25 questions in 25 minutes on the NTA computer-based test interface, all from Thermodynamics. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Thermodynamics, of which 17 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Thermodynamics mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 17 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Thermodynamics?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Chemistry
- Q1What is the entropy change (in J K⁻¹ mol⁻¹) when one mole of ice is converted into water at 0°C? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol⁻¹ at 0°C)
- 2.198 J K⁻¹ mol⁻¹
- 2.013 J K⁻¹ mol⁻¹
- 21.98 J K⁻¹ mol⁻¹
- 20.13 J K⁻¹ mol⁻¹
Answer: (C) 21.98 J K⁻¹ mol⁻¹
Δ S = (Δ H_f)/(T) = (6000 J mol⁻¹)/(273 K) = 21.98 J K⁻¹ mol⁻¹.
- Q2For a reaction at equilibrium, the standard Gibbs energy change ΔG° is related to the equilibrium constant K by ΔG° = -RT ln K. If K = 1, then ΔG° is:
- Positive
- Infinite
- Negative
- Zero
Answer: (D) Zero
ΔG° = -RT ln K. When K = 1, ln(1) = 0, so ΔG° = 0.
- Q3The values of Δ H and Δ S for the reaction C(graphite) + CO₂(g) arrow 2CO(g) are 170 kJ and 170 J K⁻¹ respectively. This reaction will be spontaneous at:
- 1110 K
- 710 K
- 510 K
- 910 K
Answer: (A) 1110 K
Spontaneous when Δ G = Δ H – TΔ S < 0 ⇒ T > Δ H/Δ S = 170×10³ / 170 = 1000 K. Among the options only 1110 K exceeds 1000 K.
- Q4One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 L to 10 L. The Δ E for this process is (R = 2 cal mol⁻¹ K⁻¹):
- 9 L·atm
- 163.7 cal
- zero
- 1381.1 cal
Answer: (C) zero
Isothermal process of an ideal gas means temperature is constant. Internal energy of an ideal gas depends only on temperature, so ΔE = 0.
- Q5According to the second law of thermodynamics, for a spontaneous process:
- ΔSₜₒₜₐₗ = 0
- ΔSₜₒₜₐₗ (universe) > 0
- ΔS_surroundings > 0 always
- ΔS_system > 0 always
Answer: (B) ΔSₜₒₜₐₗ (universe) > 0
The second law states that the total entropy of the universe (system + surroundings) increases in any spontaneous process. The system’s own entropy may decrease if surroundings increase more.
- Q6Which one of the following is the correct relationship between Cₚ and C_v for one mole of an ideal gas?
- Cₚ + C_v = R
- Cₚ – C_v = R
- C_v = R Cₚ
- Cₚ = R C_v
Answer: (B) Cₚ – C_v = R
For an ideal gas Cₚ – C_v = nR. For one mole n = 1, so Cₚ – C_v = R. Cₚ exceeds C_v because at constant pressure some heat is used to do expansion work.
- Q7Assertion (A): Breaking a chemical bond is an endothermic process. Reason (R): Energy must be supplied to overcome the attractive forces holding atoms together.
- Both A and R are true but R is not the correct explanation of A
- A is false but R is true
- A is true but R is false
- Both A and R are true and R is the correct explanation of A
Answer: (D) Both A and R are true and R is the correct explanation of A
Breaking bonds requires energy input (endothermic), precisely because energy must be supplied to overcome the attractive forces. R correctly explains A.
- Q8For which one of the following equations is Δ H°_(reaction) equal to Δ H°_f for the product?
- Xe(g) + 2F₂(g) arrow XeF₄(g)
- 2CO(g) + O₂(g) arrow 2CO₂(g)
- CH₄(g) + 2Cl₂(g) arrow CH₂Cl₂(l) + 2HCl(g)
- N₂(g) + O₃(g) arrow N₂O₃(g)
Answer: (A) Xe(g) + 2F₂(g) arrow XeF₄(g)
Enthalpy of formation is the heat change forming 1 mole of a compound from its elements in their standard states. Only (A) forms exactly 1 mole of product from elements (Xe and F₂). (B) and (C) have non-element reactants; (D) forms more than one product.
- Q9The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0°C is:
- 10.52 cal/mol K
- 5.260 cal/mol K
- 0.526 cal/mol K
- 21.04 cal/mol K
Answer: (B) 5.260 cal/mol K
Δ S = (Δ H_(fus))/(T) = (1435 cal/mol)/(273 K) = 5.26 cal/mol K.
- Q10Heat of combustion Δ H° for C(s), H₂(g) and CH₄(g) are −94, −68 and −213 kcal/mol respectively. Then Δ H° for C(s) + 2H₂(g) arrow CH₄(g) is:
- -111 kcal/mol
- -170 kcal/mol
- -17 kcal/mol
- -85 kcal/mol
Answer: (C) -17 kcal/mol
Δ H_f(CH₄) = Δ H_c(C) + 2Δ H_c(H₂) – Δ H_c(CH₄) = -94 + 2(-68) – (-213) = -94 – 136 + 213 = -17 kcal/mol.
- Q11Which of the following are NOT state functions? I. q + W II. q III. W IV. H − TS
- II, III and IV
- II and III
- I, II and III
- I and IV
Answer: (B) II and III
q + W = ΔU is a state function and H − TS = G is a state function. Heat (q, II) and work (W, III) are path functions. So II and III are not state functions.
- Q12The enthalpy of combustion of H₂, cyclohexene (C₆H₁₀) and cyclohexane (C₆H₁₂) are −241, −3800 and −3920 kJ per mol respectively. The heat of hydrogenation of cyclohexene (C₆H₁₀ + H₂ arrow C₆H₁₂) is:
- +121 kJ per mol
- -121 kJ per mol
- -242 kJ per mol
- +242 kJ per mol
Answer: (B) -121 kJ per mol
Δ H_(hyd) = [Δ H_c(C₆H₁₀) + Δ H_c(H₂)] – Δ H_c(C₆H₁₂) = [-3800 + (-241)] – (-3920) = -4041 + 3920 = -121 kJ/mol.
- Q13When 1 mole of a gas is heated at constant volume, the temperature is raised from 298 K to 308 K. Heat supplied to the gas is 500 J. Which statement is correct?
- q = ΔE = 500 J, W = 0
- ΔE = 0, q = W = -500 J
- q = -W = 500 J, ΔE = 0
- q = W = 500 J, ΔE = 0
Answer: (A) q = ΔE = 500 J, W = 0
At constant volume ΔV = 0, so W = -pΔV = 0. From the first law ΔE = q + W = q = 500 J.
- Q14Which of the following statements is correct for the spontaneous adsorption of a gas?
- ΔS is negative and therefore ΔH should be highly negative
- ΔS is positive and therefore ΔH should also be highly positive
- ΔS is negative and therefore ΔH should be highly positive
- ΔS is positive and therefore ΔH should be negative
Answer: (A) ΔS is negative and therefore ΔH should be highly negative
On adsorption a gas loses freedom, so ΔS < 0 and −TΔS is positive. For ΔG = ΔH − TΔS to be negative (spontaneous), ΔH must be highly negative to overcome the positive −TΔS term.
- Q15A gas expands against a constant external pressure of 2 atm from 1 L to 5 L. The work done by the gas (1 L·atm ≈ 101.3 J) is approximately:
- +810 J
- -810 J
- -405 J
- +405 J
Answer: (B) -810 J
w = -pₑₓₜ ΔV = -(2 atm)(5-1 L) = -8 L·atm = -8 × 101.3 ≈ -810 J. Negative because the gas does work on the surroundings.
- Q16For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at 300 K, the value of Δn_g is:
- -1
- -2
- +1
- +2
Answer: (B) -2
Δn_g = (moles gaseous products) – (moles gaseous reactants) = 2 – (1+3) = 2 – 4 = -2.
- Q17If ΔH is the change in enthalpy and ΔE the change in internal energy accompanying a gaseous reaction, then:
- ΔH < ΔE only if the number of moles of products is less than that of reactants
- ΔH is always less than ΔE
- ΔH is always greater than ΔE
- ΔH < ΔE only if the number of moles of products is greater than that of reactants
Answer: (A) ΔH < ΔE only if the number of moles of products is less than that of reactants
Δ H = Δ E + Δ n_g RT. Δ H < Δ E requires Δ n_g < 0, i.e. fewer moles of gaseous products than reactants.
- Q18At standard conditions the enthalpy change for H₂(g) + Br₂(g) arrow 2HBr(g) is −109 kJ/mol. Given that bond energies of H₂ and Br₂ are 435 and 192 kJ/mol respectively, the bond energy (in kJ/mol) of HBr is:
- 368
- 518
- 736
- 259
Answer: (A) 368
Δ H = Σ(bonds broken) – Σ(bonds formed): -109 = (435 + 192) – 2 × BE_(HBr) = 627 – 2 BE_(HBr). So 2 BE_(HBr) = 736, BE_(HBr) = 368 kJ/mol.
- Q19If the bond energies of H−H, Br−Br and H−Br are 433, 192 and 364 kJ mol⁻¹ respectively, then Δ H° for the reaction H₂(g) + Br₂(g) arrow 2HBr(g) is:
- -103 kJ
- +261 kJ
- +103 kJ
- -261 kJ
Answer: (A) -103 kJ
Δ H = (bonds broken) – (bonds formed) = (433 + 192) – 2(364) = 625 – 728 = -103 kJ.
- Q20For the reaction C₃H₈(g) + 5O₂(g) arrow 3CO₂(g) + 4H₂O(l) at constant temperature, the value of Δ H – Δ E is:
- -RT
- +3RT
- +RT
- -3RT
Answer: (D) -3RT
Δ H = Δ E + Δ n_g RT, so Δ H – Δ E = Δ n_g RT. Counting only gases: Δ n_g = 3 – (1+5) = -3. Hence Δ H – Δ E = -3RT.
- Q21Given: C(s) + O₂(g) → CO₂(g), ΔH = -393 kJ; CO(g) + ½O₂(g) → CO₂(g), ΔH = -283 kJ. The ΔH for C(s) + ½O₂(g) → CO(g) is:
- -283 kJ
- -676 kJ
- -110 kJ
- +110 kJ
Answer: (C) -110 kJ
By Hess’s law, subtract the second equation from the first: ΔH = (-393) – (-283) = -110 kJ. The target is formation of CO from C and ½O₂.
- Q22Which of the following statements is correct for a reversible process in a state of equilibrium?
- Δ G = -2.30 RTlog K
- Δ G = 2.30 RTlog K
- Δ G° = 2.30 RTlog K
- Δ G° = -2.30 RTlog K
Answer: (D) Δ G° = -2.30 RTlog K
At equilibrium Δ G = 0 and Q = K, so Δ G = Δ G° + 2.303 RTlog Q gives 0 = Δ G° + 2.303 RTlog K, i.e. Δ G° = -2.303 RTlog K.
- Q23Change in enthalpy for the reaction 2H₂O₂(l) arrow 2H₂O(l) + O₂(g) if the heats of formation of H₂O₂(l) and H₂O(l) are −188 and −286 kJ/mol respectively, is:
- -948 kJ/mol
- +196 kJ/mol
- -196 kJ/mol
- +948 kJ/mol
Answer: (C) -196 kJ/mol
Δ H = [2(-286) + 0] – [2(-188)] = -572 + 376 = -196 kJ/mol.
- Q24The entropy of a perfectly crystalline substance at absolute zero (0 K) is taken to be zero. This is a statement of the:
- Second law
- Zeroth law
- First law
- Third law
Answer: (D) Third law
The Third Law of Thermodynamics states that the entropy of a perfectly ordered crystalline substance at 0 K is zero, providing an absolute reference for entropy.
- Q25For an ideal gas, the relation between molar heat capacities is Cₚ – C_v = R. The value of Cₚ – C_v for one mole of an ideal gas is:
- 8.314 J K⁻¹ mol⁻¹
- Zero
- 16.6 J K⁻¹ mol⁻¹
- Equal to C_v
Answer: (A) 8.314 J K⁻¹ mol⁻¹
For one mole of an ideal gas, Cₚ – C_v = R = 8.314 J K⁻¹ mol⁻¹ (≈ 2 cal K⁻¹ mol⁻¹). Cₚ exceeds C_v because at constant pressure some heat does expansion work.