Question 1

Quality issue in urea production: Which compound is the well-known undesirable by-product formed when urea is overheated or held too long, especially during prilling or granulation?

Accepted Answer

Introduction / Context: Urea (46% N) is the most widely used nitrogen fertilizer. During manufacturing, especially in high-temperature hold-ups like melt conveyance, prilling, or granulation, secondary reactions can occur. Biuret formation is a central quality concern because elevated biuret levels can be phytotoxic, affecting seed germination and foliar applications. Given Data / Assumptions: We consider thermal side reactions of urea. The by-product in question is “undesirable” for agronomic quality. Plant stages include synthesis (carbamate), decomposition to urea, then finishing (prills/granules). Concept / Approach: Biuret forms by condensation of two urea molecules with release of ammonia, yielding H2N–CO–NH–CO–NH2. Its concentration in the product rises with higher melt temperature and longer residence time. Fertilizer-grade specifications therefore limit biuret (often below about 1.0–1.5% for general use, tighter for certain crops) and emphasize careful temperature control and residence-time minimization in finishing equipment. Step-by-Step Solution: Recognize thermal pathway: urea → biuret + NH3 under overheating/long hold.Identify agronomic impact: biuret inhibits growth at elevated levels.Thus, among the listed choices, “Biuret” is the characteristic undesirable by-product. Verification / Alternative check: Process and agronomy references consistently flag biuret, not carbonate or carbamate, as the product-quality issue during finishing. Ammonium carbamate and carbon dioxide are intermediates/by-products of synthesis, not the main quality-limiting impurity in finished urea. Why Other Options Are Wrong: Ammonium carbonate/carbamate are upstream species; CO2 is a normal synthesis by-product; cyanuric acid is associated with different condensation/polymerization chemistries and is not the primary fertilizer impurity of concern. Common Pitfalls: Assuming any nitrogenous by-product is equally harmful—biuret is the key specification driver in urea fertilizers. Final Answer: Biuret

Question 2

Equilibrium thinking in Haber–Bosch: Which set of operating conditions favors the maximum equilibrium yield of ammonia (N2 + 3 H2 ⇌ 2 NH3)?

Accepted Answer

Introduction / Context: The Haber–Bosch synthesis of ammonia is governed by equilibrium and kinetics. Because 4 moles of reactants form 2 moles of product, and the reaction is exothermic, Le Chatelier’s principle guides the selection of pressure, temperature, and reactant partial pressures to maximize ammonia equilibrium yield at the reactor outlet. Given Data / Assumptions: Stoichiometry: N2 + 3 H2 ⇌ 2 NH3; Δn(gas) = −2. Reaction is exothermic (heat-releasing). Question targets equilibrium yield, not rate alone. Concept / Approach: Since the reaction reduces gas moles, higher total pressure shifts equilibrium toward ammonia. Because the reaction is exothermic, lower temperature favors products, although kinetics slow at too-low temperatures; industrial practice balances both by using catalysts and moderate temperatures. High reactant partial pressures (i.e., high reactant concentration in the feed) also push equilibrium toward NH3. Therefore, the best combination is high pressure, high reactant concentration, and low temperature. Step-by-Step Solution: Apply Le Chatelier: increase pressure to favor the side with fewer moles (NH3).Lower temperature to favor exothermic product formation.Increase reactant partial pressures to drive the equilibrium toward ammonia.Combine these to select: high P, high reactants, low T. Verification / Alternative check: Plant practice uses elevated pressure (often 100–200 bar historically, lower in modern designs), catalyst beds, interbed cooling, and recycle to approach favorable equilibrium while retaining adequate reaction rates. Why Other Options Are Wrong: Low pressure works against equilibrium; high temperature lowers equilibrium NH3 even if rate improves; low reactant concentration reduces NH3 partial pressure driving force. Common Pitfalls: Confusing rate and equilibrium: very low temperature is favorable to equilibrium but too low harms kinetics—industrial conditions are a compromise, but for equilibrium yield alone, low temperature is preferred. Final Answer: High pressure, high reactants concentration, low temperature

Question 3

Wet-process acidulation: Leaching phosphate rock with strong ________ acid produces phosphoric acid in industrial practice.

Accepted Answer

Introduction / Context: Phosphoric acid, the backbone of phosphate fertilizers, is most commonly made by wet-process acidulation of phosphate rock (apatite). Although sulphuric acid is the dominant industrial reagent, hydrochloric acid can also leach phosphate rock to produce phosphoric acid, with different downstream separation steps. Recognizing that more than one strong acid can be used is important in exam contexts and process awareness. Given Data / Assumptions: Feed is apatite-rich rock phosphate. Strong acids considered: sulphuric (H2SO4) and hydrochloric (HCl). Goal: produce phosphoric acid (H3PO4) in liquid phase. Concept / Approach: In the classic wet process, rock phosphate reacts with sulphuric acid to yield monocalcium phosphate in an acid liquor and gypsum (CaSO4·2H2O) as by-product; the liquor is concentrated to phosphoric acid. Alternatively, hydrochloric acid can leach phosphate to form soluble calcium chloride and phosphate species; after appropriate separations, phosphoric acid is recovered. Hence, both sulphuric and hydrochloric acids can serve as strong-acid leachants to produce phosphoric acid from rock. Step-by-Step Solution: Identify target: H3PO4 via rock leaching.Map common route: H2SO4 wet process.Recognize alternate route: HCl leaching with different by-products and purification needs.Therefore, choose “either (a) or (b)”. Verification / Alternative check: Process literature documents H2SO4-based plants worldwide and HCl-based niche processes where by-product handling (e.g., CaCl2) and impurity profiles justify the route. Why Other Options Are Wrong: “Neither” contradicts established practice; nitrous acid is irrelevant to phosphate rock leaching. Common Pitfalls: Assuming only sulphuric acid is possible; exams may test recognition that HCl is also technically viable, albeit less common industrially. Final Answer: either (a) or (b)

Question 4

Finishing step control in urea plants: Urea prilling should occur just above its melting point with minimal residence time; otherwise, the product quality deteriorates mainly due to ________.

Accepted Answer

Introduction / Context: In urea finishing (prilling or granulation), melt temperature and residence time must be carefully controlled. If the urea melt is held too hot for too long before solidification, thermal condensation reactions proceed, forming biuret—an impurity with agronomic drawbacks. Understanding this linkage helps operators set prilling tower conditions and troubleshoot off-spec material. Given Data / Assumptions: Prilling spray occurs just above urea’s melting point (~132–135 °C), ideally with short hold-up. Overheating or long residence promotes side reactions. We need the primary product-quality risk from such conditions. Concept / Approach: The key side reaction is 2 urea → biuret + NH3. Biuret formation increases with melt temperature and time, and higher biuret percentages can harm sensitive crops (e.g., in foliar feeding). Other defects like shape or stickiness can result from different operating issues (droplet formation, humidity), but the textbook risk tied to overly hot/long prilling is biuret growth. Step-by-Step Solution: Recognize thermal chemistry: prolonged hot melt → condensation to biuret.Relate to prilling control: keep temperature near melting point; minimize residence.Therefore, biuret formation is the principal deterioration mode. Verification / Alternative check: Quality control limits in fertilizer standards specify maximum biuret; operating manuals emphasize minimizing hot-hold times to control biuret content. Why Other Options Are Wrong: Low bulk density, non-sphericity, or stickiness relate to atomization, tower dynamics, and moisture—not specifically to thermal condensation; carbamate carryover is upstream, not a prilling melt-time effect. Common Pitfalls: Confusing product-shape issues with chemical-spec issues; the latter (biuret) is the critical thermal side product. Final Answer: biuret formation

Question 5

Thermochemistry check: The oxidation of ammonia (as in the Ostwald process for nitric acid) is fundamentally ________.

Accepted Answer

Introduction / Context: Oxidation of ammonia to nitric oxide is the first step of the Ostwald process for nitric acid, a cornerstone of nitrogen fertilizer production. The thermochemistry—whether the reaction releases or absorbs heat—controls reactor design, temperature profiles, and heat recovery schemes. Given Data / Assumptions: Main reaction: 4 NH3 + 5 O2 → 4 NO + 6 H2O (with platinum-group catalyst). Industrial operation uses gauze catalysts at high temperatures (∼800–900 °C). We focus on thermal character, not side reactions. Concept / Approach: The oxidation of ammonia is strongly exothermic, liberating substantial heat that sustains high catalyst temperatures after light-off and enables waste-heat recovery (e.g., raising steam). While the process is catalytic (Pt-Rh gauzes), the core thermochemical identity is exothermicity, not endothermic or autocatalytic behavior. Step-by-Step Solution: Consider bond energies: oxidation of a reduced nitrogen species to NO/H2O releases heat.Observe practice: after ignition, the bed remains hot due to the exotherm and gas preheating.Therefore, classify the reaction as exothermic. Verification / Alternative check: Plant heat balances show significant heat recovery from converter effluent; this would be impossible for an endothermic reaction without external energy input. Why Other Options Are Wrong: Endothermic contradicts the known heat release; “non-catalytic” is false—Pt gauze is required; “autocatalytic” is not the standard description; “athermal” is incorrect. Common Pitfalls: Confusing high operating temperature with endothermicity; the bed is hot because the reaction releases heat and the catalyst must withstand it. Final Answer: exothermic

Question 6

Urea synthesis stoichiometry: Industrial urea (a nitrogenous fertilizer) is produced by reacting carbon dioxide with ________.

Accepted Answer

Introduction / Context: Urea, labeled 46-0-0 in fertilizer analysis (N-P2O5-K2O), is produced in massive volumes for agriculture. Its synthesis couples the upstream ammonia plant with carbon dioxide recovered from reforming/shift units, thereby integrating carbon management with nitrogen fixation in a single complex. Given Data / Assumptions: Inputs available on-site: ammonia (from Haber–Bosch) and carbon dioxide (from reforming and CO2 removal). Process steps: carbamate formation followed by dehydration to urea. We identify the direct chemical counterpart to CO2 in urea production. Concept / Approach: In the high-pressure synthesis section, ammonia reacts with carbon dioxide to form ammonium carbamate (NH2COONH4). Under elevated temperature and pressure, this intermediate dehydrates to urea (CO(NH2)2) and water. Unconverted carbamate is recycled after decomposition and recovery. Thus, the immediate reactant pairing for CO2 is ammonia, not nitric acid or other nitrogen oxides. Step-by-Step Solution: CO2 + 2 NH3 → NH2COONH4 (ammonium carbamate).NH2COONH4 → CO(NH2)2 + H2O (dehydration to urea).Recycle unconverted species and recover product via prilling or granulation. Verification / Alternative check: All commercial urea technologies (e.g., Stamicarbon, Saipem/Snamprogetti) implement the same chemical sequence of carbamate formation and dehydration using NH3 and CO2 as reactants. Why Other Options Are Wrong: Nitric acid and nitric oxide are used in nitric acid manufacturing and downstream nitrate fertilizers; ammonium nitrate is itself a fertilizer product formed from NH3 and HNO3, not a urea precursor; hydrogen is a reactant for ammonia synthesis, not urea formation. Common Pitfalls: Confusing “ammonia to nitrates” with “ammonia to urea”; they are separate product lines in a nitrogen complex. Final Answer: ammonia

Question 7

Thermal (electric-furnace) route for elemental phosphorus: When a mixture of phosphate rock, coke (carbon) and silica sand is heated in an electric furnace at high temperature, what principal product is formed?

Accepted Answer

Introduction / Context: The electric-furnace (thermal) process is a classic high-temperature route for producing elemental phosphorus (P4) from phosphate rock. Unlike wet processes that yield phosphoric acid or superphosphate via acidulation, the thermal route relies on carbothermic reduction in the presence of silica, followed by condensation of phosphorus vapor. This question checks whether you can distinguish the thermal reduction product from the acidulation products used in fertiliser manufacture. Given Data / Assumptions: Feed mixture: phosphate rock (fluorapatite or related Ca phosphates), coke (carbon), and silica sand (SiO2). Equipment: electric furnace operating at very high temperature (well above 1200 °C). Objective: identify the principal chemical product of the reduction. Concept / Approach: In the furnace, carbon reduces the phosphate to elemental phosphorus, while silica combines with calcium to form a fusible calcium-silicate slag. Phosphorus is volatilized as P4 and later condensed under controlled conditions. This is a reductive route, not an acid leach; therefore the products are entirely different from orthophosphoric acid or superphosphate obtained by sulphuric or phosphoric acid digestion of rock. Step-by-Step Solution: Recognize the reagents: rock + coke + silica → carbothermic reduction environment.Recall the role of silica: combines with CaO to make slag (ease of tapping).Realize the target: volatilized P4 that is condensed downstream.Hence, the principal product is elemental (white) phosphorus, P4. Verification / Alternative check: Standard furnace reactions are summarized as Ca3(PO4)2 + SiO2 + C → P4 (vapor) + Ca-silicates + CO/CO2. Process flow diagrams show electrothermal furnaces, condenser trains, and slag handling units in place of acid reactors and gypsum filtration used in wet processes. Why Other Options Are Wrong: Orthophosphoric acid: product of wet-process digestion with strong H2SO4, not high-temperature reduction. Ammonium phosphate: made by neutralizing phosphoric acid with ammonia, not in the furnace. Single superphosphate (SSP): formed by direct acidulation of rock with dilute H2SO4. Calcium phosphate glass only: slag does form, but the valuable product is P4, not just glassy calcium silicates. Common Pitfalls: Confusing thermal reduction (elemental P) with acidulation routes (fertilisers or H3PO4). Also, overlooking silica’s role as a flux for CaO can obscure the overall reaction logic. Final Answer: Elemental phosphorus (white phosphorus, P4)

Question 8

Fertiliser vs. non-fertiliser classification: Which of the following is not classified as a standard fertiliser supplying a primary N, P or K plant nutrient?

Accepted Answer

Introduction / Context: Commercial fertilisers are primarily categorized by the primary nutrients they supply: nitrogen (N), phosphorus (P) and potassium (K). Some materials are soil amendments or micronutrient sources rather than primary fertilisers. Distinguishing between these groups helps avoid mislabeling and ensures correct agronomic recommendations. Given Data / Assumptions: Primary fertilisers deliver N, P, or K as their main function. Products like CAN, urea, ammonium sulphate, and anhydrous ammonia are standard nitrogen fertilisers. Ferrous sulphate is mainly a micronutrient/soil amendment supplying iron (Fe), not a primary N, P, or K fertiliser. Concept / Approach: Urea (~46% N), ammonium sulphate (~21% N), calcium ammonium nitrate (~26–28% N), and anhydrous ammonia (~82% N) are straight nitrogen fertilisers. Ferrous sulphate is used to correct iron chlorosis and to adjust soil conditions in turf/horticulture, but it does not supply N, P, or K in significant quantities. Therefore, among the choices, ferrous sulphate is not a standard primary fertiliser. Step-by-Step Solution: Identify which items deliver N, P, or K as principal nutrients.Mark urea, ammonium sulphate, CAN, and anhydrous ammonia as nitrogen fertilisers.Recognize ferrous sulphate as a micronutrient amendment (Fe source), not a primary fertiliser.Select “Ferrous sulphate (iron(II) sulphate)”. Verification / Alternative check: Fertiliser regulations and labels list N–P–K grades for primary fertilisers; ferrous sulphate is sold with Fe percentage and is categorized under micronutrients or soil conditioners, not NPK fertilisers. Why Other Options Are Wrong: CAN, ammonium sulphate, liquid ammonia, and urea are all standard nitrogen fertilisers widely used in field crops. Common Pitfalls: Assuming any nutrient source is a fertiliser in the NPK sense; micronutrient salts correct deficiencies but are not NPK carriers. Final Answer: Ferrous sulphate (iron(II) sulphate)

Question 9

Compare nitrogen analyses: Among common products, which fertiliser typically has the lowest percentage of nitrogen by mass?

Accepted Answer

Introduction / Context: Knowing the nominal nitrogen content of common fertilisers is essential for calculating application rates and transport economics. Some products (e.g., ammonia, urea) are nitrogen-rich, while NP grades such as monoammonium phosphate (MAP) and diammonium phosphate (DAP) deliver phosphorus with modest nitrogen. This item asks you to pick the fertiliser with the lowest nitrogen percentage among typical commercial products. Given Data / Assumptions: Typical analyses used for comparison: anhydrous ammonia ≈ 82% N; urea ≈ 46% N; ammonium sulphate ≈ 21% N; DAP ≈ 18% N and 46% P2O5; MAP ≈ 11–12% N and 52% P2O5. We ignore minor grade variations and moisture for simplicity. Concept / Approach: The lowest nitrogen percentage among the listed mainstream products belongs to MAP because its formulation prioritizes phosphorus (high P2O5) with only one ammonium per phosphate. DAP contains two ammonium groups, raising its nitrogen percentage above MAP. Ammonium sulphate, although a straight N fertiliser, still carries ~21% N—greater than MAP’s ~11–12%. Urea and anhydrous ammonia contain much higher nitrogen percentages than either MAP or DAP. Step-by-Step Solution: List nominal N% values for each option.NH3 ≈ 82% N; urea ≈ 46% N; ammonium sulphate ≈ 21% N; DAP ≈ 18% N; MAP ≈ 11–12% N.Identify the smallest value: MAP.Choose “Monoammonium phosphate (MAP)”. Verification / Alternative check: Standard fertiliser datasheets and agronomic handbooks report MAP as 11–12% N and 52% P2O5, confirming it has the lowest nitrogen among the listed options. Why Other Options Are Wrong: Liquid ammonia: the highest N% (~82%). Urea: a high-N solid (~46%). Ammonium sulphate: ~21% N, still higher than MAP. DAP: ~18% N, higher than MAP’s ~11–12%. Common Pitfalls: Confusing “least nitrogen” with “least total nutrients”; NP products deliver significant phosphorus even when their nitrogen percentage is low. Final Answer: Monoammonium phosphate (MAP)

Question 10

Determinants of fertiliser effectiveness: The field effectiveness of an applied fertiliser depends on soil nature, crop type, and soil pH. It is independent of which factor, if any?

Accepted Answer

Introduction / Context: “Fertiliser effectiveness” means how much of the applied nutrient becomes available for plant uptake and translates into yield. This depends on interactions between the product and the agro-ecosystem. The question asks if there is any listed factor on which effectiveness is independent. In practice, the three cited factors all matter strongly. Given Data / Assumptions: Soil properties include texture, organic matter, cation exchange capacity (CEC), moisture, and biological activity. Crop species and growth stage alter nutrient demand and root architecture. Soil pH controls nutrient solubility (e.g., P fixation), ammonia volatilisation, and microbial transformations. Concept / Approach: Nutrient fate processes—adsorption, precipitation/dissolution, nitrification/denitrification, volatilisation, and leaching—are all governed by soil chemistry and biology. Crop type determines uptake kinetics and critical tissue concentrations. Soil pH shifts equilibria (for example, ammonium/ammonia balance and phosphate availability). Therefore, fertiliser effectiveness is not independent of any of these; it depends on all of them together. Step-by-Step Solution: Evaluate soil nature: affects retention, movement, and bioavailability → critical.Evaluate crop: different species demand different NPK ratios and timing → critical.Evaluate pH: controls solubility/volatilisation and microbial steps → critical.Conclude: none of the listed factors can be ignored; choose “None of these”. Verification / Alternative check: Extension guidelines tailor fertiliser rates to soil tests (pH, P, K, OM) and crop requirements; universal “one-rate-fits-all” recommendations are discouraged. Why Other Options Are Wrong: Each of (a), (b), (c) exerts major influence; claiming independence from any one is incorrect. “Average annual rainfall only”: rainfall matters for leaching but is not the sole determinant. Common Pitfalls: Assuming product label rates apply identically to all soils and crops; ignoring pH-dependent losses (e.g., ammonia volatilisation on high-pH soils after urea application). Final Answer: None of these (it depends on all listed factors)

Question 11

Ammonia synthesis operating temperature: Although equilibrium favours low temperature, industrial ammonia converters run at moderately high temperatures (around 400–500 °C). What is the main reason low temperature is avoided?

Accepted Answer

Introduction / Context: The synthesis of ammonia (N2 + 3 H2 ⇌ 2 NH3) is exothermic and equilibrium-favoured at low temperature and high pressure. However, converters typically operate near 400–500 °C (not “very low” temperature) because the reaction is kinetically slow. This question targets the classic trade-off between thermodynamic equilibrium and reaction kinetics in reactor design and operation. Given Data / Assumptions: Fixed-bed iron-based or promoted catalysts are used. Pressures are high (typically 100–250 bar), but finite. Single-pass conversion is enhanced by interbed cooling and recycle, not by running at arbitrarily low temperature. Concept / Approach: While lowering temperature increases equilibrium NH3 content, it also reduces the rate constant dramatically (Arrhenius relationship). Industrial design finds an optimum temperature window where the rate is high enough to produce practical space-time yields while maintaining acceptable equilibrium levels. At too low a temperature, even with high pressure and good mixing, the space-time yield collapses, making the process uneconomic despite favourable equilibrium. Step-by-Step Solution: Note: Equilibrium prefers low T, high P — but rate falls with T.Apply Arrhenius: k ∝ exp(−Ea/RT); lowering T sharply lowers k.Industrial compromise: operate ~400–500 °C with heat removal and recycle.Therefore, low T is avoided primarily because the reaction rate is too low. Verification / Alternative check: Performance curves show that decreasing bed temperature below the optimal region reduces per-pass production severely; plants use quench or heat-exchange between beds to manage exotherms while staying in the kinetic “sweet spot”. Why Other Options Are Wrong: Very high pressure (b) does improve equilibrium but is constrained by mechanics; it is not specifically “required” by low T alone to make the process feasible. Space velocity (c) is an operational choice; the root cause at low T is slow intrinsic kinetics. Poisoning (e) is not specifically a low-temperature phenomenon; many poisons are problematic regardless of T. Common Pitfalls: Confusing equilibrium improvement with practical productivity; overlooking Arrhenius control of rates in heterogeneous catalysis. Final Answer: The intrinsic rate of reaction becomes very low at low temperature

Question 12

Single superphosphate assay: Approximately what percentage of P2O5 does commercial single superphosphate (SSP) contain?

Accepted Answer

Introduction / Context: Fertiliser grades are commonly expressed in terms of oxide equivalents for phosphorus and potassium. Single superphosphate (SSP) is the product of acidulating phosphate rock with dilute sulphuric acid, yielding monocalcium phosphate plus gypsum. Recognizing its typical P2O5 assay is a basic agronomic and process-knowledge checkpoint. Given Data / Assumptions: SSP is a “single” superphosphate, not triple (TSP). Commercial grades vary slightly by moisture and production route. We use nominal textbook values. Concept / Approach: Typical SSP contains roughly 16% P2O5 (often cited as 16–18% P2O5), along with significant calcium sulphate (gypsum) as a matrix. TSP, by contrast, is around 44–46% P2O5 because it is produced by reacting rock phosphate with phosphoric acid rather than sulphuric acid, concentrating the plant-available phosphate fraction. Therefore, the closest range given for SSP is 15–20% P2O5. Step-by-Step Solution: Identify product: SSP (not TSP).Recall nominal assay: ~16% P2O5.Match against options: 15–20% fits SSP.Select 15–20% as correct. Verification / Alternative check: Manufacturers’ datasheets and fertiliser manuals list SSP grades near 16% P2O5, sometimes slightly higher depending on quality control and granulation. Why Other Options Are Wrong: 30–35% and 65–70% and 85–90%: far too high for SSP; the 44–46% region corresponds to TSP, not SSP. 5–10% is below typical commercial specification. Common Pitfalls: Confusing SSP and TSP assays; assuming all “superphosphates” have similar P2O5 content. Final Answer: 15–20%

Question 13

Role of potash in crops: Which class of fertilisers chiefly promotes development of plant fibres and enhances sugar accumulation in vegetables and fruits?

Accepted Answer

Introduction / Context: Potassium (K) is a primary macronutrient that regulates osmotic balance, enzyme activation, and transport of photosynthates. Its agronomic effects include improved fibre development in crops like cotton and enhanced sugar translocation and accumulation in fruits and vegetables. This question asks which fertiliser class most directly supports those outcomes. Given Data / Assumptions: Nitrogen promotes vegetative growth and chlorophyll formation. Phosphorus supports energy transfer (ATP) and root/flowering development. Potassium regulates stomatal function, carbohydrate transport, and stress tolerance. Concept / Approach: High K availability improves the movement of sugars from leaves to storage tissues, increases turgor for fibre elongation, and supports enzyme systems involved in carbohydrate metabolism. Consequently, potassic fertilisers (e.g., KCl, K2SO4), graded by % K2O equivalent, are recommended to improve fibre quality and sugar content where soil tests indicate K deficiency or crop demand is high. Step-by-Step Solution: Connect desired outcomes (fibre development and sugar content) with K’s physiological roles.Recognize that N primarily drives foliage; P drives energy/root/flowering; K drives transport and quality.Therefore, select potassic fertilisers.Examples: muriate of potash (MOP, KCl) and sulphate of potash (SOP, K2SO4). Verification / Alternative check: Crop guides for sugarcane, fruit trees, and cotton consistently link adequate K nutrition with higher sugar recovery, better fibre strength/length, and improved stress tolerance. Why Other Options Are Wrong: Nitrogenous: may boost vegetative growth but can dilute sugar concentrations if excessive. Phosphatic: important for roots and energy but not the primary lever for fibre/sugar accumulation. None/micronutrients only: contradicts well-established K role as a primary driver. Common Pitfalls: Overapplying N in pursuit of yield while neglecting K for quality parameters; ignoring soil test K and crop-specific K removal rates. Final Answer: Potassic fertilisers (K sources)

Question 14

Wet-process phosphoric acid at high temperature: In strong-acid leaching of phosphate rock, maintaining the reactor above ~100 °C tends to form which undesirable calcium sulphate phases or related by-products?

Accepted Answer

Introduction / Context: In the wet process for orthophosphoric acid, phosphate rock is digested with sulphuric acid, precipitating calcium sulphate and leaving H3PO4 in solution. The operating temperature influences which CaSO4 hydrate forms and thus the filtration/handling characteristics of the cake. This question probes the effect of running the reactor above about 100 °C under strong-acid conditions. Given Data / Assumptions: Dihydrate process: ~70–80 °C yields CaSO4·2H2O (gypsum), easier to filter. Hemihydrate process: ~90–110 °C tends to form CaSO4·0.5H2O (hemihydrate), more difficult to handle unless the plant is designed for it. Above ~100 °C and strong acid, anhydrite (CaSO4) may form, which can be dense and less filterable. Concept / Approach: Operating at elevated temperature shifts calcium sulphate formation away from dihydrate and toward hemihydrate or anhydrite, which often have inferior filtration properties in a flowsheet designed for dihydrate. Although polyphosphoric species (pyro/metaphosphoric acids) can appear at very high concentrations and dehydration severities, the primary processing concern in standard wet-process operation is the solid CaSO4 phase type, which directly affects filtration and overall plant performance. Step-by-Step Solution: Identify target product: orthophosphoric acid liquor with a filterable CaSO4 cake.Note temperature effect: raising T above ~100 °C favors hemihydrate/anhydrite.Recognize these phases can be undesirable if the plant is not configured for them.Select option (a) as the key undesired outcome. Verification / Alternative check: Process descriptions distinguish dihydrate, hemihydrate, and anhydrite routes, each with dedicated filtration strategies. Cross-operating a dihydrate design at hemihydrate/anhydrite temperatures reduces filter performance. Why Other Options Are Wrong: (b) and (c): Polyphosphoric acids become important in dehydration/concentration steps, not as the primary “undesirable” solid phase in wet digestion. (d): Overstates the outcome; the major practical issue is CaSO4 phase. (e): Dihydrate formation is typical below ~80 °C; it is not expected above ~100 °C in strong acid. Common Pitfalls: Assuming higher temperature is always beneficial for kinetics; overlooking the knock-on effect on solid phase and filterability. Final Answer: Calcium sulphate hemihydrate (CaSO4·0.5H2O) and/or anhydrite (CaSO4) crystals

Question 15

Identify the minimum nitrogen content: Among the following common nitrogen fertilisers, which has the lowest percentage of nitrogen by mass?

Accepted Answer

Introduction / Context: Fertiliser selection frequently requires quick comparison of nitrogen analyses to meet target application rates. Among mainstream nitrogen carriers, percentages vary widely from ~21% to ~82% N. This question asks you to choose the product with the lowest nitrogen percentage among widely used solid nitrogen fertilisers. Given Data / Assumptions: Nominal nitrogen analyses: urea ≈ 46% N; ammonium nitrate ≈ 33–35% N; ammonium chloride ≈ 25–26% N; calcium ammonium nitrate ≈ 26–28% N; ammonium sulphate ≈ 21% N. We compare on a mass-percentage basis and ignore moisture variation. Concept / Approach: Ammonium sulphate carries the lowest nitrogen percentage among the listed mainstream solids because a significant portion of its mass is sulphate sulphur. While urea is the highest-N solid, the ammonium salts vary by anion mass contribution, with sulphate being heavier than nitrate or chloride, thereby diluting the nitrogen fraction in ammonium sulphate to about 21% N. Step-by-Step Solution: List the typical N% for each option.Identify the minimum value among these: 21% N for ammonium sulphate.Select ammonium sulphate as the correct answer.Note that liquid ammonia (≈82% N) is not among the options here. Verification / Alternative check: Product datasheets and extension bulletins consistently list ammonium sulphate at about 20–21% N, confirming it as the lowest among these standard solids. Why Other Options Are Wrong: Urea: substantially higher N (≈46%). Ammonium nitrate: ≈33–35% N. Ammonium chloride: ≈25–26% N. Calcium ammonium nitrate: ≈26–28% N. Common Pitfalls: Confusing nitrogen percentage with agronomic efficiency; although lower in N%, ammonium sulphate supplies sulphur, which may be beneficial on S-deficient soils. Final Answer: Ammonium sulphate (~21% N)

Question 16

Why purge the synthesis loop? In an ammonia plant, what is the primary purpose of a small purge from the synthesis gas recycle loop?

Accepted Answer

Introduction / Context: Ammonia synthesis loops recycle unconverted hydrogen and nitrogen at high pressure over a catalyst bed. Inevitably, small amounts of inert gases (argon from air, methane from feed impurities, residual neon/helium traces) enter the loop. Because these species do not react to form NH3, they accumulate. A controlled purge is therefore required to stabilize loop composition and operating pressure drop. Given Data / Assumptions: The loop includes compressor, converter, heat recovery, condenser/separator, and recycle circuitry. Upstream purification removes CO, CO2, H2O, and sulphur; remaining inerts are mainly Ar and CH4. Purge gas is often sent to a recovery unit (e.g., hydrogen recovery by PSA) to minimize losses. Concept / Approach: Without purging, inert gases build up, diluting reactive components, lowering partial pressures of H2 and N2, and reducing per-pass conversion. Accumulation can also increase compressor load and off-design operation. A small steady purge bleeds inerts from the loop, restoring the intended composition. The H2:N2 ratio is maintained by upstream blending and controls, not by purging reactants wholesale. Ammonia vapour is condensed and separated, not purged intentionally with the loop gas. Step-by-Step Solution: Identify unwanted species: Ar, CH4 accumulate because they do not react.Recognize effect: dilution of reactants and performance loss.Apply remedy: implement a controlled purge to cap inert concentration.Therefore, the purpose of the purge is to maintain inert levels within limits. Verification / Alternative check: Standard loop material balances show inert mole fraction approaching an asymptote determined by purge fraction; designers select a purge that balances hydrogen loss against acceptable inert loading. Why Other Options Are Wrong: Removing “poisonous catalysts” is nonsensical; catalysts are fixed in beds, not in the gas stream. Exact 3:1 H2:N2 ratio is controlled by feed blending, not by purge alone. Uncondensed NH3 is minimized by refrigeration/absorption, not purged deliberately. CO2 and CO should be removed upstream; methanator protection does not rely on loop purging. Common Pitfalls: Confusing purge duty with ammonia condensation or with upstream purification; assuming purge is a crude control for feed ratios rather than an inert-management tool. Final Answer: To maintain the concentration of inert gases within acceptable limits

Question 17

Nitrogen fertilizers comparison: Why is urea generally considered a better nitrogen fertilizer than ammonium sulfate for many field applications?

Accepted Answer

Introduction / Context: Farmers often compare common nitrogen (N) fertilizers such as urea and ammonium sulfate to choose the most economical source of plant-available nitrogen. The key differentiator in many agronomic and economic calculations is the percentage of nitrogen by mass, because it governs transport, storage, application rates, and cost per kilogram of nutrient supplied. Given Data / Assumptions: Typical urea analysis ≈ 46% N (by mass). Typical ammonium sulfate analysis ≈ 21% N and ≈ 24% S (as sulfate). Decision is about “better as a nitrogen fertilizer,” not sulfur nutrition or soil chemistry side benefits. Concept / Approach: When the goal is to supply nitrogen alone, a material with higher N concentration usually reduces logistical costs per unit N. Urea's 46% N allows lower application mass for the same amount of N compared with ammonium sulfate at 21% N. This affects freight, storage volume, and spreading time. While ammonium sulfate adds valuable sulfur where needed, for pure N supplementation the higher analysis of urea is the primary advantage and is why urea is widely used globally. Step-by-Step Solution: Identify the metric: %N drives “better” for N delivery.Compare analyses: urea ≈ 46% N vs ammonium sulfate ≈ 21% N.Infer logistics: less product mass needed per kg N with urea.Conclude: higher N content makes urea generally better as an N source. Verification / Alternative check: Cost-per-kg-N calculations routinely favor urea in regions where sulfur is not limiting and volatilization losses are managed with timing, incorporation, or inhibitors. Why Other Options Are Wrong: Cheaper in every market: prices fluctuate; not always true. “Not poisonous”: both require safe handling; this is not the core reason. “Easier to manufacture”: process complexity does not define field “better.” “Contains N and P”: urea contains no phosphorus. Common Pitfalls: Ignoring sulfur needs; in S-deficient soils, ammonium sulfate may be preferable despite lower %N. Also, manage urea to minimize ammonia volatilization. Final Answer: Nitrogen content is higher.

Question 18

Electric furnace phosphorus route: A mixture of phosphate rock with which additional solid(s) is heated in an electric furnace to produce elemental phosphorus?

Accepted Answer

Introduction / Context: Elemental phosphorus (P4) can be manufactured by the thermal (electric furnace) process. This classic route reduces calcium phosphates in phosphate rock using a carbon reductant and a silica flux at very high temperatures to volatilize P4, which is then condensed under water. Knowing the required solid reactants helps understand both the chemistry and the slag formation that removes calcium as a silicate. Given Data / Assumptions: Feed: phosphate rock (chiefly fluorapatite-like calcium phosphates). Furnace: electric, high-temperature (typically > 1400°C). Additives: a carbon source (coke) and a silica source (sand). Concept / Approach: The net reaction reduces phosphate to elemental phosphorus while binding calcium with silica to form a molten calcium silicate slag. Carbon (as coke) provides the reducing environment, generating carbon monoxide. Silica (sand) reacts with calcium oxide to form stable slag, thereby driving phosphorus liberation. Omitting either the reductant or the silica flux prevents efficient phosphorus production or creates handling problems with basic slags. Step-by-Step Solution: Identify roles: coke as reductant; sand (SiO2) as flux.Recognize that both are required alongside phosphate rock.Select the mixture “sand & coke” as correct. Verification / Alternative check: Industrial flowsheets show phosphate rock + silica + coke charged to submerged-arc furnaces; off-gas condenses to white phosphorus; slag tapped as calcium silicate. Why Other Options Are Wrong: Salt & coke: sodium chloride is not used in the standard thermal route. “Rock and coke only”: lacks silica to form calcium silicate slag. “Rock and sand only”: lacks reductant to produce P4. Gypsum & coke: not the conventional flux for P4 production. Common Pitfalls: Confusing the thermal process with “wet-process” phosphoric acid manufacture, which uses sulfuric acid and does not produce elemental phosphorus directly. Final Answer: Sand & coke.

Question 19

Plant physiology and fertilizer burn: Excessive application of chemical fertilizers can cause root shrivelling and plant wilting primarily because the

Accepted Answer

Introduction / Context: “Fertilizer burn” is a common field symptom appearing as scorched leaf margins, wilt, and root injury after over-application of soluble salts. The underlying mechanism is osmotic stress: when salts accumulate around roots, water movement reverses, dehydrating plant tissues and impairing nutrient uptake. Understanding this helps calibrate application rates and irrigation practices to avoid damage. Given Data / Assumptions: Excess readily soluble chemical fertilizers applied to the root zone. Limited immediate leaching or rainfall to dilute salts. Typical plant sap osmotic potential determined by cellular solutes. Concept / Approach: Water moves across semi-permeable root membranes from regions of higher water potential (lower osmotic pressure) to lower water potential (higher osmotic pressure). Heavy fertilizer application increases dissolved ion concentration in the soil solution, raising its osmotic pressure (more negative water potential). If the soil solution's osmotic pressure exceeds that of the plant sap, water flows out of the roots, causing plasmolysis, root shrivelling, and wilting even if the soil is wet. Step-by-Step Solution: Recognize that adding salts raises soil solution osmotic pressure.Compare to plant sap: if soil solution has higher osmotic pressure, water exits root cells.Result: root dehydration and wilting despite available water in soil pores. Verification / Alternative check: Electrical conductivity (EC) of the soil solution correlates with soluble salt load; high EC zones after heavy fertilization are linked to osmotic stress and burn symptoms, reversed by adequate leaching irrigation. Why Other Options Are Wrong: “Less than plant sap”: would favor water uptake, not wilting. “Soil becomes too alkaline/acidic in all cases”: pH shifts can occur but are not the primary, immediate cause of wilting after salt over-application. “Temperature drop”: not the mechanism here. Common Pitfalls: Assuming drought when wilting is due to osmotic stress; irrigation without leaching can worsen salinity at the root surface unless salts are flushed below the root zone. Final Answer: Osmotic pressure of the soil solution becomes higher than that of the plant sap.

Question 20

Role of potassic (K) fertilizers: Which statements best describe agronomic effects of potash on crop development and quality?

Accepted Answer

Introduction / Context: Potassium (K) is a key macronutrient that regulates osmotic balance, enzyme activation, stomatal function, and carbohydrate translocation. While nitrogen often drives early vegetative growth, potassium is strongly associated with stress tolerance, disease resistance, and quality traits such as starch formation in tubers and grains. This question asks you to identify the most accurate agronomic effect among the options provided. Given Data / Assumptions: Balanced fertilization: nitrogen and phosphorus needs are addressed appropriately. Focus on characteristic effects of K nutrition. Crops include potatoes and cereals where starch content and filling are relevant. Concept / Approach: Potassium enhances photosynthate transport from leaves to storage organs, supporting starch synthesis in tubers (potatoes) and grains. It also aids water regulation and enzymatic systems. While K contributes to overall plant vigor at all stages, options that overstate universal effects on early growth independent of N are misleading. Therefore, the most defensible statement is that K promotes development of starches and improves grain/tuber filling, a widely observed outcome of adequate potash fertilization. Step-by-Step Solution: Link K to carbohydrate translocation and starch synthesis pathways.Identify crops where this is most visible: potato tuber bulking and grain filling.Select the option referencing starch and grain development as correct. Verification / Alternative check: Field trials consistently show improved specific gravity of potatoes and better test weight/grain filling in cereals with adequate K, particularly under water stress conditions. Why Other Options Are Wrong: “Only early stage usefulness”: K is beneficial throughout the crop cycle. “Universally accelerates early growth”: early vigor is more directly driven by nitrogen; K supports but does not replace N. “None of these”: incorrect because one statement is accurate. “Replaces the need for nitrogen”: nutrients are not interchangeable; K cannot substitute for N demand. Common Pitfalls: Assuming potash is primarily for “starter” fertilization; in reality, K supports yield formation and quality right through reproductive stages. Final Answer: They help in development and translocation of starches and sugars in potatoes and grains.

Fertiliser Technology Questions