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Difficulty: HardIndustrial Chemical Processes and Biotechnology Applications

In the industrial synthesis of trioxonitrate(V) acid (HNO3\text{HNO}_3) via the Ostwald process, ammonia (NH3\text{NH}_3) gas is oxidized by oxygen over a platinum-rhodium gauze catalyst at approximately 850∘C850^\circ\text{C} and 5 atm5\text{ atm}. Which balanced chemical equation correctly represents this initial catalytic oxidation step, and what is the precise thermodynamic role of the catalyst?

  1. 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g); the catalyst increases the forward reaction rate by lowering the activation energy without shifting the position of dynamic equilibrium.Answer
  2. B
    4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g); the catalyst shifts the position of dynamic equilibrium to the right to increase the overall equilibrium yield of NO(g)\text{NO}(g).
  3. C
    2NH3(g)+2O2(g)→N2O(g)+3H2O(g)2\text{NH}_3(g) + 2\text{O}_2(g) \rightarrow \text{N}_2\text{O}(g) + 3\text{H}_2\text{O}(g); the catalyst lowers the activation energy so that 22.4 dm322.4\text{ dm}^3 of NH3\text{NH}_3 at 850∘C850^\circ\text{C} produces exactly 1 mole1\text{ mole} of N2O(g)\text{N}_2\text{O}(g).
  4. D
    4NH3(g)+7O2(g)→4NO2(g)+6H2O(g)4\text{NH}_3(g) + 7\text{O}_2(g) \rightarrow 4\text{NO}_2(g) + 6\text{H}_2\text{O}(g); the catalyst acts as a limiting reactant that is consumed to maximize the rate of formation of NO2(g)\text{NO}_2(g).

Answer

The equation 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g) correctly represents the initial stage of the Ostwald process, where the platinum-rhodium catalyst lowers the activation energy to increase the reaction rate without altering the position of equilibrium.
In the Ostwald process for manufacturing trioxonitrate(V) acid (HNO3\text{HNO}_3), the initial reaction is the exothermic oxidation of ammonia gas to nitrogen(II) oxide gas according to the equation 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g). The platinum-rhodium wire gauze catalyst accelerates the reaction by lowering the activation energy barrier, enabling rapid formation of NO\text{NO} at high operating temperatures (850∘C850^\circ\text{C}). It does not affect the position of dynamic equilibrium or change the equilibrium yield of products.

Step-by-Step Solution

1
Identify the primary chemical reaction in stage 1 of the Ostwald process
Ammonia reacts with oxygen in a 4:5 mole ratio to yield nitrogen(II) oxide and steam: 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g).
Direct catalytic oxidation of ammonia produces NO\text{NO} gas first; subsequent cooling and oxidation converts NO\text{NO} to NO2\text{NO}_2 before absorption in water.
2
Analyze the catalytic role of the platinum-rhodium gauze
The catalyst provides an alternative reaction pathway with a lower activation energy (EaE_a).
Catalysts increase the rate of attainment of equilibrium equally in both forward and reverse directions but do not change ΔH\Delta H, ΔG\Delta G, or the position of equilibrium (KeqK_{eq}).

Key Concept

Ostwald Process and Catalytic Principles
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Topics

Industrial Chemical Processes and Biotechnology Applications