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2026 Statistics
Key Facts: ChE Board Exam
3
Board Subjects
PRC Board of Chemical Engineering
70%
Passing Average
PRC
50%
Minimum Per Subject
PRC
3 days
Exam Duration
PRC
₱900
Exam Fee
PRC Fee Schedule
RA 9297
Governing Law
Chemical Engineering Law
The ChE Board Exam is a three-day objective board exam administered by the PRC Board of Chemical Engineering under RA 9297. It covers three weighted subjects: Physical and Chemical Principles (30%), Chemical Engineering Principles (40%), and General Engineering, Ethics, and Contracts (30%). Passing requires a weighted general average of at least 70% with no rating below 50% in any subject. Handbooks (Perry's) are allowed for Day 1 and Day 2.
Sample ChE Board Practice Questions
Try these sample questions to test your ChE Board exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1A stoichiometry calculation is required for the complete combustion of propane (C3H8). How many moles of oxygen (O2) are required to completely burn 2.5 moles of propane gas?
A.5.0 moles
B.10.0 moles
C.12.5 moles
D.15.0 moles
Explanation: The balanced chemical equation for the combustion of propane is C3H8 + 5 O2 -> 3 CO2 + 4 H2O. According to the stoichiometry, 1 mole of propane requires 5 moles of oxygen. Therefore, 2.5 moles of propane require 2.5 * 5 = 12.5 moles of oxygen.
2Which of the following elements has the highest first ionization energy?
A.Sodium (Na)
B.Chlorine (Cl)
C.Argon (Ar)
D.Potassium (K)
Explanation: First ionization energy generally increases across a period from left to right due to increasing nuclear charge, and decreases down a group. Argon (Ar), being a noble gas at the end of Period 3, has a stable octet electron configuration, giving it the highest first ionization energy among the options.
3In coordination chemistry, what is the crystal field splitting d-orbital configuration for a high-spin octahedral complex of Fe(II) (d6 configuration)?
A.t2g4 eg2
B.t2g6 eg0
C.t2g3 eg3
D.t2g5 eg1
Explanation: In a high-spin octahedral field, the crystal field splitting energy (Delta_o) is less than the pairing energy. Electrons will occupy the t2g and eg orbitals singly before pairing starts. For a d6 ion like Fe(II), the first 5 electrons fill the 3 t2g and 2 eg orbitals singly, and the 6th electron pairs up in a t2g orbital, resulting in a t2g4 eg2 configuration.
4The van der Waals equation of state accounts for deviations from ideal gas behavior. What physical phenomena do the parameters 'a' and 'b' represent, respectively?
A.Intermolecular attractive forces and molecular volume
B.Molecular volume and intermolecular attractive forces
C.Gas molecular weight and temperature deviations
D.Collisional frequency and average kinetic energy
Explanation: In the van der Waals equation, (P + an^2/V^2)(V - nb) = nRT, parameter 'a' accounts for the intermolecular attractive forces that reduce the pressure exerted by the gas, and parameter 'b' represents the finite volume occupied by the gas molecules themselves (excluded volume).
5A water sample contains 120 mg/L of Ca2+ ions and 36 mg/L of Mg2+ ions. What is the total hardness of the water sample expressed in mg/L as CaCO3? (Molar masses: Ca = 40.0 g/mol, Mg = 24.3 g/mol, CaCO3 = 100.1 g/mol)
A.156 mg/L as CaCO3
B.300 mg/L as CaCO3
C.448 mg/L as CaCO3
D.600 mg/L as CaCO3
Explanation: Total hardness in mg/L as CaCO3 is calculated by converting the concentrations of Ca2+ and Mg2+ to equivalent CaCO3. Hardness = [Ca2+ in mg/L * (100.1 / 40.0)] + [Mg2+ in mg/L * (100.1 / 24.3)] = [120 * 2.5] + [36 * 4.12] = 300 + 148.3 = 448.3 mg/L as CaCO3.
6What is the IUPAC name for the organic compound CH3-CH(CH3)-CH(C2H5)-CH2-CH3?
A.2-methyl-3-ethylpentane
B.3-ethyl-2-methylpentane
C.3-isopropylpentane
D.2,3-diethylbutane
Explanation: The longest carbon chain contains 5 carbons (pentane). The substituents are a methyl group at carbon 2 and an ethyl group at carbon 3. IUPAC rules dictate listing substituents alphabetically ('ethyl' before 'methyl'), yielding 3-ethyl-2-methylpentane.
7Which of the following reaction conditions favors an SN2 mechanism over an SN1 mechanism for a nucleophilic substitution reaction?
A.Using a tertiary alkyl halide substrate
B.Using a polar protic solvent
C.Using a strong nucleophile at a high concentration
D.Conducting the reaction at low temperature with no nucleophile
Explanation: SN2 reactions are bimolecular and their rate depends on the concentration of both substrate and nucleophile (Rate = k[Substrate][Nucleophile]). Therefore, using a strong, high-concentration nucleophile promotes the SN2 pathway. Secondary or primary substrates and polar aprotic solvents also favor SN2.
8According to Hückel's Rule, which of the following monocyclic conjugated compounds is aromatic?
A.Cyclobutadiene
B.Cyclooctatetraene
C.Cycloheptatrienyl cation (tropylium ion)
D.Cyclopentadienyl cation
Explanation: Hückel's Rule states that a planar, monocyclic, fully conjugated ring system is aromatic if it contains (4n + 2) pi electrons, where n is an integer. The tropylium ion is planar and has 3 double bonds (6 pi electrons, n=1), making it aromatic. Cyclobutadiene (4 pi e-) and cyclopentadienyl cation (4 pi e-) are antiaromatic, while cyclooctatetraene (8 pi e-) is non-aromatic because it adopts a non-planar tub shape.
9In electrophilic aromatic substitution, which of the following groups acts as a meta-director?
A.-OCH3 (Methoxy group)
B.-NO2 (Nitro group)
C.-NH2 (Amino group)
D.-CH3 (Methyl group)
Explanation: Meta-directors are generally electron-withdrawing groups that deactivate the benzene ring by withdrawing electron density through resonance or induction. The nitro group (-NO2) is a strong electron-withdrawing group and directs incoming electrophiles to the meta position. Methoxy, amino, and methyl groups are electron-donating and act as ortho/para-directors.
10Tollens' reagent (ammoniacal silver nitrate) is commonly used to distinguish between which two classes of organic compounds?
A.Alcohols and Ethers
B.Aldehydes and Ketones
C.Carboxylic Acids and Esters
D.Alkanes and Alkenes
Explanation: Tollens' reagent acts as a mild oxidizing agent. It oxidizes aldehydes to carboxylic acids while reducing silver ions to metallic silver (creating a 'silver mirror' on the test tube). Ketones generally do not react because they cannot be easily oxidized without breaking carbon-carbon bonds, thus serving as a distinguishing test.
About the ChE Board Practice Questions
Verified exam format metadata for PRC Chemical Engineer Licensure Examination is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.