Question 1

Le Chatelier application — ammonia synthesis N2 + 3 H2 ⇌ 2 NH3: How does fractional conversion change with increasing pressure at constant temperature?

Accepted Answer

Introduction / Context: Ammonia synthesis is a flagship example of equilibrium manipulation in heterogeneous catalysis. Pressure, temperature, and composition are tuned to favor ammonia formation while maintaining practical rates over iron based catalysts. Given Data / Assumptions: Balanced reaction: N2 + 3 H2 ⇌ 2 NH3. Temperature is held constant. Ideal gas behavior approximation for qualitative reasoning. Concept / Approach: Total gas moles decrease from 4 to 2 upon forming ammonia. By Le Chatelier principle, increasing pressure shifts equilibrium toward the side with fewer moles, hence promoting ammonia formation and increasing fractional conversion of reactants to NH3 at a given temperature. Step-by-Step Solution: Count moles: reactants 4, products 2.Apply pressure increase: system minimizes the pressure change by shifting to fewer moles.Therefore equilibrium conversion to NH3 increases. Verification / Alternative check: Equilibrium constant Kp is fixed at a given temperature. The relation between Kp and partial pressures shows that higher total pressure allows compositions with higher ammonia partial pressure while satisfying Kp at equilibrium. Why Other Options Are Wrong: Decrease or no change contradict the effect of pressure on a reaction with a negative change in moles of gas; unpredictability is not correct since stoichiometry is sufficient for the qualitative conclusion. Common Pitfalls: Ignoring the simultaneous effect of temperature on Kp; although high pressure favors NH3, too high a temperature reduces Kp even if it speeds kinetics. Final Answer: Increases

Question 2

Role of catalysts in chemical reactions: Select the correct statement about how a catalyst functions.

Accepted Answer

Introduction / Context: Catalysts are central to modern chemical industry, from petroleum refining to green synthesis. Understanding the real function of a catalyst avoids common misconceptions in kinetics and reactor design. Given Data / Assumptions: A catalyst provides an alternative reaction pathway. Overall stoichiometry and equilibrium constant at a given temperature are unchanged by a catalyst. The catalyst emerges chemically unchanged after a cycle, although it may deactivate physically or chemically over time. Concept / Approach: Activation energy controls the rate according to Arrhenius behavior. A catalyst lowers the energy barrier for the rate determining step by stabilizing the transition state through adsorption, complexation, or surface mediation, thereby increasing rate without altering equilibrium composition at fixed temperature. Step-by-Step Solution: State the rate law sensitivity: k = A * exp(−Ea / RT).Lower Ea increases k at the same T.Catalyst participates in intermediate steps but is regenerated, so net consumption is zero. Verification / Alternative check: Equilibrium constant depends only on thermodynamics (ΔG°). Since catalyst does not change ΔG°, it does not change K. Faster approach to equilibrium confirms catalytic action. Why Other Options Are Wrong: A catalyst does not necessarily initiate an otherwise impossible reaction; it accelerates feasible ones. It is not consumed permanently and can be recovered chemically unchanged, allowing reuse until deactivation occurs. Common Pitfalls: Confusing catalytic participation in intermediates with consumption; overlooking deactivation and regeneration which are operational, not definitional, issues. Final Answer: A catalyst lowers the activation energy of reacting molecules.

Question 3

Parallel irreversible elementary reactions: X → Y with k1 and X → Z with k2. What is the rate of disappearance of species X?

Accepted Answer

Introduction / Context: Parallel reactions are common when a reactant can follow multiple competing pathways. Knowing the overall consumption rate of the common reactant is essential for reactor sizing and selectivity analysis. Given Data / Assumptions: Two elementary and irreversible channels: X → Y with rate r1 = k1 * CX and X → Z with rate r2 = k2 * CX. Isothermal, well mixed conditions; concentration of X is CX (denoted CA in options). No other consumption or production routes for X. Concept / Approach: For parallel first order steps in the same reactant, the total consumption rate is the sum of the individual rates because each pathway removes the same reactant molecules independently. Superposition applies to rates when pathways do not interact kinetically. Step-by-Step Solution: Write r_total = r1 + r2.Substitute r1 = k1 * CX and r2 = k2 * CX.Hence r_total = (k1 + k2) * CX. Verification / Alternative check: Dimension check: k1 and k2 have units of time^(-1) for first order; multiplying by concentration gives concentration per time, which matches the disappearance rate dimension. Why Other Options Are Wrong: Halving either constant or their sum has no basis; the pathways act simultaneously and independently, so the rates add without a factor of one half. Common Pitfalls: Confusing series reactions with parallel reactions or forgetting that selectivity depends on the ratio k1/k2, while the disappearance rate depends on k1 + k2. Final Answer: CA * (K1 + K2)

Question 4

Reaction engineering — constant-volume vs variable-volume batch operation (first order) A first-order irreversible reaction A → B is carried out separately in a constant-volume batch reactor and in a variable-volume batch reactor for the same durati…

Accepted Answer

Introduction / Context: This question tests understanding of first-order kinetics in batch reactors when volume is either constant (typical liquid-phase assumption) or allowed to change (e.g., certain gas-phase systems). It probes whether conversion X and concentration C are affected the same way by volume changes for a first-order irreversible reaction. Given Data / Assumptions: Reaction: A → B, first order in A with rate rA = −k CA. Two separate batch operations, same temperature and same reaction time t. Case 1: Constant volume V = constant. Case 2: Variable volume V = V(t) (e.g., due to gas moles change). Concept / Approach: Conversion X is defined on moles of A, X = (NA0 − NA)/NA0. For first-order batch at a fixed k and fixed t, the exponential decay of NA (moles of A) is governed primarily by the integrated rate law. Differences arise in CA = NA/V. Thus, even when NA (and therefore X) matches at a given t, the concentrations will differ if the volumes differ. Step-by-Step Solution: For constant V: dCA/dt = −k CA ⇒ CA = CA0 exp(−k t); NA = V CA.For variable V(t): NA still decays based on kinetics, while CA = NA/V(t) differs because V changes.At the same t with the same k, the fraction of A reacted (conversion) can be the same; however, CA values are different as V differs. Verification / Alternative check: If volume expands, the same NA corresponds to a smaller CA; if volume contracts, CA is larger. Conversion depends on NA, not directly on V. Why Other Options Are Wrong: (a) Concentrations cannot be identical when volumes differ. (c) Conversion need not differ if time and k are the same. (d) There is a specific correct relationship (option b). Common Pitfalls: Equating concentration behavior with conversion; they are related but not identical in variable-volume systems. Final Answer: Conversion is the same in both cases, but the concentrations are different

Question 5

Gas-phase stoichiometry — fractional volume change For the isothermal gas-phase reaction A → 3B, what is the fractional volume change of the reacting system between zero conversion and complete conversion (assume ideal-gas behavior and constant T, P…

Accepted Answer

Introduction / Context: Fractional volume change is a key parameter in gas-phase reactor design because it affects concentrations and thus rates. For ideal gases at constant temperature and pressure, volume is proportional to total moles. Given Data / Assumptions: Reaction: A → 3B (gas phase). Isothermal, isobaric, ideal-gas behavior. Fractional volume change ε is defined between zero and complete conversion. Concept / Approach: At constant T and P for an ideal gas, V ∝ n(total). For the reaction νr = 1 mole of A to νp = 3 moles of B, the change in total moles per initial mole of A is (3 − 1) = +2. The fractional volume change ε (relative to initial) equals the fractional change in total moles. Step-by-Step Solution: Initial moles per mole A fed at zero conversion: n0 = 1 (taking a 1-mole basis).Final moles at complete conversion: n = 3.Fractional change in moles (and hence volume) = (n − n0)/n0 = (3 − 1)/1 = 2. Verification / Alternative check: General formula for isothermal, isobaric gas reactions: ε = (Σνproducts − Σνreactants)/Σνreactants. Here, ε = (3 − 1)/1 = 2. Why Other Options Are Wrong: 0.5, 1, 3 are not consistent with stoichiometric mole change for A → 3B. Common Pitfalls: Forgetting the constant T, P assumption; under those conditions, volume tracks total moles. Final Answer: 2

Question 6

Temperature sensitivity rule of thumb If the rate of a chemical reaction doubles for every 10 °C rise, by what factor does the rate increase when temperature is raised from 30 °C to 70 °C?

Accepted Answer

Introduction / Context: A common engineering rule of thumb states that many reaction rates approximately double for each 10 °C temperature increase. This is a practical approximation derived from Arrhenius behavior for moderate temperature ranges. Given Data / Assumptions: Rate doubles for every ΔT = 10 °C. Temperature change: 30 °C to 70 °C (ΔT = 40 °C). Concept / Approach: Apply the doubling rule in discrete steps equal to 10 °C increments. Each 10 °C step multiplies the rate by 2. For 40 °C total increase, there are 4 such steps. Step-by-Step Solution: Number of 10 °C steps = (70 − 30)/10 = 4.Overall multiplicative factor = 2^4 = 16.Therefore, rate at 70 °C ≈ 16 times the rate at 30 °C. Verification / Alternative check: Arrhenius equation k = A exp(−Ea/RT) shows exponential sensitivity to 1/T; the doubling rule is an empirical simplification consistent with moderate Ea values over modest ranges. Why Other Options Are Wrong: 8 corresponds to three doublings (30 to 60 °C). 12 is not a power-of-two doubling outcome. 32 would require five doublings (30 to 80 °C). Common Pitfalls: Counting the number of temperature intervals incorrectly or including the starting temperature as a “step.” Final Answer: 16

Question 7

Why reaction rate increases with temperature The observed increase in reaction rate with increasing temperature is primarily due to which factor?

Accepted Answer

Introduction / Context: Temperature profoundly affects reaction rates. Understanding the mechanistic reason is essential for reactor design, safety, and scale-up, as temperature control directly influences productivity and selectivity. Given Data / Assumptions: Molecules must possess energy equal to or greater than the activation energy Ea to react. Collisions must also be favorably oriented and sufficiently energetic—so-called effective collisions. Concept / Approach: With higher temperature, the Maxwell–Boltzmann distribution broadens and shifts, increasing the fraction of molecules with energies ≥ Ea. As a result, the frequency of effective collisions rises, increasing the rate constant per Arrhenius: k = A exp(−Ea/RT). Activation energy itself does not usually decrease with temperature; rather, more molecules surmount Ea. Step-by-Step Solution: As T increases, average kinetic energy increases.A larger fraction of collisions exceed Ea and are effective.Therefore, observed rate increases primarily because effective collision frequency increases. Verification / Alternative check: Arrhenius plots (ln k versus 1/T) exhibit a straight line with slope −Ea/R; as T rises, k increases exponentially, consistent with more effective collisions. Why Other Options Are Wrong: (b) Activation energy is a property of the reaction path; it does not simply decrease with T. (c) While average kinetic energy increases, the key mechanistic statement is the increase in effective collisions; option (a) captures the collision–theory mechanism most directly. Common Pitfalls: Confusing increased average kinetic energy with the specific concept of effective collisions leading to product formation. Final Answer: Increase in the number of effective collisions

Chemical Reaction Engineering Questions