Question 1

Identify the compound: Which structural formula correctly represents tricresyl phosphate (TCP), an aryl phosphate triester?

Accepted Answer

Introduction / Context: Tricresyl phosphate (TCP) is an organophosphate used historically as a plasticizer and flame retardant (with safety considerations). Recognizing its structural representation helps distinguish organic phosphate esters from inorganic phosphates and fertiliser salts often encountered in process and agricultural chemistry contexts. Given Data / Assumptions: TCP is a triaryl phosphate ester derived from cresols (methylphenols) and phosphoric acid. It contains three cresyl (CH3–C6H4–O–) groups bound to phosphorus via oxygen. In shorthand, it is written as (CH3C6H4O)3PO or (C7H7O)3PO, overall formula C21H21O4P. Concept / Approach: A phosphate triester has the general form OP(OR)3, where R is an organic group. For tricresyl phosphate, R is cresyl. In contrast, options listing inorganic salts (ammonium phosphates, potassium phosphate) or minerals (fluorapatite) are unrelated to an organophosphate ester structure and are common fertiliser or mineral forms, not plasticizers. Step-by-Step Solution: Identify that TCP is an aryl phosphate triester → OP(OR)3 pattern.Choose the representation with three cresyloxy groups attached to P.Eliminate mineral/inorganic salt formulas that do not represent an organic ester.Select (CH3C6H4O)3PO as correct shorthand. Verification / Alternative check: Material safety data and chemical catalogs show TCP as C21H21O4P; structural depictions confirm three aryl–O–P linkages around a phosphoryl center. Why Other Options Are Wrong: Fluorapatite (b) is a phosphate mineral used in fertiliser manufacture. (c) and (d) are inorganic ammonium phosphates used as fertilisers. (e) is an inorganic salt, not an organic ester. Common Pitfalls: Assuming any “phosphate” name implies inorganic salt; overlooking organic phosphate esters in industrial chemistry. Final Answer: (CH3C6H4O)3PO (commonly written as (C7H7O)3PO)

Question 2

Effect of pressure on urea formation step: As reactor pressure increases, how does the equilibrium conversion of ammonium carbamate to urea change, all else equal?

Accepted Answer

Introduction / Context: Urea synthesis proceeds via two reactions: rapid formation of ammonium carbamate (favoured by high pressure and excess NH3) and slower dehydration of carbamate to urea and water. Understanding how pressure influences the second, equilibrium-limited step helps explain why process licensors optimise not only pressure but also temperature and NH3/CO2 ratio. Given Data / Assumptions: Key step: NH2COONH4 ⇌ (NH2)2CO + H2O. Number of moles increases from 1 to 2 in the dehydration step. Temperature and composition effects are held constant for this comparison. Concept / Approach: By Le Chatelier’s principle, increasing pressure favours the side with fewer total moles. Since the carbamate-to-urea step increases the number of moles, higher pressure shifts equilibrium back toward ammonium carbamate, decreasing urea formation at fixed temperature and composition. In practice, high pressure is still used overall to facilitate carbamate formation and maintain a liquid phase, but the intrinsic equilibrium of the dehydration step itself is disfavoured by pressure. Step-by-Step Solution: Write the dehydration reaction: 1 mol → 2 mol total.Apply pressure effect: higher pressure favours fewer moles → left side.Conclude: equilibrium conversion to urea decreases as pressure increases.Note: industrial optimisation balances kinetics, phase behaviour, and recycle. Verification / Alternative check: Thermodynamic analyses and design texts discuss opposing effects: pressure aids carbamate formation but disfavors dehydration equilibrium. Process schemes manage this with temperature control and NH3 stripping/recirculation. Why Other Options Are Wrong: Increase/unaltered: contradict equilibrium mole-count logic. “Depends only on biuret”: biuret is a side reaction at high T; not the governing factor for pressure dependence. “Increases then decreases”: not the standard monotonic equilibrium response to pressure for this stoichiometry. Common Pitfalls: Conflating overall plant pressure benefits with the specific equilibrium of the dehydration step; overlooking mole-count effects. Final Answer: It decreases

Question 3

Thermochemistry of Haber–Bosch: The catalytic synthesis of ammonia from nitrogen and hydrogen is best described as which type of reaction?

Accepted Answer

Introduction / Context: The Haber–Bosch process synthesizes ammonia by reacting nitrogen with hydrogen over an iron-based catalyst at high pressure. Its thermochemistry and reversibility govern operating choices such as temperature, pressure, and recycle, which together determine per-pass conversion and energy efficiency. Given Data / Assumptions: Reaction: N2 + 3 H2 ⇌ 2 NH3. Standard enthalpy change is negative (heat released). Equilibrium is strongly dependent on temperature and pressure. Concept / Approach: The reaction is exothermic and reversible. Lower temperatures and higher pressures favour ammonia at equilibrium; however, low temperature slows the reaction rate, so industrial plants compromise at moderate temperatures with high pressures and extensive recycle. The reversibility is evident in the need to separate ammonia (via condensation/absorption) and recycle unconverted gases to drive overall yield upward. Step-by-Step Solution: Classify the reaction enthalpy: negative → exothermic.Classify the reaction type: ⇌ indicates reversibility in practice.Connect thermodynamics to operation: high P/controlled T with condensation and recycle.Therefore select “Exothermic and reversible”. Verification / Alternative check: Plant data show heat removal between catalyst beds and use of waste heat for steam generation—typical for exothermic systems. Why Other Options Are Wrong: Endothermic: contradicts the known heat release. Irreversible: kinetics allow both forward and reverse; equilibrium limits per-pass conversion. Neither exo/endothermic: incorrect for this reaction. Common Pitfalls: Confusing kinetic limitations with irreversibility; ignoring the reverse reaction significance when temperature rises. Final Answer: Exothermic and reversible

Question 4

Catalysis in urea synthesis: In the industrial production of urea from NH3 and CO2, which catalyst is employed in the synthesis reaction?

Accepted Answer

Introduction / Context: Urea manufacture couples two steps: ammonium carbamate formation and its dehydration to urea. Unlike many petrochemical reactions, these proceed without a solid catalyst. Instead, pressure, temperature, and composition management—along with effective stripping and recycle—control performance and yield. Given Data / Assumptions: Feedstocks: ammonia and carbon dioxide. First step (carbamate formation) is rapid and exothermic under high pressure. Second step (dehydration to urea) is slower and equilibrium-limited but still non-catalytic. Concept / Approach: No heterogeneous catalyst is employed in the synthesis reactor. Reaction rates and equilibria are governed by thermodynamic and kinetic factors managed through operating conditions. Downstream, physical separation (stripping, condensation, evaporation) drives overall plant efficiency. Any “catalyst” answers typically refer to other processes (e.g., NH3 oxidation to NO uses Pt–Rh; hydrogenation uses Ni), not to urea formation itself. Step-by-Step Solution: Identify the two-step mechanism and note absence of catalytic surface.Confirm that licensor designs rely on T–P control and phase management.Exclude catalysts known from unrelated processes.Select “No catalyst is used; the synthesis proceeds thermally”. Verification / Alternative check: Process texts and licensor information consistently state non-catalytic synthesis with equipment focused on corrosion-resistant materials and efficient separation rather than catalyst internals. Why Other Options Are Wrong: V2O5, Ni, Pt–Rh: catalysts for other reactions (SO2 oxidation, hydrogenations, NH3 oxidation), not urea synthesis. “Alumina-supported cobalt”: no role in standard urea synthesis. Common Pitfalls: Assuming all high-temperature chemical syntheses use catalysts; carrying over catalyst knowledge from Ostwald/Haber to the urea loop. Final Answer: No catalyst is used; the synthesis proceeds thermally

Question 5

Use-case identification: Which material serves as a nitrogenous fertiliser, acts as a weed killer in onion fields, and helps correct acidic soils when applied properly?

Accepted Answer

Introduction / Context: Some agrochemicals play multiple roles depending on application method and soil conditions. Calcium cyanamide is a classic example: it is a nitrogen source, exhibits selective herbicidal action in certain crops (such as onions), and can reduce soil acidity when managed correctly. Identifying it among common fertilisers prevents misuse and supports integrated nutrient–weed management. Given Data / Assumptions: We consider mainstream field materials with known agronomic uses. Onion fields often benefit from selective weed suppression strategies compatible with crop sensitivity. Soil acidity correction involves materials that raise pH or buffer acidity. Concept / Approach: Calcium cyanamide (CaCN2) hydrolyses in moist soil, gradually releasing plant-available nitrogen while transiently producing alkaline conditions that suppress certain weeds and pathogens. With appropriate timing before planting or at recommended intervals, it can act as a weed killer and soil amendment. Urea, CAN, and ammonium sulphate are valuable nitrogen fertilisers but do not deliver the same combined weed-suppressing and liming action as CaCN2 under typical use patterns. Step-by-Step Solution: Match the three required roles: N fertiliser + weed control + acidity correction.Recognize CaCN2’s hydrolysis behaviour and agronomic recommendations.Exclude urea, CAN, ammonium sulphate, and DAP: primarily nutrient carriers without the same weed control profile.Select calcium cyanamide. Verification / Alternative check: Agronomic guides cite CaCN2 for pre-plant soil conditioning, weed suppression, and supplying nitrogen; safety and timing guidelines are emphasised to protect crop seedlings. Why Other Options Are Wrong: Urea/CAN/AS/DAP: effective nutrients but not standard weed killers nor liming agents. Common Pitfalls: Applying CaCN2 too close to seeding/transplanting; underestimating pH and timing effects critical to crop safety. Final Answer: Calcium cyanamide

Question 6

Feedstocks for nitric acid (Ostwald process): Which raw materials are required to manufacture nitric acid at industrial scale?

Accepted Answer

Introduction / Context: Nitric acid is produced by the Ostwald process, which begins with the catalytic oxidation of ammonia to nitric oxide, followed by further oxidation and absorption in water to yield aqueous HNO3. Recognising the true raw materials clarifies mass balances and ties ammonia production to downstream fertiliser manufacture (e.g., ammonium nitrate, nitrates, and nitrophosphates). Given Data / Assumptions: Ammonia feed is typically anhydrous and vaporised. Air provides oxygen for oxidation reactions and diluent for temperature control on the catalyst. Water is required to absorb nitrogen oxides to form nitric acid. Concept / Approach: Process sequence: NH3 + O2 (over Pt–Rh) → NO; NO + O2 → NO2; absorption of NO2 and related species in water → HNO3 with NOx recycle in the absorber. Thus, the essential raw materials are anhydrous ammonia, air, and water. Hydrogen peroxide is not a feedstock for the Ostwald process; “wet ammonia only” without oxidation cannot produce nitric acid; nitrous oxide is not used as the principal oxidised nitrogen feed for HNO3 manufacture. Step-by-Step Solution: List the required stages and corresponding inputs.Associate NH3 and air with oxidation steps; associate water with absorption.Eliminate distractors unrelated to the established Ostwald flowsheet.Select “Anhydrous ammonia, air and water”. Verification / Alternative check: Standard flowsheets show ammonia vaporizer, air preheater, Pt–Rh converter, NOx cooler/oxidiser, and an absorption tower with water feed—confirming the three raw materials. Why Other Options Are Wrong: Hydrogen peroxide route (a): not industrial for nitric acid. Ammonia and air only (b): lacks the absorber feed (water). Wet ammonia only (d): no oxidation means no HNO3. N2O + steam (e): not the Ostwald route. Common Pitfalls: Omitting water because its role seems “downstream”; forgetting that absorption is integral to acid formation. Final Answer: Anhydrous ammonia, air and water

Question 7

Industrial nitric acid (HNO3): Concentrated nitric acid often appears pale to light yellow in color. This coloration is primarily due to the presence of which dissolved nitrogen oxide species?

Accepted Answer

Introduction / Context: Nitric acid used in laboratories and industry may appear water-clear when very pure, but it frequently develops a pale to light yellow coloration during storage. This visible tint is a classic indicator of dissolved nitrogen oxides formed by slow decomposition or contamination. Recognizing the responsible species is useful in quality control and safety handling of oxidizing acids. Given Data / Assumptions: Chemical: HNO3 (typically 65–70% w/w commercial grade). Observed feature: pale/light yellow color in the liquid phase. Environment: storage in light/heat may accelerate decomposition. Concept / Approach: Concentrated HNO3 slowly decomposes to nitrogen dioxide (NO2), oxygen, and water. NO2 is brownish-red as a gas but imparts a yellow to amber coloration when dissolved in nitric acid, even at modest concentrations. Other nitrogen compounds listed (NO, NH3, N2O, hydrazinium species) either do not persist in strongly oxidizing acidic media or do not produce the characteristic yellow hue under normal storage conditions. Step-by-Step Solution: Recognize slow decomposition: 4 HNO3 → 4 NO2 + 2 H2O + O2 (overall representation).Identify chromophore: dissolved NO2 gives yellow to brown tint depending on level.Exclude alternatives: NH3 is neutralized by strong acid; NO is unstable and colorless; N2O is colorless; hydrazinium species are not stable in oxidizing nitric acid.Conclude NO2 is responsible for the yellow coloration. Verification / Alternative check: Freshly distilled, fuming nitric acid is nearly colorless but rapidly yellows on exposure to light/air as NO2 dissolves; degassing under vacuum or cooling can reduce the tint, supporting the NO2 explanation. Why Other Options Are Wrong: NO: colorless and not persistent in strong oxidizer. NH3: protonated/consumed; does not cause yellow color. N2H5 species: incompatible with strong oxidizing acid. N2O: colorless, low solubility; no yellow tint. Common Pitfalls: Confusing the brown gas above warm acid (NO2) with the liquid's pale tint; both arise from the same oxide but differ in appearance due to phase and concentration. Final Answer: NO2 (nitrogen dioxide).

Question 8

Fertilizer grade notation “5–10–5”: In the standard N–P–K labeling system for fertilizers, what do the numbers 5–10–5 specifically indicate about nutrient analysis by mass percent?

Accepted Answer

Introduction / Context: Bag labels such as 5–10–5, 10–26–26, or 20–20–0 follow a global convention for reporting fertilizer analysis. Understanding this convention ensures correct nutrient budgeting, pricing per unit nutrient, and proper field application rates across crops and soils. Given Data / Assumptions: The sequence is nitrogen (N) – available phosphate (as P2O5) – soluble potash (as K2O). Values are weight percentages of the bag contents. “Available phosphate” and “soluble potash” are expressed as oxide-equivalents by long-standing industry convention. Concept / Approach: The first number is the mass percent of elemental nitrogen. The second and third are not elemental P or K but their oxide forms P2O5 and K2O as defined in fertilizer regulations. A 5–10–5 grade therefore guarantees at least 5% N, 10% P2O5, and 5% K2O by mass. The remainder of the product comprises other nutrients, conditioners, or carrier materials depending on formulation. Step-by-Step Solution: Read the label convention: N – P2O5 – K2O.Substitute the values: 5–10–5 → 5% N, 10% P2O5, 5% K2O.Confirm these are mass percentages, not mole fraction nor elemental P/K.Select the option that matches this interpretation exactly. Verification / Alternative check: Regulatory definitions and fertilizer standards worldwide (e.g., AOAC/FAO) align with the oxide-equivalent convention, ensuring consistent marketplace labeling. Why Other Options Are Wrong: “5–10% active nutrients overall”: misreads the notation; each number is a specific nutrient percentage. “None of these”: incorrect because option (a) is correct. “N2 as a gas”: nitrogen is not reported as molecular nitrogen; the analysis is based on total N content. Common Pitfalls: Confusing P2O5 and K2O with elemental P and K; conversion factors are needed if elemental values are required for scientific calculations. Final Answer: 5% N, 10% P2O5, and 5% K2O.

Question 9

Steam reforming of naphtha for ammonia synthesis gas: Thermochemical nature of the primary reforming step (with nickel catalyst and steam) is best described as

Accepted Answer

Introduction / Context: Ammonia plants generate hydrogen by reforming hydrocarbons. When naphtha is the feed, it is vaporized, mixed with steam, and passed over a nickel-based catalyst in heated tubes. Recognizing that this stage consumes heat is critical for furnace design, energy integration, and catalyst life management. Given Data / Assumptions: Feed: light naphtha hydrocarbons (approximate paraffinic mix). Reactor: tubular, fired primary reformer with Ni/Al2O3 catalyst. Process: hydrocarbons + H2O → CO + H2 (with shift downstream). Concept / Approach: Steam reforming breaks C–C and C–H bonds using steam, forming H2 and CO/CO2. The net chemistry is strongly endothermic, requiring continuous high-temperature heat input from a dedicated furnace. Catalysis is essential to achieve practical rates and selectivity; without Ni catalysts, the process would be too slow and would coke rapidly. Subsequent secondary reforming/shift steps have different heat effects, but the primary reforming step remains endothermic and catalytic. Step-by-Step Solution: State the representative paraffin reaction: CnH2n+2 + n H2O → n CO + (2n+1) H2 (endothermic).Note operating reality: tubes heated externally to maintain high wall/bed temperatures.Identify catalyst role: nickel promotes reforming while suppressing coking.Select “Endothermic (heat absorbed) and catalytic.” Verification / Alternative check: Energy balances show large radiant duty to sustain reformer outlet temperatures; flue gas heat recovery confirms substantial heat input. Why Other Options Are Wrong: Autocatalytic/non-catalytic: contradicts established Ni-catalyzed practice. Exothermic: opposite of observed heat requirement. Athermal/self-sustaining: not physically accurate for steam reforming. Common Pitfalls: Confusing steam reforming with partial oxidation (exothermic); many modern plants use autothermal reforming combining both, but the pure steam reformer is endothermic. Final Answer: Endothermic (heat absorbed) and catalytic.

Fertiliser Technology Questions