Question 1

For a reversible endothermic reaction at equilibrium, how does the equilibrium conversion change as temperature decreases?

Accepted Answer

Introduction / Context: The temperature dependence of equilibrium conversion is governed by Le Chatelier's principle and the van’t Hoff relation. Recognizing how endothermicity affects conversion is essential for reactor design and optimization. Given Data / Assumptions: Reaction is reversible and endothermic (ΔH > 0). System is at equilibrium. Concept / Approach: For endothermic reactions, increasing temperature favors products to absorb heat; decreasing temperature favors reactants. The van’t Hoff relation shows that ln K increases with T when ΔH > 0. Step-by-Step Solution: 1) Endothermic: ΔH > 0.2) van’t Hoff: d(ln K)/dT = ΔH/(R*T^2) > 0 ⇒ K increases with T.3) Lowering T decreases K, shifting equilibrium toward reactants, reducing conversion to products. Verification / Alternative check: Le Chatelier's principle: remove heat (lower T) and the equilibrium shifts to the exothermic direction (here, the reverse), decreasing product conversion. Why Other Options Are Wrong: Option b inverts the temperature effect for endothermic reactions; c and d ignore the thermodynamic relation; e is unsupported without specific complex mechanisms. Common Pitfalls: Confusing endothermic with exothermic behavior; assuming kinetics rather than equilibrium control. Final Answer: It decreases.

Question 2

For which type of application is a batch reactor especially suitable from an operations and control perspective?

Accepted Answer

Introduction / Context: Reactor selection depends on reaction phase, scale, heat effects, and product slate. Batch reactors are widely used in fine chemicals, pharma, and specialty applications. Given Data / Assumptions: We compare batch with continuous systems. Focus on practicality: flexibility, control, and typical phases. Concept / Approach: Batch reactors provide flexibility, tight recipe control, and high-quality assurance, particularly for liquid-phase reactions and multiproduct campaigns. Large-scale gas-phase commodity production typically favors continuous plug-flow or CSTR systems. Step-by-Step Solution: 1) Assess phase: many liquid-phase syntheses in pharma/fine chemicals are batch.2) Consider control needs: batch offers staged additions, temperature profiling, and sampling.3) Throughput: gas-phase large volume is usually continuous for efficiency.4) Highly exothermic polymerisations often require specialized reactor and control strategies; blanketly stating batch ensures “uniform products” is unsafe. Verification / Alternative check: Industry practice aligns: batch dominates for liquid-phase specialty products, where flexibility and traceability are critical. Why Other Options Are Wrong: Option a: 100% conversion is kinetic/thermodynamic, not guaranteed by batch mode. Option b: Large gas-phase production is rarely batch. Option d: Exothermic polymerisations demand careful heat removal; mode alone doesn’t ensure uniformity. Option e: High-throughput cracking is continuous. Common Pitfalls: Equating reactor mode with guaranteed conversion or product uniformity; ignoring scale and heat transfer limits. Final Answer: Liquid-phase reactions, including specialty and multiproduct operations.

Question 3

For the reaction 3A → B, the dimensions of the rate constant are given as (L per gm-mole) per minute, i.e., (L·(gm-mole)^{-1})·min^{-1}. What is the reaction order?

Accepted Answer

Introduction / Context: Units of the rate constant reveal the reaction order when concentration units are specified. This is a staple skill in reaction engineering. Given Data / Assumptions: Rate law form: r = k C^n for a single key species basis. Units of concentration: gm-mole per liter (gm-mole·L^-1). Units of k given: L·(gm-mole)^{-1}·min^{-1} = (concentration)^{-1}·time^{-1}. Concept / Approach: For an nth order reaction, k has units (concentration)^{1-n} / time. Match the given units to solve for n. Step-by-Step Solution: 1) General: [k] = (conc)^{1-n} / time.2) Given: [k] = (conc)^{-1} / time.3) Equate exponents: 1 - n = -1 ⇒ n = 2. Verification / Alternative check: Check typical patterns: first order has 1/time; second order has 1/(conc·time). The given matches second order. Why Other Options Are Wrong: Zero/first/third/half orders have different unit patterns; only second order yields (conc)^{-1}/time. Common Pitfalls: Using mass units instead of molar concentration; forgetting that the stoichiometric coefficient (3A → B) does not mandate the kinetic order. Final Answer: Second order.

Question 4

If the time required to reach any fixed fractional conversion varies inversely with the initial concentration of reactants, what is the order of the reaction?

Accepted Answer

Introduction / Context: Time-to-fractional-conversion criteria are practical diagnostics for reaction order. Recognizing these signatures is valuable in lab-scale kinetics. Given Data / Assumptions: Initial concentration C0 varies between runs. Measured time to a fixed conversion X is inversely proportional to C0. Concept / Approach: For a second-order reaction with rate r = k C^2, the integrated form yields t at fixed conversion proportional to 1/C0. For a first-order reaction, the time to a fixed fraction is independent of C0. Step-by-Step Solution: 1) Second order integrated law: 1/C - 1/C0 = k t.2) At a fixed conversion X, C = C0*(1 - X).3) Substitute to find t = (1/(k*C0)) * (X/(1 - X)). Thus t ∝ 1/C0.4) Therefore, inverse dependence means second order. Verification / Alternative check: First order gives t = (1/k) * ln(1/(1 - X)), independent of C0—contrary to the observation. Why Other Options Are Wrong: Zero/first/third orders have different C0 dependencies; only second order yields the observed inverse proportionality. Common Pitfalls: Confusing half-life patterns (first-order half-life independent of C0) with second-order behavior (half-life ∝ 1/C0). Final Answer: Second order.

Question 5

Why does the presence of a catalyst increase the observed rate of a chemical reaction?

Accepted Answer

Introduction / Context: This question targets the core kinetic reason a catalyst accelerates reactions: barrier lowering. It is foundational for understanding reactor performance and catalyst design. Given Data / Assumptions: No external temperature change; only catalyst addition. Catalyst is regenerated and not consumed. Concept / Approach: Rate constant k depends exponentially on activation energy or free energy. A catalyst provides an alternative pathway with smaller ΔE or ΔG‡, thus increasing k at a given temperature. Step-by-Step Solution: 1) Arrhenius form: k = A * exp(-Ea/(RT)); transition-state form: k = (kBT/h) * exp(-ΔG‡/(R*T)).2) Lower Ea or ΔG‡ ⇒ larger exponent ⇒ higher k ⇒ faster rate.3) Equilibrium (ΔG°) stays the same; catalyst does not alter final composition, only the speed toward it. Verification / Alternative check: Experiments show catalysts change rates without affecting equilibrium constants. This directly supports barrier lowering rather than thermodynamic driving force changes. Why Other Options Are Wrong: Option b contradicts observed acceleration; higher barrier would slow rates. Option c is not the general mechanism; collision geometry matters but barrier lowering is key. Option d changes equilibrium, which catalysts do not do. Option e violates energy conservation. Common Pitfalls: Confusing kinetic effects (rate) with thermodynamic effects (equilibrium) and assuming catalysts are consumed. Final Answer: It lowers the activation barrier.

Question 6

Chemical reaction engineering – characteristics of a zero-order reaction For a zero-order chemical reaction, which of the following statements are true?

Accepted Answer

Introduction / Context: Zero-order reactions are an important special case in chemical reaction engineering where the rate is independent of reactant concentration. Such behavior is often observed in catalytic systems at saturation, photochemical reactions, or enzyme kinetics at very high substrate levels. Understanding time–concentration profiles and half-life scaling helps with reactor sizing and conversion prediction. Given Data / Assumptions: Overall rate law: r = k (constant), independent of concentration. Isothermal, well-mixed (batch) conditions for interpreting time profiles. Products start at zero concentration unless otherwise stated. Concept / Approach: For a zero-order reaction consuming A to form products P, the material balance in a batch reactor is dC_A/dt = −k, which integrates to C_A = C_A0 − k t. Product concentration C_P then grows as k t (assuming stoichiometric 1:1), a linear relation with time. The half-life t_1/2 is the time required for C_A to drop to C_A0/2, which gives t_1/2 = C_A0/(2k); it scales directly with the initial concentration C_A0. Step-by-Step Solution: Start with dC_A/dt = −k.Integrate: C_A(t) = C_A0 − k t.Define product build-up (for 1:1): C_P(t) = k t. This is a straight line through the origin at t = 0.Half-life: set C_A = C_A0/2 → C_A0/2 = C_A0 − k t_1/2 → t_1/2 = C_A0/(2k) ∝ C_A0. Verification / Alternative check: Dimensional check confirms k has units of concentration/time for zero order; linear changes in both reactant consumption and product formation with time are consistent with constant rate. Why Other Options Are Wrong: Options (a), (b), and (c) are each true individually; the most complete correct choice is “All (a), (b) and (c)”. Common Pitfalls: Confusing zero-order with first-order kinetics; first order shows exponential decay and constant half-life independent of initial concentration, which is not the case here. Final Answer: All (a), (b) and (c).

Question 7

Catalyst deactivation – primary causes in heterogeneous catalysis A catalyst loses its activity due to which of the following phenomena?

Accepted Answer

Introduction / Context: Catalyst activity and selectivity determine reactor performance and economics. Over time, real catalysts deactivate, necessitating regeneration or replacement. Recognizing the major mechanisms of deactivation helps in designing operating windows and protection strategies (e.g., guards, sulfur traps, optimized temperatures). Given Data / Assumptions: Heterogeneous catalysts with active surface sites. Thermal and chemical environments typical of industrial reactors. Concept / Approach: Deactivation mechanisms include (1) loss of surface area via support collapse or phase changes; (2) agglomeration/sintering where metal nanoparticles grow, reducing dispersion and the number of active sites per mass; and (3) poisoning/fouling in which contaminants (S, Cl, metals, coke) block sites or plug pores, cutting off access to the active phase. Step-by-Step Solution: Thermal sintering → particle growth → lower active surface area.Support changes → pore collapse → diminished external/internal area.Poisoning/fouling → chemisorbed species or deposits cover sites → rate drops. Verification / Alternative check: Post-run characterizations (BET surface area, TEM particle size, TPO/TPR of spent catalyst) routinely show these deactivation signatures. Why Other Options Are Wrong: Each individual mechanism can occur; in practice they often act together. Therefore the comprehensive choice is correct. Common Pitfalls: Confusing reversible fouling (e.g., coke burn-off) with irreversible poisoning (e.g., mercury amalgamation on noble metals). Not all deactivation is permanent, but all listed mechanisms reduce activity. Final Answer: All (a), (b) and (c).

Question 8

Batch reactor kinetics – first-order irreversible, isothermal, variable volume For an isothermal batch reactor with a first-order irreversible reaction, which relation between conversion (X_A) and time (t) is applicable (even if volume changes with …

Accepted Answer

Introduction / Context: Relating conversion to time is fundamental for reactor design and performance prediction. A common doubt is whether changing volume during reaction alters the classic first-order exponential behavior in batch systems. Given Data / Assumptions: Isothermal batch reactor. Elementary first-order irreversible reaction: r_A = −k C_A. Volume may vary with conversion due to stoichiometry (ε_A ≠ 0). Concept / Approach: For a batch system, the total moles of A, N_A, satisfy dN_A/dt = r_A V = −k C_A V. Because C_A = N_A/V, the product C_A V equals N_A. Hence dN_A/dt = −k N_A regardless of how V changes with time. This decouples the kinetics from the volume change for first-order reactions in batch mode and leads to an exponential decay of N_A and the familiar logarithmic conversion relation. Step-by-Step Solution: Write mole balance: dN_A/dt = −k N_A.Separate variables: dN_A/N_A = −k dt.Integrate: ln(N_A/N_A0) = −k t → N_A = N_A0 e^(−k t).Define conversion: X_A = 1 − N_A/N_A0 = 1 − e^(−k t).Rearrange: −ln(1 − X_A) = k t. Verification / Alternative check: Check limiting behavior: as t → 0, X_A → 0; as t grows, X_A approaches 1 asymptotically, consistent with first-order kinetics. Why Other Options Are Wrong: X_A = k t: Linear in time is characteristic of zero-order, not first-order. ε_A * ln(1 − X_A) = k t: ε_A affects concentration–conversion relationships in flow reactors, not the batch first-order N_A decay law. Common Pitfalls: Assuming variable volume changes the exponential decay of N_A for first order in batch; it does not, because C_A V = N_A cancels V(t). Final Answer: -ln(1 - X_A) = k t

Question 9

Nuclear/chemical kinetics – order of radioactive decay Radioactive decay processes follow which overall kinetic order?

Accepted Answer

Introduction / Context: Radioactive decay is a stochastic, unimolecular process in which unstable nuclei transform spontaneously. In chemical engineering and nuclear science, its time dependence is modeled as a kinetic rate law that is independent of concentration of other species. Given Data / Assumptions: Each nucleus decays independently. No external reactant is required; decay probability per unit time is constant. Concept / Approach: For first-order processes, the rate of decrease in the number of nuclei N is proportional to N itself: dN/dt = −λ N, where λ is the decay constant. Integration yields an exponential law and a constant half-life independent of initial amount, matching observed radionuclide behavior. Step-by-Step Solution: Write decay law: dN/dt = −λ N.Integrate: ln(N/N0) = −λ t → N = N0 e^(−λ t).Half-life t_1/2: set N = N0/2 → t_1/2 = (ln 2)/λ (constant). Verification / Alternative check: Measured activities (A = λ N) decay exponentially; plotting ln A vs. time gives a straight line with slope −λ, confirming first-order kinetics. Why Other Options Are Wrong: Second- or third-order kinetics require bimolecular/termolecular interactions, which are not involved in spontaneous nuclear decay. Zero order would imply a constant decay rate independent of N, contrary to experiment. Common Pitfalls: Confusing activity decay (A) with dose rate or detector counts affected by geometry and shielding; the underlying nuclear decay remains first order. Final Answer: First order

Question 10

Diffusion in small pores – definition and dependence of Knudsen diffusivity When catalyst pore diameter is much smaller than the mean free path of gas molecules, wall collisions dominate. The effective diffusivity under this regime (Knudsen diffusiv…

Accepted Answer

Introduction / Context: In porous catalysts, mass transfer may occur in different regimes. When the pore diameter is small relative to the gas mean free path, molecule–wall collisions dominate over molecule–molecule collisions. This is the Knudsen regime, critical in interpreting effectiveness factors and internal diffusion limitations. Given Data / Assumptions: Mean free path ≫ pore diameter → Knudsen regime. Gas diffusion inside cylindrical pores. Ideal-gas behavior at process pressures. Concept / Approach: Knudsen diffusivity D_K for species i is proportional to the pore radius and to the square root of absolute temperature, and is independent of total pressure: D_K ≈ (2/3) * r_pore * sqrt(8 R T / (π M_i)). Thus, temperature increases D_K via √T, while pressure does not appear explicitly (unlike molecular diffusion D_ij ∝ T^(3/2)/P). Step-by-Step Solution: Identify regime: Knudsen if λ_mean_free_path ≫ pore diameter.Use scaling: D_K ∝ r_pore * √T / √M_i.Conclude dependence: increases with T; independent of P (for ideal gas). Verification / Alternative check: Dimension and units check confirm D_K has dimensions of area/time and the lack of pressure dependence aligns with kinetic theory for molecule–wall collision dominance. Why Other Options Are Wrong: Pressure: Does not enter D_K; pressure mainly affects molecular (bulk) diffusion, not Knudsen. Both (a) and (b) / Neither: Only temperature applies here. Common Pitfalls: Using gas-phase binary diffusivity correlations inside tiny pores; in Knudsen regime, use D_K, not D_ij, for the controlling mechanism. Final Answer: Temperature

Question 11

Energy changes – identifying exothermic reactions In an exothermic chemical reaction, how do the energies of reactants and products compare?

Accepted Answer

Introduction / Context: Classifying reactions as exothermic or endothermic is fundamental to reactor design and safety. Energy balance, heat removal, and temperature control hinge on knowing the enthalpy change and relative energy levels of reactants and products. Given Data / Assumptions: Exothermic reaction: releases heat to surroundings. Comparison at the same reference temperature and pressure. Concept / Approach: For an exothermic reaction, the standard enthalpy change ΔH is negative. This means products lie at a lower enthalpy (energy) level than reactants. The energy difference is liberated as heat. Temperature of the mixture may rise if heat is not removed, but “higher temperature” is an outcome, not a definition of energy level comparison. Step-by-Step Solution: State: Exothermic → ΔH < 0.Implication: H_products < H_reactants → reactants have more energy.Heat release equals (H_reactants − H_products) at the reference condition. Verification / Alternative check: Potential energy diagrams show reactants at a higher enthalpy plateau transitioning to products at a lower plateau, with heat released to maintain energy conservation. Why Other Options Are Wrong: Higher temperature (option a): Not a reliable comparison; temperature depends on heat removal and heat capacity. Less or same energy (options c, d): Contradicts ΔH < 0 definition. Common Pitfalls: Equating temperature directly with energy content; enthalpy depends on both temperature and composition. Always compare enthalpies at a common reference to classify exo/endothermicity. Final Answer: Reactants possess more energy than products

Question 12

Reactor selection – very high-pressure gas-phase reactions Which reactor configuration is most suitable for conducting very high-pressure gas-phase reactions?

Accepted Answer

Introduction / Context: Operating at very high pressure imposes stringent mechanical and safety constraints. Reactor geometry, wall thickness, stress distribution, and mixing/heat transfer characteristics all influence the preferred design for gas-phase processes (e.g., polymerization, hydrogenation). Given Data / Assumptions: Gas-phase reaction requiring very high pressure. Continuous operation is feasible. Mechanical integrity and pressure containment are primary concerns. Concept / Approach: For pressure vessels, the required wall thickness is proportional to diameter (for a given pressure and allowable stress). Long, small-diameter tubes can withstand high pressures with comparatively thinner walls and offer high area/volume for heat transfer. Consequently, tubular (plug-flow) reactors are favored over large-diameter stirred vessels or fluidised beds under very high pressure. Step-by-Step Solution: Assess mechanical design: small diameter tubes minimize hoop stress → practical at very high pressure.Assess heat removal: large surface/volume in tubes aids temperature control.Assess operability: continuous flow, predictable residence time → suitable for high-pressure duty. Verification / Alternative check: Industrial practice: high-pressure synthesis (e.g., LDPE tubular reactors, certain hydrogenations) commonly use tubular configurations for pressure-holding efficiency. Why Other Options Are Wrong: Batch/CSTR: Large diameters demand very thick walls and heavy construction at high pressure. Fluidised bed: Gas–solid contacting and bed dynamics complicate ultra-high-pressure operation and vessel design. Common Pitfalls: Ignoring pressure-vessel design rules and focusing only on kinetics. Mechanical constraints often dictate reactor choice at extreme pressures. Final Answer: Tubular flow reactor

Question 13

Temperature control strategy – highly exothermic solid-catalysed reactions For achieving high conversion in a highly exothermic, solid-catalysed reaction, which reactor sequence is preferred?

Accepted Answer

Introduction / Context: Highly exothermic catalytic reactions pose hot-spot and runaway risks. Proper temperature management improves selectivity, avoids catalyst sintering, and enables high overall conversion. Combining different reactor types can exploit their strengths in heat removal and approach to equilibrium. Given Data / Assumptions: Gas-phase, solid-catalysed, strongly exothermic reaction. Objective: high overall conversion with safe temperature control. Industrial-scale heat removal considerations are critical. Concept / Approach: Fluidised beds provide excellent heat transfer and near-isothermal operation, limiting hot spots during the most vigorous part of the reaction (high driving force). After the major exotherm is tamed and reactant concentration drops, a polish fixed bed can drive conversion higher with lower heat release rates per volume, simplifying temperature control. Step-by-Step Solution: Stage 1: Use a fluidised bed to remove heat efficiently and maintain near-uniform temperature during the high-rate region.Stage 2: Send partially converted stream to a fixed bed for final conversion “push,” benefiting from simpler hardware and lower heat flux.Result: High conversion with mitigated thermal risks and improved catalyst life. Verification / Alternative check: Analogous industrial strategies include multi-bed adiabatic reactors with intercooling or fluidised-bed front-ends followed by fixed beds for finishing steps. Why Other Options Are Wrong: Fixed bed only: Prone to hot spots at the inlet where reaction rate is highest. Fixed → fluidised: Places the more temperature-sensitive step first; less optimal for hot-spot control. Fluidised only: Good temperature control but may not maximize final conversion economically without a polishing stage. Common Pitfalls: Ignoring heat removal limits and focusing solely on kinetics; for highly exothermic reactions, thermal management is often the bottleneck. Final Answer: Fluidised bed reactor followed by a fixed bed reactor

Question 14

Rate constants – temperature and reactant dependence for second-order reactions For a second-order reaction, which statements about the specific rate constant are correct?

Accepted Answer

Introduction / Context: The specific rate constant k characterizes intrinsic kinetics for a given reaction mechanism and order. Correct understanding of its dependencies is essential for reactor design and temperature control strategies. Given Data / Assumptions: Second-order overall reaction (e.g., A + B → products or 2A → products). Arrhenius behavior applies over the temperature range of interest. Concept / Approach: According to the Arrhenius equation, k = A * exp(−E_a/(R T)), the rate constant depends strongly on absolute temperature and activation energy E_a. The pre-exponential factor A and E_a are properties of the specific reaction system, which are determined by the nature of the reacting species and mechanism. Therefore, k varies with T and depends on the reactants involved. Step-by-Step Solution: Recognize temperature dependence: increasing T typically increases k exponentially.Recognize chemical identity dependence: different reactant pairs have different A and E_a → different k values at the same T.Hence statements (b) and (c) are both correct. Verification / Alternative check: Experimental Arrhenius plots (ln k vs. 1/T) for different reagent pairs yield distinct slopes (−E_a/R) and intercepts (ln A), confirming dependence on both T and reactant identity. Why Other Options Are Wrong: Independent of temperature: Contradicts Arrhenius behavior observed for most reactions. Common Pitfalls: Confusing reaction order (which affects units of k) with temperature dependence; order does not eliminate k’s Arrhenius dependence. Final Answer: Both (b) and (c).

Question 15

Chemical reaction engineering — space velocity (SV) in flow reactors: Identify the incorrect statement about the definition and interpretation of space velocity.

Accepted Answer

Introduction / Context: Space velocity (SV) is a key sizing and performance metric for flow reactors in chemical reaction engineering. It normalizes the volumetric feed rate by the reactor volume and therefore indicates how many reactor volumes of feed pass through per unit time. Given Data / Assumptions: SV is defined at specified temperature and pressure of the feed. Liquid Hourly Space Velocity (LHSV) and Gas Hourly Space Velocity (GHSV) are common forms. Reactor is assumed to be of constant holdup at the stated conditions. Concept / Approach: By definition, SV = volumetric feed rate / reactor volume. Hence its SI sense is time^(-1). An SV of 3 hr^(-1) literally means that, each hour, a volume of feed equal to three times the reactor volume is processed under the stated conditions. Step-by-Step Solution: Write SV = Q / V_R.If SV = 3 hr^(-1), then Q = 3 * V_R per hour.Therefore, the interpretation as one third reactor volume per hour is incorrect. Verification / Alternative check: Residence time τ for a well mixed system is τ = V_R / Q. Therefore τ = 1 / SV. If SV = 3 hr^(-1), then τ = 1/3 hr, which confirms the above interpretation. Why Other Options Are Wrong: Time^(-1) is the correct unit, and three reactor volumes per hour is the correct reading for SV = 3 hr^(-1). Hence only the one third statement is wrong. Common Pitfalls: Confusing SV with 1/SV. Mixing feed conditions with reactor conditions can also mislead when compressibility is significant. Final Answer: SV = 3 hr^(-1) means one third of a reactor volume of feed is fed each hour.

Question 16

Surface area characterization — regarding the BET adsorption apparatus, which statement is correct?

Accepted Answer

Introduction / Context: The Brunauer–Emmett–Teller (BET) method estimates specific surface area of porous and nonporous solids by nitrogen or other gas adsorption at cryogenic temperatures. The apparatus and the principle are often misunderstood in entry level labs. Given Data / Assumptions: Adsorptive gas is usually nitrogen at 77 K or krypton at lower coverages. Pressures employed are subambient, typically in the relative pressure P/P0 range 0.05 to 0.3. Apparatus includes glass or metal manifolds, vacuum system, calibrated volumes, and detectors. Concept / Approach: BET does not measure area directly. It infers monolayer capacity from measured adsorption isotherms and converts that to area using known molecular cross sections. The method uses low absolute pressures and high vacuum, not high pressure. Construction materials vary; many systems use glass dewar and sample tubes with metallic manifolds, not necessarily all stainless steel. Step-by-Step Solution: Collect adsorption data at several P/P0 points.Fit the BET equation to obtain monolayer capacity n_m.Compute area: A = n_m * N_A * molecular cross section. Verification / Alternative check: Cross validate area with mercury porosimetry or microscopy for appropriate materials; results differ when micro porosity dominates and t plot or NLDFT is needed. Why Other Options Are Wrong: Area is inferred, not directly measured; pressures are low; and there is no strict requirement that every wetted part is stainless steel only. Common Pitfalls: Using an incorrect P/P0 fitting range or ignoring degassing quality yields significant errors in calculated surface area. Final Answer: None of these.

Question 17

Reactor selection for high yield when high reactant concentration is beneficial: Which reactor type is most unsuitable if high concentration favors conversion and yield?

Accepted Answer

Introduction / Context: For positive order reactions or systems where rate and selectivity improve with higher reactant concentration, the choice of reactor affects yield and size significantly. Understanding mixing patterns is essential for proper selection. Given Data / Assumptions: Reaction order is positive or rate increases with higher reactant concentration. Isothermal conditions for fair comparison. Steady state, single pass reactors. Concept / Approach: In a backmixed reactor, the entire contents are at the outlet concentration, which is the lowest along the reaction path, reducing rate and overall conversion for a given volume. A plug flow reactor maintains the highest concentration at the entrance and decreases along the length, generally achieving higher conversion for the same volume for positive order kinetics. Step-by-Step Solution: Consider r = k * C^n with n > 0.CSTR: uniform concentration equals outlet; rate is limited by the lowest concentration.PFR: concentration is highest at inlet; integral of rate over volume is larger for same size.Thus, CSTR is least suitable when high concentration enhances yield. Verification / Alternative check: Levenspiel plots (1/−r vs. conversion) show larger shaded area for CSTR than PFR at the same conversion for positive orders, confirming the larger required volume and poorer suitability. Why Other Options Are Wrong: Single PFR and PFRs in series keep concentration high near the inlet or in stages; multiple CSTRs in series approach plug flow behavior and are superior to a single CSTR. Common Pitfalls: Assuming backmixing is always desirable because it simplifies control; it can be detrimental to selectivity and conversion when rate is concentration sensitive. Final Answer: Backmixed reactor (CSTR)

Question 18

Flow modeling and reactor comparison: Identify the incorrect statement about vessel dispersion, space time, and ideal flow patterns.

Accepted Answer

Introduction / Context: Ideal reactor models help engineers estimate sizes and performance. Plug flow and complete mixing are two extremes, characterized by dispersion and mixing behavior. Recognizing correct statements avoids design errors. Given Data / Assumptions: D is axial dispersion coefficient, U is superficial velocity, L is reactor length. Constant density system for space time equivalence with residence time. Positive overall reaction order. Concept / Approach: Plug flow implies no axial mixing, which mathematically yields a vessel dispersion number approaching zero. A CSTR is totally mixed, analogous to infinite axial dispersion. For positive orders, PFR requires less volume than a CSTR at the same conversion, not the other way around. Step-by-Step Solution: Evaluate statement a: correct limit behavior for dispersion number.Evaluate b: with constant density, space time = V/Q equals residence time; correct.Evaluate d: ideal tubular reactor has radial mixing only; correct by definition.Therefore c is the incorrect claim. Verification / Alternative check: Levenspiel plot method shows area for CSTR is larger than PFR for positive orders, meaning a larger volume is needed for the CSTR to reach the same conversion. Why Other Options Are Wrong: They are not wrong; they state standard truths about dispersion, space time, and ideal tubular flow. Common Pitfalls: Applying the CSTR advantage in negative order or autocatalytic systems to all kinetics; the comparison reverses depending on order and selectivity goals. Final Answer: For positive reaction orders, a mixed (CSTR) is always smaller than a plug flow reactor for the same duty.

Question 19

Industrial catalysis: in which of the following processes is sulfuric acid commonly used as the catalyst?

Accepted Answer

Introduction / Context: Choosing the correct catalyst is central to refinery and petrochemical operations. Different reactions demand specific catalytic functions: acid catalysis, redox catalysis, or metal surface catalysis. Given Data / Assumptions: Hydrogenation needs metal catalysts such as Ni, Pd, or Pt. Gas phase oxidation of SO2 to SO3 uses V2O5 based catalysts in the contact process. Alkylation is an acid catalyzed reaction. Concept / Approach: Refinery alkylation units produce high octane alkylate by reacting isobutane with olefins. This requires a strong acid catalyst; both sulfuric acid and hydrofluoric acid are industrial choices. Sulfuric acid provides a liquid acid medium to generate carbocations that drive the alkylation mechanism. Step-by-Step Solution: Identify process mechanisms: hydrogenation is metal surface driven, not acid.SO2 oxidation is a redox process, catalyzed by V2O5, not H2SO4.Alkylation relies on strong acids; sulfuric acid is standard in many units. Verification / Alternative check: Process design texts list H2SO4 alkylation units alongside HF units; catalyst circulation and spent acid regeneration practices are well documented in refining practice. Why Other Options Are Wrong: Hydrogenation does not use H2SO4; SO2 to SO3 uses V2O5; hence only alkylation matches sulfuric acid catalysis. Common Pitfalls: Confusing the presence of sulfuric acid in absorption or drying steps with its role as a true catalyst in a particular reaction system. Final Answer: Alkylation of hydrocarbons (e.g., isobutane with olefins)

Question 20

Reactor terminology: when is a reactor commonly termed an autoclave?

Accepted Answer

Introduction / Context: The term autoclave is widely used in laboratory and pilot plants, yet its precise meaning is sometimes blurred. Proper terminology helps avoid safety and design misunderstandings. Given Data / Assumptions: Batch operation, pressure vessel with closure. Capability to operate above atmospheric pressure and often at elevated temperature. Materials and seals selected for pressure service. Concept / Approach: An autoclave is a sealed high pressure batch reactor or sterilizer. It is not a tubular reactor and not limited to atmospheric service. Pressure ratings and code compliance (e.g., ASME) distinguish it from ordinary open tanks. Step-by-Step Solution: Identify batch mode with a pressure tight closure.Note pressure and temperature design beyond ambient conditions.Conclude that the standard usage corresponds to a high pressure batch reactor. Verification / Alternative check: Vendor catalogs and standards describe autoclaves as bench top or floor mounted pressure vessels for reactions, polymerizations, hydrothermal synthesis, or sterilization cycles. Why Other Options Are Wrong: Atmospheric tank or CSTR is not an autoclave; a high pressure tubular reactor is a different configuration (plug flow), not typically labeled autoclave. Common Pitfalls: Assuming every sealed vessel is an autoclave; the term implies pressure rated batch service, not merely a closed lid. Final Answer: When it is a high pressure batch reactor

Chemical Reaction Engineering Questions