Question

Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is \((a) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) \((b) \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2}\) \((c) \mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO}\) \((d) \mathrm{N}_{2} \rightleftharpoons 2 \mathrm{N}\) \((e)\) all of the above

Short answer

a) H2 + 0.5 O2 <=> H2O b) CO + 0.5 O2 <=> CO2 c) N2 + O2 <=> 2NO d) N2 <=> 2N Answer: c) N2 + O2 <=> 2NO

Step-by-step solution

Identify the number of moles of gaseous species on each side

For each reaction, count the number of moles of gaseous species on the reactant side and the product side. a) H2 + 0.5 O2 <=> H2O Reactant side: 1.5 moles Product side: 1 mole b) CO + 0.5 O2 <=> CO2 Reactant side: 1.5 moles Product side: 1 mole c) N2 + O2 <=> 2NO Reactant side: 2 moles Product side: 2 moles d) N2 <=> 2N Reactant side: 1 mole Product side: 2 moles

Applying Le Chatelier's Principle to each reaction

According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in pressure, the equilibrium will shift in a direction that counteracts the change in pressure. For reactions a), b), and d), an increase in pressure will favor the side with fewer moles of gaseous species, while a decrease in pressure will favor the side with more moles of gaseous species. However, for reaction c), both the reactant side and the product side have the same number of moles of gaseous species (2 moles). Therefore, changes in pressure will not affect the equilibrium composition of this reaction.

Identify the correct answer

Since reaction (c) has the same number of moles of gaseous species on both the reactant side and the product side, changes in pressure will not affect its equilibrium composition. Therefore, the correct answer is (c) N2 + O2 <=> 2NO.