For the reaction in equilibrium $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2…

For the reaction in equilibrium $\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \Delta \mathrm{H}=-\mathrm{Q}$ Reaction is favoured in forward direction by:
  1. use of catalyst
  2. decreasing concentration of $\mathrm{N}_2$
  3. low pressure, high temperature and high concentration of ammonia
  4. high pressure, low temperature and higher concentration of $\mathrm{H}_2$

Solution

$\mathrm{N}_2(\mathrm{~g})+3 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_3(\mathrm{~g}), \Delta \mathrm{H}=-\mathrm{Q}$ According to Le Chatelier's principle. - Exothermic reactions are favoured at low temperature. - Increase in pressure shifts the reaction in direction having lesser number of moles. Hence, the given reaction shifts forward on increasing pressure. - Increasing the concentration of reactants shifts the reaction in forward direction. So high concentration of $\mathrm{H}_2$ shifts reaction in forward direction.

Asked in: NEET 2024 (Re-NEET)

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