In the reaction N₂ + O₂ ⇌ 2NO, the moles/litre of N₂, O₂ and NO respectively 0.25, 0.05 and 1.0 at equilibrium, the initial concentration of N₂ and O₂ will be respectively.

• A. 0.75 mol/litre, 0.55 mole/litre.
• B. 0.50 mol/litre, 0.75 mole/litre.
• C. 0.25 mol/litre, 0.50 mole/litre.
• D. 0.25 mol/litre, 1.0 mole/litre.

To find the initial concentrations of N₂ and O₂ in the reaction N₂ + O₂ ⇌ 2NO, we can use the equilibrium concentrations provided. At equilibrium, the concentrations are as follows: N₂ = 0.25 mol/L, O₂ = 0.05 mol/L, and NO = 1.0 mol/L.

Setting Up the Reaction

The balanced equation shows that one mole of N₂ reacts with one mole of O₂ to produce two moles of NO. Therefore, we can express the changes in concentration as follows:

• Let x be the change in concentration of N₂ and O₂ that reacted.
• At equilibrium, the concentration of NO is 1.0 mol/L, which means 0.5 moles of N₂ and O₂ have reacted (since 2 moles of NO are produced from 1 mole of each reactant).

Calculating Initial Concentrations

From the equilibrium concentrations:

• For N₂: Initial concentration = Equilibrium concentration + Change = 0.25 mol/L + 0.5 mol/L = 0.75 mol/L
• For O₂: Initial concentration = Equilibrium concentration + Change = 0.05 mol/L + 0.5 mol/L = 0.55 mol/L

Final Answer

The initial concentrations of N₂ and O₂ are:

0.75 mol/L for N₂ and 0.55 mol/L for O₂.

Thus, the correct option is A. 0.75 mol/litre, 0.55 mole/litre.