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Zorluk: Çok zorIndustrial Chemical Processes and Biotechnology Applications
In the industrial Contact process for the manufacture of tetraoxosulfate(VI) acid, the conversion of sulfur(IV) oxide to sulfur(VI) oxide proceeds according to the thermochemical equation:
2SO2(g)+O2(g)⇌2SO3(g)ΔH=−197 kJ mol−12SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197 \text{ kJ mol}^{-1}
Although Le Chatelier's principle predicts a higher equilibrium yield of SO3(g)SO_3(g) at lower temperatures, the process is industrially operated at a compromise temperature of approximately 450∘C450^\circ\text{C} using a vanadium(V) oxide (V2O5V_2O_5) catalyst. Which of the following best explains why operating at a significantly lower temperature is commercially unviable?
  1. Lowering the temperature significantly reduces the kinetic energy of the reactant molecules and renders the V2O5V_2O_5 catalyst inactive, resulting in an unacceptably slow reaction rate despite the favorable equilibrium position.Cevap
  2. B
    Lowering the temperature causes the V2O5V_2O_5 catalyst to shift the position of equilibrium back toward the reactants, thereby decreasing the overall conversion efficiency of sulfur(IV) oxide.
  3. C
    Lowering the temperature increases the molar volume of gas species beyond 22.4 dm3 mol−122.4 \text{ dm}^3\text{ mol}^{-1} under operating pressures, causing reactant gases to dissociate before colliding.
  4. D
    Lowering the temperature causes sulfur(VI) oxide to condense prematurely alongside unreacted sulfur(IV) oxide, preventing effective separation based on fractional boiling points.

Cevap

Operating at a lower temperature is commercially unviable because lowering the temperature significantly reduces the kinetic energy of reactant molecules and renders the vanadium(V) oxide catalyst inactive, making the rate of reaching equilibrium extremely slow despite a favorable equilibrium yield.
The correct explanation emphasizes the crucial distinction between chemical equilibrium (yield) and reaction rate (kinetics) in industrial chemistry. For exothermic reactions like the oxidation of SO2SO_2, decreasing temperature shifts equilibrium to the right, yielding more product. However, at lower temperatures, the rate of reaction drops exponentially, and the V2O5V_2O_5 catalyst loses its catalytic activity (which requires temperatures ≥400∘C\ge 400^\circ\text{C}). Thus, 450∘C450^\circ\text{C} is chosen as an optimum compromise temperature.

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1
Analyze the thermodynamic profile of the reaction
The forward reaction 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) is exothermic (ΔH=−197 kJ mol−1\Delta H = -197 \text{ kJ mol}^{-1}). According to Le Chatelier's principle, lowering the temperature favors the exothermic direction, increasing the equilibrium concentration of SO3SO_3.
Understanding thermodynamic equilibrium factors governing yield.
2
Analyze kinetic factor and catalyst requirements
Reaction rate depends on activation energy and temperature. At lower temperatures, fewer molecules possess energy E≥EaE \ge E_a. Furthermore, vanadium(V) oxide (V2O5V_2O_5) catalyst is only active above ∼400∘C\sim 400^\circ\text{C}.
Commercial viability requires a balance between yield (thermodynamics) and rate (kinetics).
3
Evaluate the commercial compromise
Operating at 450∘C450^\circ\text{C} provides an optimum compromise: a satisfactory rate of conversion (∼98%\sim 98\% yield) in a reasonable timeframe.
Industrial chemical processes must maximize production rate alongside conversion efficiency.

Anahtar Kavram

Compromise Conditions in Industrial Chemical Synthesis (Kinetics vs. Equilibrium Yield)
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Industrial Chemical Processes and Biotechnology Applications