Le Chatelier's Principle Chemistry Example 4

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Example 4

hard
The contact process produces sulfur trioxide: 2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \rightleftharpoons 2\text{SO}_3\text{(g)}, ΔH=−198 kJ\Delta H = -198\,\text{kJ}. Explain why the industrial process uses: (a) high pressure, (b) moderate temperature (450°C450°\text{C}) rather than low temperature, and (c) a V2O5\text{V}_2\text{O}_5 catalyst.

Solution

  1. 1
    (a) High pressure: reactant side has 3 moles of gas (2+12 + 1), product side has 2 moles. By Le Chatelier's, high pressure shifts equilibrium right, increasing SO3\text{SO}_3 yield.
  2. 2
    (b) The reaction is exothermic (ΔH<0\Delta H < 0), so low temperature would favor products. However, at low temperature the reaction is too slow. A moderate temperature of 450°C450°\text{C} is a compromise between yield and rate.
  3. 3
    (c) The V2O5\text{V}_2\text{O}_5 catalyst speeds up both forward and reverse reactions equally, allowing equilibrium to be reached faster without changing the equilibrium position. It does not affect ΔH\Delta H or KeqK_{eq}.

Answer

High P (fewer gas moles), moderate T (rate/yield compromise), catalyst (speed)\text{High P (fewer gas moles), moderate T (rate/yield compromise), catalyst (speed)}
Industrial chemical processes apply Le Chatelier's principle to maximize yield while maintaining economically viable reaction rates. The contact process is a perfect example of balancing thermodynamic and kinetic considerations.

About Le Chatelier's Principle

When a system at chemical equilibrium is subjected to an external stress — such as a change in concentration, pressure, or temperature — the equilibrium.

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