Question 1

Soda ash manufacturing routes: Which named industrial process is used in the ammonia–soda route for producing sodium carbonate (soda ash)?

Accepted Answer

Introduction / Context: Sodium carbonate (Na2CO3), also called soda ash, is a bulk inorganic chemical used in glass, detergents, and pulp. The historical main production route is the ammonia–soda process, widely known by the name of its inventor, Ernest Solvay. Given Data / Assumptions: Key inputs: brine (NaCl), ammonia, and CO2. Intermediate: NaHCO3 precipitates; calcination yields Na2CO3. Ammonia is recovered and recycled using lime (CaO) produced from limestone (CaCO3). Concept / Approach: The flow scheme carbonates ammoniated brine to precipitate sodium bicarbonate selectively. After filtration and calcination, soda ash is obtained. Byproduct CaCl2 forms in the ammonia recovery loop. Step-by-Step Solution: Match the ammonia–soda description to the eponym: Solvay.Reject unrelated processes (Ostwald for nitric acid; Bosch for water–gas shift/H2; Haber for NH3 synthesis).Select “Solvay process.” Verification / Alternative check: Chemical engineering texts consistently label the route “Solvay.” Modern variants include natural trona processing but do not change the naming for the ammonia–soda process. Why Other Options Are Wrong: Each incorrect option names a different iconic process unrelated to soda ash manufacture. Common Pitfalls: Confusing Solvay with Sol-Gel (materials processing) or with downstream glass melting steps. Final Answer: Solvay process

Question 2

Glass technology: During glass manufacture, which additive is most effective for reducing the coefficient of thermal expansion (CTE) and improving thermal shock resistance?

Accepted Answer

Introduction / Context: Thermal expansion control is crucial in glass design for cookware, laboratory ware, and high-temperature service. Borosilicate glasses are renowned for low coefficient of thermal expansion (CTE) and superior thermal shock resistance compared with soda–lime glass. Given Data / Assumptions: Goal: lower CTE to resist cracking during rapid temperature changes. Common oxides in glass: network formers (SiO2, B2O3), intermediates (Al2O3, ZnO), and modifiers (Na2O, K2O, CaO, PbO). Borosilicate formulations (e.g., ~80% SiO2, ~13% B2O3) demonstrate very low CTE. Concept / Approach: B2O3 acts as a network former that increases network connectivity and reduces mobility of modifier cations, thereby decreasing thermal expansion. In contrast, alkali oxides (Na2O, K2O) depolymerize the network and increase CTE; PbO often raises density and CTE while enhancing optical properties. Step-by-Step Solution: Identify the property target: low CTE.Recall composition of borosilicate glass: significant B2O3 reduces CTE.Select “B2O3 (boric oxide)” as the most effective additive among the choices. Verification / Alternative check: Comparative data show borosilicate CTE values roughly one-third to one-half of soda–lime glass, confirming B2O3 as the key variable. Alumina can also help but was not among the provided options. Why Other Options Are Wrong: Na2O/K2O: Increase CTE by breaking the silica network. PbO: Typically increases density and often CTE; not preferred for thermal shock resistance. Common Pitfalls: Assuming any oxide addition lowers CTE; overlooking the role of network formers versus modifiers. Final Answer: B2O3 (boric oxide)

Question 3

In sugar technology, approximately what percentage of sucrose is found in the raw juice expressed from sugar cane (before clarification and evaporation)?

Accepted Answer

Introduction / Context: Understanding sucrose content in extracted cane juice is fundamental to mass balance and process control in sugar mills. It affects liming, clarification, evaporation load, crystallization yields, and molasses formation. Given Data / Assumptions: Raw juice is the immediate product of milling/pressing or diffusion. Typical cane composition includes sucrose, water, fiber (bagasse), and non-sugars. Field conditions and variety cause variation, but a well-accepted range exists. Concept / Approach: Globally, expressed cane juice commonly contains roughly 15–20 % sucrose by weight, with the balance being water and impurities (reducing sugars, ash, organic acids, color bodies). Subsequent unit operations concentrate and purify this sucrose for crystallization. Step-by-Step Solution: Identify realistic sucrose range for expressed juice.Eliminate extreme values inconsistent with plant data (1–2 % is too low; 50 % or more occurs only after evaporation).Select 15–20 % as the accepted range. Verification / Alternative check: Material balances in mills show evaporators must remove large water fractions to reach massecuite; starting juice near 15–20 % sucrose is consistent with steam economy calculations. Why Other Options Are Wrong: 1–2 %: Would make milling uneconomic. 50–60 % / 80–85 %: These correspond to syrup or mother liquor during late evaporation/crystallization, not raw juice. Common Pitfalls: Confusing cane composition (in cane) with juice composition (after expression); assuming refinery syrups reflect raw juice strength. Final Answer: 15–20 %

Question 4

Sugar refining practice: Why is wood charcoal (activated carbon) used for decolourization in sugar processing?

Accepted Answer

Introduction / Context: Decolourization is a key step in producing white sugar or refined syrups. Porous carbons such as bone char or activated wood charcoal remove color bodies formed from polyphenols, Maillard reaction products, and caramels. Given Data / Assumptions: Charcoal presents very high internal surface area and pore structure. Sugar solutions carry a spectrum of organic chromophores and trace metals. Process is typically conducted after clarification and filtration. Concept / Approach: Decolourization proceeds via physical adsorption: van der Waals forces and pore size distribution allow selective uptake of color molecules. With time, the carbon exhausts and is regenerated or replaced. The mechanism is not oxidation, reduction, or chemical conversion in standard practice. Step-by-Step Solution: Identify dominant mechanism: adsorption on porous carbon.Relate high surface area to capacity for color uptake.Choose the option that explicitly states adsorption. Verification / Alternative check: Color removal correlates with iodine number/surface area of carbons. Analytical monitoring (ICUMSA color) demonstrates performance consistent with adsorption isotherms. Why Other Options Are Wrong: Oxidation, reduction, or “conversion to salts” are not the operative mechanisms under typical refinery conditions without added reagents. Common Pitfalls: Assuming “activated carbon” always oxidizes contaminants; in sugar, it primarily adsorbs them. Final Answer: Because it adsorbs colored impurities onto its high-surface-area pores

Question 5

DDT synthesis step: In the classic preparation of DDT by condensing chloral with chlorobenzene, which catalyst is used to promote the reaction?

Accepted Answer

Introduction / Context: DDT (dichloro-diphenyl-trichloroethane) is produced by electrophilic condensation of chloral (trichloroacetaldehyde) with chlorobenzene. The historical laboratory and industrial procedures employ strong acid catalysts to generate a reactive electrophile and drive condensation. Given Data / Assumptions: Reactants: chloral and chlorobenzene. Reaction type: acid-catalyzed aromatic substitution/condensation. Strong acidic medium is required for effective conversion. Concept / Approach: Oleum (fuming sulfuric acid, containing dissolved SO3) provides a superacidic environment that protonates carbonyl and facilitates formation of the electrophilic species. This enhances rate and selectivity toward the DDT framework compared to weaker acids. Step-by-Step Solution: Recognize need for strong acid catalyst.Identify oleum as the conventional catalyst used in practice.Reject non-catalytic UV photolysis for this condensation. Verification / Alternative check: Reference preparations cite concentrated H2SO4/oleum mixtures; oleum is specifically noted for improved activity due to SO3 content. Why Other Options Are Wrong: Dilute H2SO4: Insufficient acidity for efficient conversion. UV light: Not the standard catalytic route for this electrophilic condensation. None: Inaccurate because an acid catalyst is required. Common Pitfalls: Assuming any sulfuric acid strength suffices; underestimating the role of SO3 in creating a superacidic medium. Final Answer: Oleum (fuming sulfuric acid)

Question 6

Biomolecules: Haemoglobin found in red blood cells is best classified as which type of macromolecule?

Accepted Answer

Introduction / Context: Haemoglobin is the oxygen-transport metalloprotein in vertebrate red blood cells. Knowing its biochemical class clarifies its structure–function relationship, cooperative binding behavior, and distinction from enzymes and simple amino acids. Given Data / Assumptions: Haemoglobin consists of four polypeptide chains (globins) and four heme prosthetic groups. Each heme contains an iron ion that binds oxygen reversibly. Function is transport, not catalysis. Concept / Approach: Proteins are polymers of amino acids. Haemoglobin is a conjugated globular protein: a polypeptide (protein) plus a non-protein prosthetic group (heme). Its cooperative oxygen binding (sigmoidal curve) is a hallmark of multimeric proteins, not enzymes in general, though some proteins can be catalytic; haemoglobin is not. Step-by-Step Solution: Identify macromolecule class: protein.Note presence of prosthetic group: heme makes it a conjugated protein.Select the option explicitly stating “Protein … with heme”. Verification / Alternative check: Structural biology shows α2β2 tetramer arrangement and heme–Fe2+ centers. Kinetics of O2 binding are transport-oriented, not catalytic turnover. Why Other Options Are Wrong: Amino acid: Monomer unit; haemoglobin is a polymeric protein. Biological catalyst/Enzyme: Haemoglobin is not an enzyme; it does not catalyze chemical reactions in vivo. Common Pitfalls: Equating all proteins with enzymes; overlooking the distinction between structural/transport proteins and catalytic enzymes. Final Answer: Protein (conjugated globular protein with heme)

Question 7

Materials science – ordinary (soda–lime) glass formulation Which of the following is not a standard raw material intentionally used in the manufacture of ordinary soda–lime glass (window and container glass)?

Accepted Answer

Introduction / Context: Ordinary glass used for windows and bottles is typically the soda–lime–silica composition. Understanding which ingredients are deliberately added versus which appear only as impurities helps in quality control, color control, and furnace operation. Given Data / Assumptions: Soda–lime glass is the reference composition. Common intentional batch ingredients include silica (from sand), soda ash, and limestone/dolomite. Minor coloring/decolorizing agents may also be used, but the question asks about standard raw materials for ordinary clear glass. Concept / Approach: Ordinary soda–lime glass is produced from three principal raw materials: silica (SiO2) for the network former, soda ash (Na2CO3) as the alkali flux to lower the melting temperature, and limestone (CaCO3) to improve durability (providing CaO). Iron oxides, if present in sand, are typically treated as impurities and cause green/brown coloration; they are not intentionally used for making clear, ordinary glass. Step-by-Step Solution: Identify principal batch materials: silica, soda ash, limestone → all standard.Iron oxide is not a required constituent for ordinary soda–lime glass; it is an impurity or a controlled colorant in specialty glasses.Therefore, the choice not intentionally used as a raw material is iron oxide. Verification / Alternative check: Typical batch recipes list sand (SiO2), soda ash (Na2CO3), limestone/dolomite (CaCO3/MgCO3) and sometimes minor Na2SO4 as a fining agent; Fe2O3 is kept low for high clarity, confirming it is not a standard deliberate addition for ordinary clear glass. Why Other Options Are Wrong: Soda ash: Essential flux that reduces melting temperature. Limestone: Provides CaO to increase chemical durability. Silica: The primary network former; the main component of glass. Common Pitfalls: Confusing fining agents or colorants with principal raw materials. Small iron content can be used for tinting, but that is not the standard for ordinary clear glass production. Final Answer: Iron oxide

Question 8

Industrial chemicals – identity of ‘‘salt cake’’ In chemical processing terminology, the product commonly called salt cake is chemically represented by:

Accepted Answer

Introduction / Context: Trade names like ‘‘salt cake’’ are frequently encountered in chemical industries, especially in glass, pulp and paper, and dye manufacture. Correctly identifying the chemical behind a trade name is essential for stoichiometric calculations, safety data sheets, and procurement. Given Data / Assumptions: ‘‘Salt cake’’ is a well-known commodity chemical. Options list common sulfates with varying hydration and cations. We seek the most widely accepted industrial meaning of ‘‘salt cake.’’ Concept / Approach: In industrial usage, ‘‘salt cake’’ refers to sodium sulfate (Na2SO4), historically produced in the Leblanc process and used in glass making and as a pulping aid. The term does not typically refer to calcium sulfate hemihydrate (plaster of Paris), magnesium sulfate (Epsom salt), or barium sulfate (blanc fixe). Step-by-Step Solution: Match trade name to chemical: salt cake → sodium sulfate.Check alternatives: CaSO4·½H2O is plaster of Paris; MgSO4 is Epsom salt; BaSO4 is barite/blanc fixe.Therefore, the correct identity is Na2SO4. Verification / Alternative check: Materials handbooks and glass batch recipes consistently use ‘‘salt cake’’ to denote Na2SO4, confirming the selection. Why Other Options Are Wrong: CaSO4·½H2O: Known as plaster of Paris, not salt cake. MgSO4: Epsom salt; different industrial uses. BaSO4: Barite or blanc fixe; a pigment/filler, not ‘‘salt cake.’’ Common Pitfalls: Assuming all sulfates share the same trade name. In industry, each has distinct nomenclature linked to its applications and hydration state. Final Answer: Na2SO4

Question 9

Industrial salts – starting material for alum manufacture Which substance is used as the principal starting raw material for manufacturing alum (e.g., potash alum)?

Accepted Answer

Introduction / Context: Alum (commonly potassium aluminium sulfate dodecahydrate) is used in water treatment, paper sizing, and dyeing. The route to alum industrially depends on an aluminium-bearing ore that can be dissolved and converted efficiently. Given Data / Assumptions: Commercial alum production targets Al-bearing feedstocks. Bauxite is the most abundant economical source of aluminium compounds. Options include pure chemicals and non-aluminium minerals. Concept / Approach: Bauxite (a hydrated aluminium oxide ore) is acid-leached to extract aluminium, which is then reacted (commonly with potassium sulfate) to crystallize alum. While alumina is a refined product from bauxite, industry typically references bauxite as the starting raw material for alum manufacture rather than purchasing high-purity alumina for this purpose. Step-by-Step Solution: Identify practical, economical Al source → bauxite.Recognize that gypsum and ammonium bicarbonate do not supply aluminium.Select bauxite as the correct starting raw material. Verification / Alternative check: Standard inorganic chemical technology texts describe alum preparation from bauxite via acid digestion followed by crystallization with appropriate alkali metal sulfate. Why Other Options Are Wrong: Alumina: A refined material; not the usual ‘‘starting’’ ore feed in bulk alum production context. Gypsum: Calcium sulfate; no aluminium. Ammonium bicarbonate: Provides ammonium/carbonate; no aluminium. Common Pitfalls: Assuming the refined intermediate (alumina) is the designated raw material; in commodity manufacture, the ore (bauxite) is the common starting point. Final Answer: bauxite

Question 10

Polymer materials – SBR versus natural rubber properties Compared to natural rubber, styrene–butadiene rubber (SBR) generally shows which of the following performance differences?

Accepted Answer

Introduction / Context: SBR is a widely used synthetic rubber in tires and general-purpose goods. Comparing it to natural rubber highlights trade-offs that guide material selection, compounding, and service environment decisions. Given Data / Assumptions: SBR (styrene–butadiene) versus NR (cis-1,4-polyisoprene). Unfilled, unreinforced baseline comparison under similar test conditions. Focus on tensile strength, oxidative resistance, and heat build-up (hysteresis). Concept / Approach: Natural rubber exhibits exceptional strain-induced crystallization, delivering higher tensile strength and tear resistance. SBR lacks strong crystallization under strain; its oxidative stability is also lower without antioxidants. SBR tends to have higher hysteresis, generating more heat under cyclic loading than NR, which is critical for tire temperature rise. Step-by-Step Solution: Compare tensile behavior: NR > SBR in tensile strength → SBR is poorer.Assess oxidation: NR with natural antioxidants still often outperforms base SBR; SBR needs stabilizers → poorer oxidation resistance.Evaluate heat build-up: SBR higher hysteresis → greater heat under heavy loading. Verification / Alternative check: Standard property tables for vulcanizates show NR’s superior tensile/tear and lower hysteresis versus general-purpose SBR, confirming all three statements are true simultaneously. Why Other Options Are Wrong: Options (a), (b), and (c) each capture a true statement; hence the combined option ‘‘all (a), (b) and (c)’’ is most correct. Common Pitfalls: Overgeneralizing after compounding: fillers (e.g., carbon black), oils, and antioxidants can significantly modify performance; the baseline material tendencies are as stated. Final Answer: all (a), (b) and (c).

Question 11

Organic reactions in industry – what does esterification produce? Esterification (reaction of an acid with an alcohol) is industrially used to make which of the following product classes?

Accepted Answer

Introduction / Context: Esterification is a core organic reaction forming esters from acids and alcohols. In the chemical process industries, many surfactants and specialty chemicals are esters. Distinguishing esterification from saponification, hydrogenation, or thiolation avoids common mix-ups in exam questions. Given Data / Assumptions: Esterification: acid + alcohol → ester + water (often acid-catalyzed). Industrial detergents can include nonionic types that are fatty acid or fatty alcohol esters (e.g., fatty acid esters of polyols or ethoxylates), made via esterification or transesterification. Soap is produced by saponification (base-catalyzed hydrolysis of esters), not by esterification. Concept / Approach: Among the listed choices, detergents encompass classes (notably nonionic types) that are prepared through esterification of fatty acids and polyhydric alcohols or ethoxylated alcohols. Vanaspati (hydrogenated vegetable oil) arises from catalytic hydrogenation, soaps from base hydrolysis of triglyceride esters, and mercaptans (thiols) are formed via thiolation routes, not esterification. Step-by-Step Solution: Identify reaction type: esterification forms esters.Match products: nonionic detergents and surfactants often are esters.Exclude vanaspati (hydrogenation), soap (saponification), mercaptans (thiolation). Verification / Alternative check: Surfactant manufacturing literature documents fatty acid esters (e.g., sorbitan esters, glycerol esters) used as detergents/emulsifiers, confirming the esterification route. Why Other Options Are Wrong: Vanaspati: Produced by hydrogenating edible oils; no esterification step dominates. Soap: Produced by saponification (ester hydrolysis), essentially the reverse of esterification. Mercaptans: Prepared by reactions introducing –SH; not via ester formation. Common Pitfalls: Confusing soap manufacture (saponification) with esterification. Remember: saponification breaks esters; esterification creates them. Many detergents are not esters (e.g., sulfonates), but esters are used for certain detergent/surfactant types, making ‘‘detergent’’ the best fit among the options. Final Answer: detergent

Question 12

Fertilizer and phosphorus industry – uses of phosphate rock Phosphate rock serves as a raw material for which of the following products?

Accepted Answer

Introduction / Context: Phosphate rock (apatite) is the backbone of the phosphorus value chain. From this mineral, multiple downstream products are manufactured, feeding agriculture and chemical industries. Recognizing the breadth of applications clarifies plant flowsheets and raw material linkages. Given Data / Assumptions: Feedstock: phosphate rock (fluorapatite). Common downstream routes include acidulation and thermal reduction. Target products: phosphoric acid, elemental phosphorus, and superphosphates. Concept / Approach: Phosphoric acid is produced by digesting phosphate rock with sulfuric acid (wet process). Elemental phosphorus can be produced by the electric furnace process, reducing the rock with carbon and silica. Superphosphates are made by treating rock phosphate with acids (sulfuric to make single superphosphate; phosphoric to make triple superphosphate). Thus, all the listed products derive from phosphate rock. Step-by-Step Solution: Route 1: Rock + H2SO4 → wet-process phosphoric acid + CaSO4 co-product.Route 2: Electric furnace → elemental phosphorus (then oxidized to P2O5 or further processed).Route 3: Acidulation → single and triple superphosphates. Verification / Alternative check: Fertilizer industry overviews confirm these three are the primary industrial pathways from rock phosphate. Why Other Options Are Wrong: Each single option covers only one use; the comprehensive, correct selection is all of the above. Common Pitfalls: Confusing elemental phosphorus production (thermal route) with wet-process acidulation. Both start from the same mineral but use different technologies. Final Answer: all (a), (b) and (c)

Question 13

Cement chemistry – making high alumina cement (HAC) High alumina cement is produced by fusing bauxite with which material before clinkering and grinding?

Accepted Answer

Introduction / Context: High alumina cement (HAC) is notable for rapid strength gain and chemical resistance. Its manufacture involves creating calcium aluminates, which form during fusion and subsequent cooling of specific raw materials. Given Data / Assumptions: Bauxite supplies alumina (Al2O3). We need a calcium source to form calcium aluminates. Standard raw mix practice for HAC production is assumed. Concept / Approach: HAC consists primarily of calcium aluminates such as CA and CA2. To form these, bauxite (Al2O3) is fused with a calcium source—most commonly limestone (CaCO3), which decomposes to CaO in the kiln. The fused mass (‘‘clinker’’) is then crushed and ground. Other listed materials do not provide CaO in the needed form or are not used for this purpose. Step-by-Step Solution: Identify Al source: bauxite.Identify Ca source: limestone → CaCO3 → CaO.Fuse bauxite + limestone → calcium aluminates → grind to HAC. Verification / Alternative check: Textbook HAC manufacturing routes confirm the use of bauxite and limestone as primary raw materials fused at high temperature to create calcium aluminate phases. Why Other Options Are Wrong: Alum: A sulfate salt; not a calcium source. Coke: Fuel/reductant, not a CaO source. Quartz: Silica; would form silicates, not desired aluminates. Common Pitfalls: Confusing HAC with Portland cement chemistry; HAC emphasizes aluminates rather than silicates and uses a different raw mix and heat treatment. Final Answer: limestone

Question 14

Alcohol industry by-products – identity of fusel oil Fusel oil obtained in spirit production is best described as:

Accepted Answer

Introduction / Context: During fermentation and distillation of spirits, by-products accompany ethanol. ‘‘Fusel oil’’ is a well-known fraction that has implications for beverage quality, industrial alcohol purification, and effluent handling. Given Data / Assumptions: Feedstock: molasses or other sugar substrates. Process: yeast fermentation followed by distillation. Focus: identity and composition of fusel oil. Concept / Approach: Fusel oil is the higher-boiling side stream comprised mainly of higher alcohols (e.g., amyl alcohols, propanol, butanol), small amounts of aldehydes, and esters. It is separated from spirit streams and can be processed or disposed of depending on application. It is not an essential oil nor an herbal extract; it arises from fermentation by-products and distillation cut management. Step-by-Step Solution: Recognize fermentation by-products: higher alcohols formed via amino acid metabolism.During distillation, these concentrate in tails → ‘‘fusel oil’’ fraction.Therefore, fusel oil is a mixture of higher alcohols from spirit production. Verification / Alternative check: Distillation practice and beverage technology references list isoamyl alcohol as a principal component of fusel oil; industrial alcohol plants recover or treat this by-product. Why Other Options Are Wrong: Essential oil / herbal extract: Derived from plants via steam distillation/solvent extraction, not from molasses fermentation. None of these: Incorrect since the precise definition is provided in option (c). Common Pitfalls: Assuming all distilled oily fractions are essential oils; in fermentation distillation, ‘‘oily’’ refers to higher-alcohol mixtures rather than terpene-rich essential oils. Final Answer: mixture of higher molecular weight alcohols (a by-product obtained during production of alcohol from molasses).

Question 15

Thermochemistry – identify the endothermic reaction Which of the following reactions is endothermic under standard conditions?

Accepted Answer

Introduction / Context: Knowing whether a reaction absorbs or releases heat is vital for furnace design, reactor heat balance, and process safety. Endothermic reactions require heat input; exothermic reactions release heat and may need cooling or careful temperature control. Given Data / Assumptions: Reactions are considered at standard conditions for qualitative identification. Combustion reactions are typically exothermic. Thermal decomposition of carbonates is known to be endothermic. Concept / Approach: Decomposition of calcium carbonate into quicklime and carbon dioxide consumes heat (endothermic), forming the basis of lime kiln operation. By contrast, oxidation reactions such as forming CO or CO2 and methanation (CO + 3 H2 → CH4 + H2O) are exothermic, releasing heat. Step-by-Step Solution: Identify decomposition vs. oxidation: CaCO3 → CaO + CO2 is decomposition.Recall kiln practice: continuous heat supply needed to sustain calcination.Therefore, CaCO3 → CaO + CO2 is endothermic. Verification / Alternative check: Enthalpy data show positive heat of reaction for calcination, while the combustion and methanation reactions have negative heats (exothermic), aligning with industrial observations (heat release in burners and methanators). Why Other Options Are Wrong: C + ½ O2 → CO: Exothermic partial oxidation. CO + 3 H2 → CH4 + H2O: Methanation is exothermic. CO + ½ O2 → CO2: Exothermic full/partial oxidation step. Common Pitfalls: Assuming all gas-producing reactions are exothermic; calcination is a classic endothermic process despite gas evolution. Final Answer: CaCO3 → CaO + CO2

Question 16

Explosives – composition of traditional gunpowder Classic gunpowder (black powder) is a mechanical mixture of charcoal, sulfur, and which oxidizing salt?

Accepted Answer

Introduction / Context: Gunpowder (black powder) is one of the earliest propellants/explosives. Knowing its composition is basic knowledge in chemical technology and safety. It illustrates the combination of fuel, sulfur sensitizer, and oxidizer in a granular mixture. Given Data / Assumptions: Fuel component: charcoal (carbon). Sensitizer/combustion modifier: sulfur. Oxidizer must supply oxygen rapidly upon ignition. Concept / Approach: The oxidizing component of black powder is potassium nitrate, historically called saltpetre (or salt petre). Upon ignition, KNO3 decomposes to provide oxygen, enabling rapid combustion of carbon and sulfur. The other listed substances are different high-energy materials or fuels not used in classical black powder formulation. Step-by-Step Solution: Identify historical composition: charcoal + sulfur + KNO3.Saltpetre = potassium nitrate; matches the oxidizer requirement.Therefore, the correct component is salt petre. Verification / Alternative check: Standard references cite typical weight ratios (e.g., 75% KNO3, 15% charcoal, 10% sulfur) for black powder, confirming the oxidizer identity. Why Other Options Are Wrong: Glycerene / nitroglycerene / dynamite: Unrelated to black powder; nitroglycerin and dynamite are separate, later high explosives. Common Pitfalls: Mixing up black powder with smokeless powders (nitrocellulose-based) or high explosives; compositions and behaviors are very different. Final Answer: salt petre

Question 17

In industrial chemistry, which of the following substances is an organometallic compound (i.e., an organic molecule containing a direct carbon–metal bond)?

Accepted Answer

Introduction: Organometallic chemistry focuses on compounds with at least one direct bond between a metal and a carbon atom of an organic group. Identifying such compounds is foundational for catalysis, fuel additives history, and materials synthesis. This question tests recognition of a classic organometallic: tetraethyl lead. Given Data / Assumptions: We compare four choices spanning hydrocarbons, alcohols, minerals, and a lead compound. Definition: organometallic = compound with metal–carbon sigma bond(s). No advanced spectroscopy is required; structural knowledge suffices. Concept / Approach: Tetraethyl lead, Pb(C2H5)4, contains lead–carbon bonds. Cumene (isopropylbenzene) and isopropyl alcohol are purely organic with no metal atom. Zeolite is an inorganic aluminosilicate solid without C–metal bonds; it is a catalyst/adsorbent, not an organometallic molecule. Step-by-Step Solution: Identify the metal presence: lead in tetraethyl lead.Check for C–metal bonds: ethyl groups bonded to Pb confirm organometallic nature.Eliminate purely organic (cumene, isopropyl alcohol) and inorganic framework (zeolite). Verification / Alternative check: Historical use of tetraethyl lead as an antiknock gasoline additive relied on its decomposition pathway, characteristic of organometallics undergoing homolysis and delivering lead species to quench radicals. Why Other Options Are Wrong: Cumene: hydrocarbon, no metal. Zeolite: inorganic aluminosilicate lattice, no C–metal bonds. Isopropyl alcohol: oxygenated organic, no metal atom. Common Pitfalls: Confusing “contains a metal” with “organometallic.” Many catalysts (zeolites) contain metals or metal-like centers but lack direct metal–carbon bonds and thus are not organometallics. Final Answer: Tetraethyl lead

Question 18

In the contact process (SO2 → SO3 over V2O5), which condition thermodynamically favours the oxidation of SO2 to SO3?

Accepted Answer

Introduction: Optimizing SO2 oxidation to SO3 is central to sulfuric acid manufacture. Understanding how temperature and pressure shift equilibrium is a direct application of Le Chatelier’s principle and reaction stoichiometry, key thermodynamics for process engineers. Given Data / Assumptions: Reaction: 2 SO2 + O2 → 2 SO3. Exothermic reaction (heat released). Moles decrease (3 gas moles → 2 gas moles). Concept / Approach: For exothermic reactions, lower temperature shifts equilibrium toward products. When total gas moles decrease, higher pressure also drives equilibrium to the product side. Practical operation uses a compromise temperature (about 400–450°C) to maintain good kinetics while not sacrificing equilibrium too much. Step-by-Step Solution: Apply Le Chatelier to heat: exothermic → lower T favours SO3.Apply to moles: fewer moles in products → higher pressure favours SO3.Combine: low T and high P best favour oxidation thermodynamically. Verification / Alternative check: Equilibrium constants decrease with increasing temperature for exothermic reactions. Pressure dependence follows stoichiometric reduction in total gas moles, pushing equilibrium to SO3 at higher pressure. Why Other Options Are Wrong: High temperature options shift equilibrium backward for an exothermic reaction. Low pressure disfavors the side with fewer moles (SO3). Common Pitfalls: Confusing kinetic requirements (need moderate T for rate) with pure equilibrium preference; plant conditions balance both, but the question asks about what “favours” the conversion thermodynamically. Final Answer: Low temperature and high pressure

Question 19

Food chemistry recall: among the common dietary sugars listed, which one is intrinsically the sweetest on a per-mass basis at room temperature?

Accepted Answer

Introduction: Sensory sweetness varies by sugar type due to molecular structure and receptor interaction. Identifying the sweetest common sugar is a frequent exam item in food science and nutrition fundamentals. Given Data / Assumptions: Sugars considered: glucose, fructose, sucrose, lactose. Standard conditions: aqueous solution, ~20–25°C. Relative sweetness scale using sucrose ~100 as reference. Concept / Approach: Fructose (levulose), a ketohexose, typically shows the highest relative sweetness (about 120–170 depending on conditions), exceeding sucrose and glucose. Lactose is the least sweet among the listed disaccharides/monosaccharides. Step-by-Step Solution: Compare typical relative sweetness: fructose > sucrose > glucose > lactose.Account for temperature dependence: fructose remains top under room temperature.Select fructose as the sweetest common dietary sugar. Verification / Alternative check: Food formulation references consistently rank fructose highest in sweetness intensity per gram under ambient conditions. Why Other Options Are Wrong: Sucrose: sweet standard but generally below fructose. Glucose: significantly less sweet than sucrose. Lactose: least sweet of the group. Common Pitfalls: Assuming sucrose is always sweetest because it is the table sugar reference; in fact, fructose often exceeds it in sweetness, especially at lower temperatures. Final Answer: Fructose

Question 20

Thermochemistry check: classify the reaction C (solid) + O2 (gas) → CO2 (gas) in terms of heat effect.

Accepted Answer

Introduction: Combustion reactions are central to energy engineering. Identifying whether a reaction releases or absorbs heat is a basic thermochemistry skill with direct safety and process control implications. Given Data / Assumptions: Reaction: carbon burning completely to carbon dioxide. Standard conditions for enthalpy of formation. No catalysts or intermediates specified. Concept / Approach: The standard enthalpy of formation of CO2 is negative, reflecting heat release when forming stable CO2 from elemental carbon and oxygen. Hence, the reaction is exothermic. Autocatalysis refers to reactions accelerated by one of their own products, which is not implied here. Step-by-Step Solution: Use enthalpy of formation: ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants).ΔH°f(CO2) ≈ −393.5 kJ/mol; ΔH°f(C) = 0; ΔH°f(O2) = 0.Compute: ΔH°rxn ≈ −393.5 kJ/mol ⇒ heat released ⇒ exothermic. Verification / Alternative check: Practical observation: burning carbon (e.g., coal, charcoal) releases heat readily used for power and heating applications. Why Other Options Are Wrong: Endothermic: would require heat input, contrary to data. Autocatalytic: no product-catalyzed pathway indicated. None of these: incorrect because exothermic is correct. Common Pitfalls: Confusing formation of CO with CO2; incomplete combustion to CO still releases heat but less than complete combustion to CO2. Final Answer: Exothermic

Chemical Process Questions