Question 1

Ammonium nitrate production routes: The vapor-phase neutralization of ammonia with nitric acid to form ammonium nitrate is commonly referred to as which named process?

Accepted Answer

Introduction: Ammonium nitrate (AN) is produced by reacting ammonia with nitric acid. Industrially, this neutralization can be carried out in liquid or vapor phases. Named processes help distinguish operating regimes and equipment designs used in fertilizer manufacture. Given Data / Assumptions: Reaction: NH3 + HNO3 → NH4NO3 (highly exothermic). Focus: vapor-phase neutralization terminology. Application: prill/granule production for fertilizers. Concept / Approach: The vapor-phase route, historically known as the Stengel process, feeds gaseous ammonia and nitric acid vapor (or mist) to achieve rapid neutralization and produce AN with controlled moisture content before concentration and finishing. Other names in the list denote unrelated processes: Haber for ammonia synthesis, Le Chatelier for equilibrium principle (not a process), DuPont as a company, and Ostwald for nitric acid production from ammonia oxidation—not AN neutralization. Step-by-Step Solution: Identify the target operation: vapor-phase NH3–HNO3 neutralization.Match to the correct named route: Stengel process.Exclude names associated with other chemistries (Haber, Ostwald, etc.). Verification / Alternative check: Fertilizer technology references distinguish liquid neutralizers from the Stengel vapor-phase method, emphasizing equipment differences and control of off-gases and moisture. Why Other Options Are Wrong: Haber: Ammonia synthesis, not AN production. Le-Chatelier's: A principle, not a named AN process. DuPont's: A company; not a standard process name here. Ostwald: Nitric acid production, not the neutralization step. Common Pitfalls: Conflating the nitric acid (Ostwald) and ammonia (Haber) upstream processes with the downstream neutralization step that actually makes ammonium nitrate. Final Answer: Stengel

Question 2

Phosphorus chemistry and terminology: Select the WRONG statement regarding phosphoric acid nomenclature, phosphorus allotropes, and industrial conversions.

Accepted Answer

Introduction: Phosphorus occurs in multiple allotropes, primarily white (yellow) and red, which differ in structure and reactivity. Industrial phosphoric acid production routes and trade names (e.g., “green acid”) also appear in fertilizer technology. This question checks accurate recognition of these facts and flags a common misconception. Given Data / Assumptions: Wet-process phosphoric acid contains impurities and is sometimes colloquially called green acid. Red phosphorus is polymeric and relatively stable. White phosphorus is molecular (P4), very reactive, and air-sensitive. Concept / Approach: Among the allotropes, white phosphorus is substantially more reactive and toxic, readily oxidizing and even inflaming on exposure to air; therefore it is stored under water. Red phosphorus is less reactive due to its extended polymeric network, which lowers the driving force for oxidation compared with discrete P4 units. Conversion from red to white is feasible by volatilizing phosphorus and condensing to obtain the molecular allotrope. Step-by-Step Solution: Evaluate each statement for correctness against known properties.Identify that calling red phosphorus “the most reactive” contradicts the established order (white > red).Confirm the remaining statements align with standard industrial and safety knowledge. Verification / Alternative check: Textbook comparisons of allotropes consistently show white phosphorus as the most reactive and dangerous form; wet-process phosphoric acid is indeed impure and sometimes termed green acid. Why Other Options Are Wrong: (a) True for impure wet-process phosphoric acid. (b) True: polymeric structure reduces reactivity. (d) True: conversion via vaporisation/condensation. (e) True: storage under water is standard practice. Common Pitfalls: Assuming “red” implies higher reactivity; the opposite holds in phosphorus chemistry due to structural differences. Final Answer: Red phosphorus is the most reactive allotropic form of phosphorus.

Question 3

Safety in fertilizer finishing: Why are ammonium nitrate prills or granules commonly coated with limestone powder or similar inert materials?

Accepted Answer

Introduction: Ammonium nitrate (AN) is a powerful oxidizer and, under certain conditions, can detonate, especially when contaminated or confined. Finishing steps such as coating with limestone (calcium carbonate) improve the product’s safety and handling characteristics in agriculture. Given Data / Assumptions: AN is hygroscopic and prone to caking. Thermal and impact sensitivity are significant safety concerns. Limestone coating is inert and compatible with fertilizer use. Concept / Approach: Coating AN with finely divided limestone dilutes the reactive surface, reduces dust, decreases hygroscopicity, and lowers the risk of transition to explosive behavior by absorbing heat and impurities on contact. The coating also helps prevent prill coalescence and reduces sensitivity to shock and friction, improving storage safety. Step-by-Step Solution: Recognize AN’s inherent oxidizing hazard and sensitivity.Note that an inert mineral coating reduces risk and improves stability.Select the purpose most aligned with safety: avoiding the risk of explosion. Verification / Alternative check: Safety codes and fertilizer practice documents recommend inert coatings and controlled additives to render AN safer (e.g., CAN grades contain limestone/dolomite to moderate sensitivity). Why Other Options Are Wrong: Increasing nitrogen content: Coating dilutes N fraction slightly; it does not increase it. Cutting production cost: Coating adds a step and material; cost is not the driver. Adding extra nutrient: Limestone supplies Ca but the intention here is safety, not primary nutrition. Cosmetic color: Incidental; the core driver is safety. Common Pitfalls: Assuming coatings are only for caking control; while that is helpful, the principal motivation is risk mitigation for storage and transport. Final Answer: avoid the risk of explosion.

Question 4

Ammonia synthesis catalyst (Haber–Bosch): Which catalyst material is used in finely divided, promoted form for the industrial manufacture of ammonia from nitrogen and hydrogen?

Accepted Answer

Introduction: The Haber–Bosch process synthesizes ammonia (NH3) by reacting nitrogen with hydrogen over a heterogeneous catalyst at elevated temperatures and pressures. Catalyst choice dictates activity, selectivity, and operating conditions. Given Data / Assumptions: Reaction: N2 + 3 H2 ⇌ 2 NH3. Typical conditions: 350–550°C, 100–250 bar. Need a robust, economical, and active catalyst. Concept / Approach: Promoted iron (wustite-derived) catalysts, often containing potassium, alumina, calcium, and other promoters, are standard in ammonia plants due to cost-effectiveness and proven longevity. While ruthenium-based catalysts are more active per site, they are used in specialized low-pressure designs. Alternatives listed serve other processes: nickel for methanation/hydrogenation, vanadium pentoxide for SO2 oxidation, alumina as a support, copper chromite for hydrogenations. Step-by-Step Solution: Match process to catalyst: Haber–Bosch → promoted iron.Eliminate catalysts associated with other reactions.Select “iron” as the correct answer. Verification / Alternative check: Ammonia plant licensors and textbooks consistently specify magnetite/wustite-derived iron catalysts with promoters as the workhorse for NH3 synthesis loops. Why Other Options Are Wrong: Nickel: Methanation/hydrogenation, not primary NH3 synthesis catalyst. V2O5: Catalyzes SO2 → SO3 in sulfuric acid plants. Alumina: Common support/promoter, not the active metal. Copper chromite: Hydrogenation and dehydrogenation, not NH3 synthesis. Common Pitfalls: Confusing supports/promoters with the active catalytic phase; in ammonia synthesis, elemental iron provides the active sites for N2 dissociation. Final Answer: iron

Question 5

CAN fertilizer agronomy (calcium ammonium nitrate): In calcium ammonium nitrate, which statement correctly describes the plant availability (speed of action) of the nitrogen forms present?

Accepted Answer

Introduction: Calcium ammonium nitrate (CAN) is a granular fertilizer containing a mix of nitrate nitrogen (NO3−) and ammoniacal nitrogen (NH4+), often with a liming component (calcium carbonate/dolomite). Understanding the relative speed with which plants can absorb these forms informs timing and placement decisions. Given Data / Assumptions: CAN typically comprises roughly half nitrate and half ammoniacal nitrogen (formulations vary). Soil conditions are typical agricultural field conditions. Focus is on comparative speed of plant uptake. Concept / Approach: Nitrate nitrogen is immediately available and mobile in soil water, making it fast-acting for crop uptake. Ammoniacal nitrogen is retained on cation exchange sites and generally must nitrify (biological oxidation to nitrate) before rapid plant uptake, making it slower-acting relative to nitrate under many field conditions. Step-by-Step Solution: Identify nitrogen forms in CAN: nitrate and ammonium.Recall mobility and uptake: nitrate is readily absorbed and moves with soil moisture.Conclude: nitrate nitrogen is quick acting compared with ammoniacal nitrogen. Verification / Alternative check: Agronomy references consistently describe nitrate as immediately available and ammonium as slower due to nitrification dynamics, particularly in cooler or less aerated soils. Why Other Options Are Wrong: Ammoniacal quick acting or nitrate slower: Opposite to observed field behavior. None/equally slow: Misrepresents the distinct kinetics. Common Pitfalls: Overapplication leading to nitrate leaching; while quick acting, nitrate is also more prone to losses, so split applications are often recommended. Final Answer: nitrate nitrogen is quick acting

Question 6

Allotropy of phosphorus: Conversion of yellow (white) phosphorus to red phosphorus is carried out by heating in covered retorts at what temperature range (°C) in the absence of air?

Accepted Answer

Introduction / Context: Phosphorus exists in several allotropes—white (yellow), red, and black—each with distinct structures and properties. Converting the highly reactive, toxic white form to the safer red form is an industrially important step for matches, flame retardants, and reagents. This conversion is a solid-state rearrangement triggered by controlled heating without oxygen. Given Data / Assumptions: Feed: white (yellow) phosphorus. Operation performed in covered retorts, i.e., an oxygen-free (inert) environment. Objective: obtain red phosphorus safely. Concept / Approach: White phosphorus consists of molecular P4 units with high ring strain, making it pyrophoric. Heating at moderate temperatures allows atoms to re-polymerize into a more stable, polymeric red phosphorus network. Oxygen must be excluded to prevent combustion and oxidation to phosphorus pentoxide. Step-by-Step Solution: Identify the typical industrial range for the white→red transformation.Safe and effective conversion occurs around 250–400 °C under inert/oxygen-free conditions.Higher temperatures risk sublimation or different chemistry; lower temperatures are too slow.Hence, select 250–400 °C. Verification / Alternative check: Materials processing texts and MSDS data list 250–400 °C as the practical window for converting white to red phosphorus in sealed vessels, often with inert gas sweeping to remove traces of white phosphorus vapor. Why Other Options Are Wrong: 50–80 °C: far too low; white phosphorus remains unchanged. 800–900 °C and 1000–1200 °C: unnecessarily high and potentially hazardous; may lead to volatilization or undesired products. Common Pitfalls: Allowing oxygen ingress; even traces can lead to fire. Overheating beyond recommended range, increasing pressure and hazard. Final Answer: 250–400

Question 7

Phosphoric acid manufacturing routes: Compared with the wet process, which statement correctly characterizes the electric (thermal) process for manufacturing phosphoric acid?

Accepted Answer

Introduction / Context: Two major industrial routes produce phosphoric acid: the wet process (acidulation of phosphate rock with sulphuric acid) and the electric/thermal process (production of elemental phosphorus in an electric furnace, oxidation to P2O5, and hydration to H3PO4). Understanding key differences guides process selection for fertilizer vs. high-purity acid markets. Given Data / Assumptions: Wet process yields ordinary (merchant-grade) acid with gypsum byproduct. Thermal process yields high-purity, stronger acid and is capital/energy intensive. Single best statement is required. Concept / Approach: The electric (thermal) route is less common because it consumes more power and requires high-grade rock and coke/silica in a furnace. It does not generate gypsum; that byproduct arises in the wet process. The thermal route directly makes stronger, purer acid preferred for specialty uses (food, electronics), whereas wet process acid is often concentrated and purified for fertilizers. Step-by-Step Solution: Identify attributes: less frequent industrial use; high-purity product; no gypsum.Match to options: only option (a) captures these facts succinctly.Eliminate the remaining distractors accordingly. Verification / Alternative check: Process flow diagrams and industry references distinguish wet-process acid (with CaSO4 byproduct) from thermal acid (no gypsum, higher purity), confirming the selection. Why Other Options Are Wrong: (b) Opposite of reality; thermal route prefers higher-grade rock. (c) Gypsum is a wet-process byproduct, not from the electric route. (d) Thermal route yields strong acid; wet is initially weaker/dilute. Common Pitfalls: Confusing wet-process (gypsum) with thermal (no gypsum). Assuming frequent use; thermal is niche due to cost. Final Answer: It is used less frequently and produces strong acid without a gypsum byproduct.

Question 8

Fertilizer fundamentals: Pick the single incorrect statement about common fertilizers and plant nutrition.

Accepted Answer

Introduction / Context: Fertilizers supply essential nutrients—nitrogen (N), phosphorus (P), and potassium (K)—each with distinct physiological roles. This question checks common agronomic knowledge and clarifies misconceptions about nutrient functions and product terminology. Given Data / Assumptions: Straight vs. complex fertilizers follow standard definitions. General horticultural guidance on nutrient roles is considered. Weed-suppressing behavior of calcium cyanamide is recognized in certain contexts. Concept / Approach: Phosphorus is primarily associated with root development, energy transfer (ATP), flowering, and early vigor. Stem strength and lodging resistance are far more strongly linked to potassium (K), which enhances turgor, cell wall strength, and overall plant robustness. Hence, stating that phosphorus is the primary nutrient for stem strengthening and branching is inaccurate. Step-by-Step Solution: Evaluate each statement against agronomic roles and definitions.(a) True: water solubility of nitrogenous fertilizers varies (e.g., urea vs. calcium cyanamide).(b) True: 'straight' means a single primary nutrient (e.g., urea for N).(c) True: calcium cyanamide can suppress weeds and act as a soil conditioner.(d) Incorrect: P is not primarily the 'stem-strength' nutrient; K is. Verification / Alternative check: Crop nutrition guides consistently attribute stem strength and lodging resistance to adequate potassium, while phosphorus is emphasized for roots and reproductive development. Why Other Options Are Wrong: They are factually correct within standard agronomy references. Common Pitfalls: Overgeneralizing nutrient effects; multiple nutrients interact. Assuming all N fertilizers dissolve equally quickly in water. Final Answer: Phosphorus nutrient primarily strengthens stems and increases branching.

Question 9

Hydrogen for nitrogenous fertilizers: In steam reforming of naphtha to produce hydrogen, what is the usual steam-to-carbon (S/C) molar ratio maintained in the reformer feed?

Accepted Answer

Introduction / Context: Hydrogen is a key feedstock for ammonia synthesis in the nitrogenous fertilizer industry. Steam reforming converts hydrocarbons and steam into synthesis gas. The steam-to-carbon (S/C) ratio is a critical operating parameter affecting coke formation, equilibrium, and downstream shift conversion. Given Data / Assumptions: Feedstock: naphtha (heavier than natural gas). Catalytic steam reforming process with downstream shift. Usual industrial practice, not extreme coking or water-limited cases. Concept / Approach: Adequate steam suppresses coking on the reforming catalyst and drives equilibria toward hydrogen and carbon oxides. For light feeds (natural gas), S/C around 3 is typical. For heavier feeds like naphtha, a slightly higher S/C—about 3.5—is commonly used to mitigate coking while maintaining energy efficiency. Step-by-Step Solution: Recognize heavier feed → higher coking tendency.Choose S/C to balance coke suppression and heat duty → ~3.5.Match to provided options: 3.5 : 1. Verification / Alternative check: Process design manuals list S/C ≈ 3.0 for methane and ≈ 3.5 for naphtha, consistent with field operation to protect nickel catalysts. Why Other Options Are Wrong: 1.5 : 1 risks severe coking; 10 : 1 or 15 : 1 are unnecessarily high, penalizing energy and equipment size. Common Pitfalls: Running too low S/C to save steam, which shortens catalyst life. Ignoring feed quality; heavier or sulfur-laden feeds may need higher S/C or pre-treatment. Final Answer: 3.5 : 1

Question 10

Field application in the tropics: Why is liquid anhydrous ammonia generally not applied directly as a fertilizer in hot tropical countries such as India?

Accepted Answer

Introduction / Context: Anhydrous ammonia (NH3) is widely used as a high-analysis nitrogen fertilizer in temperate regions by injecting it into soil. However, its direct application is uncommon in hot tropical climates due to handling and loss concerns tied to its physical properties. Given Data / Assumptions: High ambient temperatures typical of tropical regions. Surface or shallow application (spraying) under field conditions. No specialized injection equipment ensuring deep soil sealing. Concept / Approach: Ammonia boils at around −33 °C at 1 atm. In hot climates, any unsealed application leads to immediate volatilization, causing large nitrogen losses and safety hazards. Effective use requires pressurized tanks and deep injection with soil sealing—logistically complex for many smallholder systems. Step-by-Step Solution: Recall NH3 volatility at ambient tropical temperatures.Understand need for sub-surface injection and sealing to prevent loss.Recognize practical limitations in tropical agriculture; thus rapid evaporation is the key reason.Select option (c). Verification / Alternative check: Extension guidelines recommend alternatives like urea or ammonium sulphate for broadcast or top-dress in hot regions, reserving NH3 for mechanized systems with injection rigs. Why Other Options Are Wrong: NH3 has very high nitrogen content (~82% N), so (a) is false. Cost and availability vary, but evaporation loss is the primary technical barrier in this context. Common Pitfalls: Applying NH3 without proper sealing, leading to crop injury and N loss. Confusing aqua ammonia solutions with anhydrous ammonia practices. Final Answer: It rapidly evaporates upon application/spraying under hot conditions.

Question 11

Fertiliser analysis convention: Potassic fertilisers are graded and labeled on the basis of their equivalent potassium oxide content. Which expression fills the blank correctly?

Accepted Answer

Introduction / Context: Fertiliser labels worldwide commonly use oxide-equivalent notation: P as P2O5 and K as K2O. This historical convention persists for consistency across grades and origins, even though plants take up elemental forms (e.g., K+). The question asks which basis is used to grade potassic fertilisers. Given Data / Assumptions: Standard commercial labeling practices are assumed. Products may be KCl, K2SO4, or other potassium salts. Label lists K content as % K2O equivalent. Concept / Approach: Regardless of the chemical form (chloride, sulphate, nitrate), potassium content is converted to the oxide-equivalent K2O for labeling. Thus, “MOP 60% K2O” denotes muriate of potash with about 60% K2O equivalent. The convention allows comparison across products with different anions (Cl−, SO4^2−) by referencing a common basis. Step-by-Step Solution: Identify labeling convention: K as % K2O equivalent.Cross-check common commercial names (MOP, SOP) and their K2O grades.Recognize that elemental K or specific salts are not used directly in the label percentage.Select K2O as the correct fill-in. Verification / Alternative check: Fertiliser standards and trade literature list K grades as % K2O; similarly P grades are listed as % P2O5. Why Other Options Are Wrong: KCl/K2SO4/KNO3: real products but not the unit of grading. Elemental K only: not the standard label basis. Common Pitfalls: Confusing product identity with label basis; the oxide-equivalent system is a convention, not a chemical requirement. Final Answer: K2O

Question 12

Nitrogen fertiliser valuation: The “fertiliser value” of a nitrogenous product is expressed in terms of its percentage of which chemical quantity?

Accepted Answer

Introduction / Context: Nitrogen fertilisers are purchased and applied on the basis of how much nitrogen they supply to crops. Labels therefore report the percentage of elemental nitrogen, irrespective of whether the product is urea, ammonium nitrate, ammonium sulphate, or other forms. This question clarifies that “fertiliser value” for nitrogen means % N, not % N2, NO2−, or HNO3. Given Data / Assumptions: Standard agronomic labeling is assumed. Products release plant-available nitrogen in ammonium or nitrate form after soil processes. We seek the unit used on the label, not the chemical pathway in soil. Concept / Approach: Fertiliser value for N is expressed as mass percent of elemental nitrogen, % N. For example, urea is ~46% N, ammonium nitrate ~33–35% N, ammonium sulphate ~21% N. The gas N2 is not applied directly to soil, nitrite is an intermediate and rarely present in fertilisers, and nitric acid is a manufacturing reagent rather than the delivered nutrient basis. Step-by-Step Solution: Identify the industry standard unit: % N (elemental nitrogen).Relate typical grades: urea 46% N, AN ~34% N, AS ~21% N.Conclude that label value refers to elemental nitrogen content.Select “Elemental nitrogen (N) content”. Verification / Alternative check: National fertiliser control orders and international standards specify nitrogen content as % N on a mass basis for product certification and trade. Why Other Options Are Wrong: N2: atmospheric gas, not the label basis. NO2− and HNO3: not used for value expression. NH4NO3 equivalent: not a universal unit for nitrogenous products. Common Pitfalls: Confusing oxide-equivalent labeling used for P and K with the direct elemental % labeling used for N. Final Answer: Elemental nitrogen (N) content

Question 13

Fertilisers and phosphates—identify the wrong statement: Choose the option that is factually incorrect under standard industrial practice and agronomy for India.

Accepted Answer

Introduction / Context: This item tests practical knowledge that bridges chemistry and field agronomy. Some statements describe core process thermochemistry, while others concern fertiliser application suitability in Indian conditions. You must select the incorrect statement among generally accepted facts. Given Data / Assumptions: Industrial urea is produced via ammonium carbamate; the carbamate-to-urea step is endothermic. Wet-process phosphoric acid coproduces gypsum. Single superphosphate (SSP) results from reacting rock phosphate with dilute sulphuric acid. Indian agriculture spans hot, humid regions with significant volatilisation risks. Concept / Approach: Direct soil injection of anhydrous (liquid) ammonia is common in temperate countries with suitable equipment and soils. In tropical climates like much of India, high temperatures, smallholder field practices, and limited injection equipment make direct NH3 use impractical and prone to large volatilisation and safety issues. Hence the statement endorsing it as “suitable” is wrong. The other statements align with standard chemistry and process practice. Step-by-Step Solution: Evaluate (a): Carbamate → urea + H2O is endothermic → correct.Evaluate (c): Wet-process H3PO4 coproduces gypsum → correct.Evaluate (d): Rock + dilute H2SO4 → SSP → correct.Evaluate (b): Claims liquid NH3 use is suitable in India → incorrect under typical conditions. Verification / Alternative check: Process texts show energy input for the carbamate dehydration section; fertiliser manuals describe SSP production via H2SO4 and widespread Indian use of urea/complex fertilisers rather than anhydrous NH3 injection. Why Other Options Are Wrong (i.e., not the wrong statement): (a) True thermochemistry. (c) True wet-process by-product. (d) True for SSP manufacture. (e) True: AN neutralisation is exothermic. Common Pitfalls: Assuming practices from temperate mechanised agriculture transfer directly to tropical smallholder systems; overlooking safety and equipment constraints. Final Answer: Direct field application of liquid ammonia is suitable for a tropical country like India

Question 14

Nitrogen analysis comparison: Among the listed fertilisers, which contains the maximum percentage of nitrogen by mass under standard specifications?

Accepted Answer

Introduction / Context: Different nitrogen fertilisers supply widely varying percentages of nitrogen, influencing transport, storage, and field application economics. Recognising which has the highest N% helps in quick selection and troubleshooting dosing errors. Given Data / Assumptions: Typical grades: urea ≈ 46% N; ammonium nitrate ≈ 33–35% N; ammonium sulphate ≈ 21% N; CAN ≈ 26–28% N. Anhydrous (liquid) ammonia contains about 82% N by mass. Nominal values are used for comparison. Concept / Approach: Anhydrous ammonia (NH3) is essentially nitrogen-rich, containing ~14/17 by mass ≈ 82% N. No other common solid nitrogen fertiliser exceeds this percentage. Urea at 46% N is the highest among typical solids but still far below anhydrous ammonia. Therefore, among the options provided, liquid ammonia has the maximum nitrogen content. Step-by-Step Solution: List N% for each product.Compare values: 82% (NH3) > 46% (urea) > ~34% (AN) > ~26–28% (CAN) > ~21% (AS).Select the maximum.Answer: liquid ammonia. Verification / Alternative check: Specification sheets and fertiliser handbooks confirm the above typical analyses. Why Other Options Are Wrong: Urea: highest solid N %, but less than ammonia. Ammonium nitrate/CAN/ammonium sulphate: significantly lower N contents. Common Pitfalls: Confusing volumetric concentration with mass percentage; forgetting ammonia’s very high N fraction due to its small molecular mass. Final Answer: Liquid ammonia

Question 15

Electric furnace (thermal) phosphorus route: In producing elemental white phosphorus from phosphate rock in an electric furnace, which additional materials are used?

Accepted Answer

Introduction / Context: The thermal (electric furnace) process reduces phosphate rock to elemental phosphorus (P4). Understanding the required reactants clarifies stoichiometry and separates this route from acidulation processes that make fertilisers or phosphoric acid. Given Data / Assumptions: Feed: phosphate rock (fluorapatite or related). High-temperature electric furnace operation. Goal: produce elemental white phosphorus. Concept / Approach: In the thermal process, silica (SiO2) acts as a flux to form calcium silicate slags, and coke (carbon) provides the reducing atmosphere that converts phosphate to phosphorus and CO/CO2. Sulphuric or phosphoric acids are not used in this high-temperature reduction route; those acids are relevant to wet-process phosphoric acid or superphosphate production, not to elemental phosphorus manufacture. Step-by-Step Solution: Identify reduction requirement: phosphate must be reduced to P.Supply carbon (coke) as reductant; include silica to capture CaO as silicate.Operate the electric furnace at high temperature to volatilise P4.Therefore, the correct combination is phosphate rock with silica and coke. Verification / Alternative check: Classical reaction set shows Ca3(PO4)2 + SiO2 + C → P4 (vapor) + Ca-silicate + CO/CO2. Why Other Options Are Wrong: Phosphoric/sulphuric acid options refer to wet processes and superphosphate manufacture. Rock + coke only omits silica necessary for slag formation. Gypsum and air are not reagents in the thermal phosphorus route. Common Pitfalls: Mixing up thermal reduction (elemental P) with acidulation (fertiliser or H3PO4). Final Answer: Phosphate rock with silica and coke

Question 16

Concentrated sulphuric acid reaction: Phosphate rock treated with about 98% H2SO4 primarily yields which product under wet-process conditions?

Accepted Answer

Introduction / Context: Phosphate processing follows different routes depending on the acid used. The wet-process route to phosphoric acid treats phosphate rock with strong sulphuric acid, forming phosphoric acid while precipitating calcium sulphate (gypsum or anhydrite) as a by-product. The concentration of H2SO4 influences by-product hydration state but the primary product is H3PO4. Given Data / Assumptions: Reagent: Concentrated H2SO4 (~98%). Feed: Phosphate rock (fluorapatite). Industrial wet-process conditions assumed. Concept / Approach: Under these conditions the rock is digested to release phosphoric acid. Calcium from the rock is converted to calcium sulphate, which precipitates (dihydrate or anhydrite depending on temperature/acid strength), and the filtrate contains orthophosphoric acid (H3PO4). Superphosphate refers to direct acidulation of rock with dilute sulphuric acid to form monocalcium phosphate and gypsum within a solid product matrix; white phosphorus is obtained via the thermal electric furnace process, not via acidulation. Step-by-Step Solution: Recognize 98% H2SO4 as strong acid digestion.Predict formation of phosphoric acid plus calcium sulphate by-product.Exclude superphosphate and elemental phosphorus routes.Select orthophosphoric acid as the product. Verification / Alternative check: Wet-process flowsheets show reactor–filter systems producing H3PO4 liquor and gypsum/anhydrite solids at higher acid strengths. Why Other Options Are Wrong: Superphosphate: produced by dilute H2SO4 directly on rock, not strong-acid digestion for H3PO4. White phosphorus: requires thermal reduction with silica and coke. Triple superphosphate: produced by reacting rock with phosphoric acid (H3PO4), not H2SO4. Common Pitfalls: Confusing SSP/TSP manufacture with wet-process acid; overlooking the role of acid concentration in determining the by-product phase (gypsum vs anhydrite). Final Answer: Orthophosphoric acid

Question 17

Fertilizer analysis: A fertilizer is reported to contain 82% nitrogen (N) by mass. Which common nitrogen fertilizer could match approximately 82% N content?

Accepted Answer

Introduction / Context: Nutrient analysis of fertilizers is typically expressed as mass percent of the nutrient element. For nitrogenous fertilizers, knowing the theoretical nitrogen fraction from chemical formulae allows quick identification. This question tests whether you can connect a stated 82% nitrogen content to the correct fertilizer material used widely in agriculture and ammonia-based industries. Given Data / Assumptions: Labeled nitrogen content is 82% by mass. Consider common fertilizers: anhydrous ammonia, urea, ammonium nitrate, and blends like CAN. Assume pure materials (no inert fillers or coatings). Concept / Approach: The mass percent of an element in a compound is calculated as (mass of the element in the formula / molar mass of the compound) * 100. For NH3, molar mass = 17 (N = 14; H = 3); percent N = 14/17 ≈ 82.35%. Urea contains about 46% N; ammonium nitrate contains about 35% N; calcium ammonium nitrate is lower still due to added limestone. Therefore, only anhydrous ammonia matches 82% N closely. Step-by-Step Solution: Compute %N for NH3: (14 / 17) * 100 ≈ 82.35%.Compute %N for urea: N mass 28; molar mass 60 → (28 / 60) * 100 ≈ 46.7%.Compute %N for ammonium nitrate: N mass 28; molar mass 80 → (28 / 80) * 100 = 35%.CAN mixes ammonium nitrate with limestone, further lowering %N below 35%.Match the given 82% to anhydrous ammonia; select liquid NH3. Verification / Alternative check: Any fertilizer handbook lists NH3 at ~82% N, urea ~46% N, ammonium nitrate ~33–35% N, and CAN ~26–28% N depending on grade. The given value can only correspond to anhydrous ammonia among the options. Why Other Options Are Wrong: Urea: about 46% N, not close to 82%. Ammonium nitrate: roughly 35% N. Calcium ammonium nitrate: lower than ammonium nitrate due to limestone. None of these: incorrect because NH3 fits. Common Pitfalls: Confusing percent nitrogen with percent ammonia; remembering that “82” is a quick mnemonic for anhydrous ammonia. Final Answer: Anhydrous ammonia (liquid NH3).

Question 18

Hydrogen for ammonia synthesis (Haber–Bosch): among common commercial routes, which production method is generally the costliest per unit of hydrogen produced?

Accepted Answer

Introduction / Context: Ammonia plants require large volumes of hydrogen, traditionally sourced from fossil feedstocks. Selection of the hydrogen route depends on feedstock price, capital cost, and energy efficiency. This question checks your awareness of relative cost among standard industrial options, ignoring special cases like stranded renewable electricity or by-product hydrogen. Given Data / Assumptions: Conventional large-scale ammonia production. Typical energy prices; no special subsidies. Comparison across standard technologies: coke oven gas separation, steam reforming (naphtha or natural gas), and water electrolysis. Concept / Approach: Steam methane reforming (SMR) of natural gas is the dominant, lower-cost route at scale. Steam reforming of naphtha is similar but typically costlier where naphtha is pricier than pipeline gas. Recovery from coke oven gas can be economical at integrated steel plants. Electrolysis of water, while clean at point of use, requires substantial electricity; without very low-cost power, its levelized hydrogen cost is usually highest among these choices. Step-by-Step Solution: List routes and principal energy drivers: SMR (natural gas), reforming naphtha, recovery from COG, electrolysis (electric power).Note current industrial economics: electricity cost dominates electrolysis OPEX.Conclude electrolysis is generally the costliest for conventional ammonia production. Verification / Alternative check: Benchmark reports show hydrogen from SMR has historically lower cost per kilogram than electrolysis at typical grid prices; electrolysis becomes competitive only with very cheap renewable power. Why Other Options Are Wrong: Coke oven gas: often an available by-product stream; separation can be cost-effective. Steam reforming of naphtha or natural gas: established, typically cheaper at scale. Partial oxidation of heavy oil: also typically cheaper than electrolysis for H2 per unit energy. Common Pitfalls: Assuming the “greenest” route is the cheapest; in practice, electricity price is decisive for electrolysis economics. Final Answer: Electrolysis of water.

Question 19

Thermochemistry of ammonia synthesis: For the reaction N2(g) + 3 H2(g) → 2 NH3(g), the overall process under Haber–Bosch conditions is best described as

Accepted Answer

Introduction / Context: Knowing whether a reaction releases or absorbs heat guides temperature selection, heat integration, and equilibrium management in reactors. Ammonia synthesis is a flagship industrial reaction with strong temperature and pressure sensitivities dictated by both thermodynamics and kinetics. Given Data / Assumptions: Reaction: N2 + 3 H2 → 2 NH3. Typical industrial conditions: elevated pressure, iron- or ruthenium-based catalyst. Standard thermochemical data indicate heat release. Concept / Approach: Ammonia formation reduces the number of gas moles (4 → 2) and forms strong N–H bonds, releasing heat. The standard enthalpy change at 25°C is negative (exothermic). Consequently, lower temperatures shift equilibrium toward NH3, while higher temperatures accelerate kinetics; industrial operation balances these via moderate temperatures and high pressures with interstage cooling and recycle. Step-by-Step Solution: Consider bond energetics: forming N–H bonds releases energy.Mole reduction favors high pressure by Le Châtelier’s principle.Combine: exothermic reaction; choose operating temperature accordingly with heat removal. Verification / Alternative check: Handbooks report ΔH° ≈ −92 kJ per 2 mol NH3 (about −46 kJ per mol NH3), confirming exothermicity. Why Other Options Are Wrong: Endothermic: incorrect—would require heat input to proceed. Autocatalytic: catalysts are external (Fe, Ru), not a product-driven catalysis. None/Athermal: contradicted by measured heat of reaction. Common Pitfalls: Confusing kinetic requirements (needing temperature) with thermodynamic heat effect; exothermic does not mean it proceeds rapidly without a catalyst. Final Answer: Exothermic.

Question 20

Allotropy of phosphorus: White phosphorus shows maximum stability under which of the following general conditions?

Accepted Answer

Introduction / Context: Phosphorus exists in several allotropes (white, red, black) with differing stability and reactivity. White phosphorus (P4) is the most reactive form and must be handled carefully. The question asks under what common laboratory conditions white phosphorus is comparatively most stable to storage and handling. Given Data / Assumptions: Focus is on practical stability against oxidation and conversion to other allotropes. Options include pressure, temperature, and environmental exposure. Assume simple laboratory storage practices. Concept / Approach: White phosphorus is pyrophoric in air near ambient temperatures and is light-sensitive; it oxidizes and can ignite spontaneously. It is typically stored under water, in the dark, at room temperature and atmospheric pressure to exclude oxygen and light, which suppress reactivity and slow conversion to red phosphorus. High temperatures or illumination accelerate transformation to red or black phosphorus and enhance oxidation hazards; very high pressure/high temperature favor other allotropes rather than P4 stability. Step-by-Step Solution: Recall handling: store P4 under water to exclude air.Avoid light and heat to prevent conversion/oxidation.Thus, the “most stable” routine condition is atmospheric pressure, dark, under water at room temperature. Verification / Alternative check: Safety data sheets and inorganic chemistry texts recommend storage of white phosphorus under water in light-proof containers, confirming best-practice stability conditions. Why Other Options Are Wrong: Very high pressure: not a practical stabilization method; can promote other phases. Room temperature in dry air with light: unsafe; risk of ignition and rapid oxidation. > 600°C in air or vacuum at high temperature: conditions would convert/oxidize P4. Common Pitfalls: Assuming “room temperature” alone is sufficient; exposure to air and light makes white phosphorus hazardous even at room temperature unless submerged. Final Answer: Atmospheric pressure in the dark under water.

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