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Key Facts: NEB Class 12 Chemistry Exam

75

Theory Marks

25

Practical Examination Marks

3 hours

Theory Paper Duration

35%

Minimum Theory Score

NEB

Conducting Board

The NEB Class 12 Chemistry examination is conducted by the National Examinations Board (NEB) of Nepal. It consists of a 75-mark written theory paper (3 hours) plus 25 marks of practical examination, and students need at least 35% in theory to avoid a Non-Graded (NG) result. OpenExamPrep offers free independent practice questions on physical, inorganic, organic and applied chemistry. OpenExamPrep's questions are an independent English-language multiple-choice study adaptation of the curriculum topics. They are not an official translation of the board paper or a simulation of its written format.

Sample NEB Class 12 Chemistry Practice Questions

Try these sample questions to review concepts for the NEB Class 12 Chemistry exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1What volume of 0.50 M sulfuric acid (H2SO4) is required to completely neutralize 25.0 mL of 0.40 M sodium hydroxide (NaOH) solution?
A.5.0 mL
B.20.0 mL
C.25.0 mL
D.10.0 mL
Explanation: Sulfuric acid is dibasic (n-factor = 2), so its normality N1 = 2 × 0.50 M = 1.0 N. Sodium hydroxide is monoacidic (n-factor = 1), so N2 = 1 × 0.40 M = 0.40 N. By the law of chemical equivalence: N1 × V1 = N2 × V2 → 1.0 N × V1 = 0.40 N × 25.0 mL → V1 = 10.0 mL.
2An ideal gas expands from an initial volume of 2.0 L to a final volume of 6.0 L against a constant external pressure of 3.0 atm. During the expansion, the gas absorbs 500 J of heat from the surroundings. What is the change in internal energy (ΔU) of the gas? (1 L atm = 101.3 J)
A.-715.6 J
B.+1715.6 J
C.+715.6 J
D.-1215.6 J
Explanation: Work done during expansion w = -P_ext × ΔV = -3.0 atm × (6.0 L - 2.0 L) = -12.0 L atm. Converting to joules: w = -12.0 × 101.3 J = -1215.6 J. Heat absorbed q = +500 J. From the First Law of Thermodynamics: ΔU = q + w = +500 J + (-1215.6 J) = -715.6 J.
3Given the following standard combustion enthalpies at 298 K: C(graphite) + O2(g) → CO2(g), ΔH° = -393.5 kJ/mol H2(g) + 0.5 O2(g) → H2O(l), ΔH° = -285.8 kJ/mol CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), ΔH° = -890.3 kJ/mol What is the standard enthalpy of formation (ΔfH°) of methane, CH4(g)?
A.+74.8 kJ/mol
B.-211.0 kJ/mol
C.-74.8 kJ/mol
D.-965.1 kJ/mol
Explanation: Target formation reaction: C(graphite) + 2 H2(g) → CH4(g). By Hess's Law: ΔfH° = ΔH°_comb(C) + 2 × ΔH°_comb(H2) - ΔH°_comb(CH4) = -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -965.1 + 890.3 = -74.8 kJ/mol.
4The standard enthalpies of formation at 298 K are: ΔfH°[CO2(g)] = -393.5 kJ/mol, ΔfH°[H2O(l)] = -285.8 kJ/mol, and ΔfH°[C2H5OH(l)] = -277.7 kJ/mol. What is the standard enthalpy of combustion for liquid ethanol? C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)
A.-1366.7 kJ/mol
B.+1366.7 kJ/mol
C.-1922.1 kJ/mol
D.-1089.0 kJ/mol
Explanation: Standard enthalpy of combustion ΔH°_rxn = Σ ΔfH°(products) - Σ ΔfH°(reactants). Here, ΔfH°[O2(g)] = 0. Products = 2(-393.5) + 3(-285.8) = -787.0 - 857.4 = -1644.4 kJ/mol. Reactants = -277.7 kJ/mol. ΔH°_rxn = -1644.4 - (-277.7) = -1366.7 kJ/mol.
5Calculate the enthalpy change (ΔH) for the gas-phase hydrogenation of ethene: C2H4(g) + H2(g) → C2H6(g). Given the average bond enthalpies: C=C = 614 kJ/mol, C-C = 348 kJ/mol, C-H = 413 kJ/mol, and H-H = 436 kJ/mol.
A.+124 kJ/mol
B.-248 kJ/mol
C.+348 kJ/mol
D.-124 kJ/mol
Explanation: ΔH = Σ Bond enthalpies of broken bonds - Σ Bond enthalpies of formed bonds. Broken: 1 C=C (614) + 1 H-H (436) = 1050 kJ/mol (the four C-H bonds in C2H4 remain intact). Formed: 1 C-C (348) + 2 C-H (2 × 413 = 826) = 1174 kJ/mol. ΔH = 1050 - 1174 = -124 kJ/mol.
6The standard enthalpy of vaporization of water (ΔvapH) at its normal boiling point (373.15 K) is 40.66 kJ/mol. What is the standard entropy change of vaporization (ΔvapS) of water?
A.10.9 J K⁻¹ mol⁻¹
B.85.2 J K⁻¹ mol⁻¹
C.109.0 J K⁻¹ mol⁻¹
D.152.0 J K⁻¹ mol⁻¹
Explanation: At the normal boiling point, liquid and vapor are in thermodynamic equilibrium, so ΔG = 0. Therefore, ΔvapS = ΔvapH / Tb = 40,660 J/mol / 373.15 K = 108.96 J K⁻¹ mol⁻¹ ≈ 109.0 J K⁻¹ mol⁻¹.
7For a certain endothermic chemical reaction, ΔH° = +45.0 kJ/mol and ΔS° = +150.0 J K⁻¹ mol⁻¹. At what temperature range does this reaction become spontaneous under standard conditions?
A.At temperatures above 300 K
B.At temperatures below 300 K
C.At all temperatures
D.It is non-spontaneous at all temperatures
Explanation: A reaction is spontaneous when ΔG° = ΔH° - TΔS° < 0. For both positive ΔH° and ΔS°, spontaneity requires TΔS° > ΔH°, so T > ΔH° / ΔS° = 45,000 J/mol / 150.0 J K⁻¹ mol⁻¹ = 300 K.
8For a reversible chemical equilibrium at 298 K, the standard Gibbs free energy change ΔG° is -11.41 kJ/mol. What is the value of the equilibrium constant (K) at this temperature? (R = 8.314 J K⁻¹ mol⁻¹; 2.303 × R × 298 = 5705 J/mol)
A.0.01
B.2.0
C.100
D.1000
Explanation: The relationship between standard free energy and equilibrium constant is ΔG° = -2.303 R T log10(K). Substituting: -11,410 J/mol = -5705 J/mol × log10(K) → log10(K) = -11,410 / -5705 = 2.00 → K = 10^(2.00) = 100.
9Two moles of an ideal gas undergo a reversible isothermal expansion at 300 K from an initial volume of 10.0 L to a final volume of 100.0 L. What is the work done on the system (w)? (R = 8.314 J K⁻¹ mol⁻¹; 2.303 × 8.314 × 300 = 5744 J/mol)
A.-5.74 kJ
B.+11.49 kJ
C.-22.98 kJ
D.-11.49 kJ
Explanation: For an isothermal reversible expansion: w = -2.303 n R T log10(V2 / V1). Substituting: w = -2.303 × 2 mol × 8.314 J K⁻¹ mol⁻¹ × 300 K × log10(100 / 10) = -2 × 5744 J × log10(10) = -11,488 J = -11.49 kJ.
10Which of the following thermodynamic quantities is a path-dependent function rather than a state function?
A.Enthalpy (H)
B.Internal energy (U)
C.Work (w)
D.Gibbs free energy (G)
Explanation: Work (w) and heat (q) depend on the specific path taken between initial and final states, making them path functions. In contrast, enthalpy (H), internal energy (U), entropy (S), and Gibbs free energy (G) depend solely on the thermodynamic state of the system.

About the NEB Class 12 Chemistry Exam

The Nepal NEB Class 12 Chemistry Examination is the national secondary school leaving board examination administered by the National Examinations Board (NEB), Bhaktapur. The theory curriculum encompasses Physical Chemistry, Inorganic Chemistry, Organic Chemistry, and Applied Chemistry, evaluated via a 3-hour 75-mark written paper alongside practical laboratory assessment.

Exam sponsor: National Examinations Board (NEB), Sanothimi, Bhaktapur. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Written board examination: a 75-mark written theory paper (3 hours) plus 25 marks of practical examination.

Time Limit

3 hours (theory paper)

Passing Score

At least 35% in the theory paper (below that: Non-Graded, NG)

Exam / Certification Fees

Set annually by NEB; paid through the school

Exam sponsor website

Reported exam pass rate: Not published by subject. Under NEB's letter-grading rules, a theory score below 35% is recorded as Non-Graded (NG) for that subject. Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

30% of practice bank

Physical Chemistry

Kinetics, thermodynamics, equilibrium, electrochemistry and solutions.

21% of practice bank

Inorganic Chemistry

Transition and coordination chemistry, metallurgy, and main-group elements.

41% of practice bank

Organic Chemistry

Functional groups, reaction mechanisms, named reactions, and organic conversions.

8% of practice bank

Applied Chemistry

Chemistry in industry, medicine, agriculture and the environment.

Preparing for the NEB Class 12 Chemistry Exam

What You Need to Know

  • Passing score: At least 35% in the theory paper (below that: Non-Graded, NG)
  • Assessment: Written board examination: a 75-mark written theory paper (3 hours) plus 25 marks of practical examination.
  • Time limit: 3 hours (theory paper)
  • Exam / certification fees: Set annually by NEB; paid through the school Official sources

Using Our Practice Resources

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

NEB Class 12 Chemistry: Suggested Study Strategy

1Master core numerical problem types in physical chemistry: buffer pH calculations, solubility products (Ksp), Nernst cell potentials, and first-order chemical kinetics equations.
2Memorize key inorganic extraction equations and flowsheets for copper from copper pyrites, zinc from zinc blende, and iron from hematite.
3Practice organic named reactions, conversions, and mechanism pathways (SN1 vs SN2, Aldol condensation, Cannizzaro reaction, and Hoffmann bromamide reaction).
4Pay attention to applied chemistry definitions, heavy metal toxicology mechanisms, and pharmaceutical classifications to secure reliable marks.

Frequently Asked Questions

What is the format of the NEB Class 12 Chemistry examination?

The official assessment is a 75-mark written theory paper (3 hours) plus 25 marks of practical examination. It is a written board examination, not a multiple-choice test.

What score is needed to pass NEB Class 12 Chemistry?

Under NEB's letter-grading rules, students need at least 35% of the theory marks. A lower theory score is recorded as Non-Graded (NG) for the subject.

What happens if a student is Non-Graded (NG) in Chemistry?

Students who are NG in up to two subjects can sit the grade-increment (supplementary) examination announced by NEB after the results; others take the subject again in a later regular examination.

Are these practice questions official NEB papers?

No. OpenExamPrep's questions are an independent English-language multiple-choice study adaptation of the curriculum topics. They are not an official translation of the board paper or a simulation of its written format. OpenExamPrep is independent and is not affiliated with or endorsed by NEB or the Curriculum Development Centre (CDC).