Difficulty: Medium
Correct Answer: Yield of B increases as temperature increases
Explanation:
Introduction / Context:
In consecutive reaction networks, optimizing selectivity to the intermediate product B is a classic design problem. Temperature alters relative rates via Arrhenius sensitivities, especially when activation energies differ markedly.
Given Data / Assumptions:
Concept / Approach:
Rate constants: k1 ∝ exp(-E1/RT), k2 ∝ exp(-E2/RT). The selectivity toward B depends on the ratio k1/k2 and on residence time. Because E1 is much larger, increasing temperature amplifies k1 more strongly than k2. Thus k1/k2 increases with T, favoring formation of B relative to its consumption to C and improving achievable yield/selectivity of B for a given space time.
Step-by-Step Solution:
Consider ratio: k1/k2 = exp(-(E1 - E2)/(RT)).With E1 > E2, the exponent increases (becomes less negative) as T rises.Therefore, k1/k2 increases with T, improving B formation relative to loss.Hence, yield of B increases with temperature (up to constraints like side reactions or thermal limits).
Verification / Alternative check:
Optimal temperature-residence time strategies show higher selectivity windows at elevated T when the desired step has higher activation energy than the undesired subsequent step.
Why Other Options Are Wrong:
Common Pitfalls:
Ignoring the need to limit residence time to prevent over-conversion of B to C at very high temperatures; selectivity involves both kinetics and contact time.
Final Answer:
Yield of B increases as temperature increases
Discussion & Comments