale with amount
  2. (b)Internal energy cannot be compared unless the assembly path is known
  3. (c)Volume and internal energy double, while temperature and pressure remain unchanged
  4. (d)All four properties remain unchanged
Show answer

(c) Volume and internal energy double, while temperature and pressure remain unchanged

Volume and internal energy are extensive; temperature and pressure are intensive state properties.

Question 2

A gas reaches the same final state by two different paths. What must be true?

  1. (a)The two paths have the same ΔU, although their q and w values may differ
  2. (b)The final temperature doubles on the longer path because temperature is extensive
  3. (c)Both q and w must separately be identical for the two paths
  4. (d)The two paths can have different ΔU because internal energy depends on path
Show answer

(a) The two paths have the same ΔU, although their q and w values may differ

Internal energy is a state function, while heat and work are path functions.

Heat, pressure-volume work, sign conventions, and the first law

Question 3

A system absorbs 500 J of heat and does 200 J of work on the surroundings. What is ΔU?

  1. (a)−200 J-200\,\text{J} because a process with work must have nonzero cyclic energy loss
  2. (b)+700 J+700\,\text{J} from using ΔU=q−w\Delta U=q-w with w already signed as work on the system
  3. (c)−700 J-700\,\text{J}
  4. (d)+300 J+300\,\text{J}
Show answer

(d) +300 J+300\,\text{J}

Here q = +500 J and w = -200 J, so ΔU = q + w = +300 J.

Question 4

A system completes a cycle, absorbing 80 J of net heat. What net work follows from the first law?

  1. (a)w=+80 Jw=+80\,\text{J} because the first law is ΔU=q−w\Delta U=q-w
  2. (b)w=0w=0 because a cycle has a nonzero internal-energy change
  3. (c)w=−80 Jw=-80\,\text{J}
  4. (d)w=−160 Jw=-160\,\text{J}
Show answer

(c) w=−80 Jw=-80\,\text{J}

For a cycle ΔU = 0, so q + w = 0 and w = -80 J.

Enthalpy, internal energy, and process conditions

Question 5

At constant pressure with only pressure-volume work, which equality applies?

  1. (a)qp=ΔUq_p=\Delta U
  2. (b)qp=ΔHq_p=\Delta H
  3. (c)qp=0q_p=0
  4. (d)ΔT=0\Delta T=0 because constant pressure is adiabatic
Show answer

(b) qp=ΔHq_p=\Delta H

Enthalpy is defined so that constant-pressure heat equals ΔH under the stated work condition.

Question 6

Which statement about an adiabatic compression is correct?

  1. (a)qp=ΔUq_p=\Delta U for every adiabatic process
  2. (b)q=0q=0, but work can increase U and temperature
  3. (c)ΔT=0\Delta T=0 by the definition of adiabatic
  4. (d)w=0w=0 because no heat crosses the boundary
Show answer

(b) q=0q=0, but work can increase U and temperature

Adiabatic means no heat transfer; compression work can raise internal energy and temperature.

Heat capacity, specific heat, and calorimetry

Question 7

What thermodynamic quantity is obtained directly from a rigid bomb calorimeter?

  1. (a)Only the reacting sample absorbs heat; the calorimeter constant is irrelevant
  2. (b)qv=ΔHq_v=\Delta H because a cup calorimeter is also constant volume
  3. (c)qv=ΔUq_v=\Delta U
  4. (d)qp=ΔHq_p=\Delta H
Show answer

(c) qv=ΔUq_v=\Delta U

The sealed bomb has constant volume, so pressure-volume work is zero and q_v equals ΔU.

Question 8

In a constant-pressure cup calorimeter, the solution warms by 4 K. Which energy balance is correct?

  1. (a)qrxn=qsolutionq_{\text{rxn}}=q_{\text{solution}}
  2. (b)qrxn=−(qsolution+qcal)q_{\text{rxn}}=-(q_{\text{solution}}+q_{\text{cal}})
  3. (c)qcal=0q_{\text{cal}}=0 for every insulated calorimeter
  4. (d)The measured heat is q_v because cup calorimetry is constant volume
Show answer

(b) qrxn=−(qsolution+qcal)q_{\text{rxn}}=-(q_{\text{solution}}+q_{\text{cal}})

Energy conservation requires the reaction heat to be the negative of all heat gained by solution and apparatus.

Thermochemical equations and standard enthalpy changes

Question 9

What is ΔfH∘\Delta_fH^\circ of O2(g)\text{O}_2(g) at 298 K in its standard state?

  1. (a)An arbitrary value because phase labels do not matter
  2. (b)0 kJ mol−10\,\text{kJ mol}^{-1}
  3. (c)−285.8 kJ mol−1-285.8\,\text{kJ mol}^{-1}
  4. (d)A positive value because combustion is exothermic
Show answer

(b) 0 kJ mol−10\,\text{kJ mol}^{-1}

Standard formation enthalpy of an element in its reference standard state is defined as zero.

Question 10

Which expression gives standard reaction enthalpy from formation enthalpies?

  1. (a)Stoichiometric sum for products minus stoichiometric sum for reactants
  2. (b)A positive number for every exothermic reaction
  3. (c)Stoichiometric sum for reactants minus products
  4. (d)Ignore physical states and subtract any listed values
Show answer

(a) Stoichiometric sum for products minus stoichiometric sum for reactants

The product-minus-reactant sum follows the Hess cycle from elements to both sides.

Hess's law, bond enthalpy, atomisation, and lattice cycles

Question 11

Using mean bond enthalpies, how is gaseous reaction enthalpy estimated?

  1. (a)Use only the lattice enthalpy and omit ionisation and electron gain
  2. (b)Sum of bonds broken minus sum of bonds formed
  3. (c)Use the full Cl-Cl dissociation enthalpy to make one mole of Cl atoms
  4. (d)Sum of bonds formed minus sum of bonds broken
Show answer

(b) Sum of bonds broken minus sum of bonds formed

Bond breaking costs energy and bond formation releases it, giving broken minus formed.

Second and third laws, surroundings entropy, and stability

Question 12

At constant pressure a system releases heat with ΔHsys<0\Delta H_{sys}<0. What is the sign of ΔSsurr\Delta S_{surr} at temperature T?

  1. (a)Negative because absolute entropy can be below zero
  2. (b)Zero because every solid has zero entropy at 0 K
  3. (c)Positive, because ΔSsurr=−ΔHsys÷T\Delta S_{surr}=-\Delta H_{sys}\div T
  4. (d)Negative because surroundings entropy has the same sign as system heat
Show answer

(c) Positive, because ΔSsurr=−ΔHsys÷T\Delta S_{surr}=-\Delta H_{sys}\div T

The surroundings receive the heat released by the system, so their entropy increases.

Spontaneity, entropy, and total-entropy criteria

Question 13

For a reversible isothermal transfer of heat q_rev into a system at temperature T, what is the system entropy change?

  1. (a)ΔS=qrev÷T\Delta S=q_{rev}\div T
  2. (b)ΔS=T÷qrev\Delta S=T\div q_{rev}
  3. (c)Entropy is only a visual idea and cannot be calculated
  4. (d)For an isolated spontaneous process ΔS must be zero
Show answer

(a) ΔS=qrev÷T\Delta S=q_{rev}\div T

The thermodynamic definition for reversible isothermal heat transfer is ΔS = q_rev/T.

Gibbs energy, temperature, and spontaneity

Question 14

At 300 K, a reaction has ΔH=40.63 kJ mol−1\Delta H=40.63\,\text{kJ mol}^{-1} and ΔS=108.8 J K−1mol−1\Delta S=108.8\,\text{J K}^{-1}\text{mol}^{-1}. What is ΔG?

  1. (a)−7.99 kJ mol−1-7.99\,\text{kJ mol}^{-1} because positive ΔH and ΔS favour low temperature
  2. (b)0 kJ mol−10\,\text{kJ mol}^{-1} because any negative ΔG would mean equilibrium
  3. (c)+7.99 kJ mol−1+7.99\,\text{kJ mol}^{-1}
  4. (d)+73.27 kJ mol−1+73.27\,\text{kJ mol}^{-1} because negative ΔH and ΔS favour high temperature
Show answer

(c) +7.99 kJ mol−1+7.99\,\text{kJ mol}^{-1}

Convert TΔS to 32.64 kJ mol⁻¹, then subtract it from 40.63 kJ mol⁻¹.

Standard Gibbs energy and equilibrium constants

Question 15

At equilibrium, when the composition is not the standard one, which quantity must be zero?

  1. (a)The actual reaction Gibbs energy, ΔG
  2. (b)ΔG° for every possible equilibrium mixture
  3. (c)The temperature dependence of K
  4. (d)Both ΔG and ΔG° regardless of K
Show answer

(a) The actual reaction Gibbs energy, ΔG

At equilibrium ΔG = 0; ΔG° remains tied to K through -RT ln K.

Common misconception in Thermodynamics

Open full misconception guide →

Does an exothermic reaction always proceed spontaneously?

Wrong

A negative enthalpy change alone guarantees spontaneity.

Why it feels right. Releasing heat sounds like a favourable change.

Right

At constant temperature and pressure, Gibbs energy determines the thermodynamic direction.

Worked example

ΔG in kJ/mol
−10+15
+5
At 300 K, the +15 contribution from −TΔS outweighs the −10 contribution from ΔH.

If ΔH = −10 kJ/mol and ΔS = −50 J/(mol K), at 300 K the change ΔG = −10 + 15 = +5 kJ/mol.