Question

7. Gas Law Problem Nitric oxide (NO) reacts with molecular oxygen 2NO(g) + O2(g) → 2NO2(g). Initially, equal volumes of the reactant gases are separated in a 2.0 L container. The pressure of the NO is 1.25 atm, while the pressure of the O2 is 0.85 atm. What are the partial pressures of the gases once the reaction has gone to completion. Assume the temperature remains constant at 0°C.

Answer 1

The reaction is

2 NO (g) + O₂ (g) --------> 2 NO₂ (g)

Number of moles of NO gas = PV/RT = (1.25 atm * 2.0 L)/(0.0821 L.atm.mol^–1.K^–1 * 273.15 K) = 0.11148 mol

Number of moles of O₂ gas = PV/RT = (0.85 atm * 2.0 L)/(0.0821 L.atm.mol^–1.K^–1 * 273.15 K) = 0.075806 mol

According to balanced equation,

2 moles of NO reacts with 1 mole of O₂ gas

So, 0.11148 mol of NO would react with 0.11148 mol* ½ = 0.05574 mol of O₂ gas

But, the number of moles of O₂ gas present in excess. So, O₂ is the excess reagent.

NO is the Limiting reagent.

According to balanced equation,

2 moles of NO produces 2 moles of NO₂

So, 0.11148 mol of NO would produce 0.11148 mol of NO₂ gas

Number of moles of NO₂ gas = 0.11148 mol

Number of moles of NO gas remaining = 0 mol ( as it is the Limiting reagent)

Number of moles of O₂ gas remaining= 0.075806 – 0.05574 = 0.020066 mol

Partial pressure of NO₂ gas = nRT/V = (0.11148 mol * 0.0821 L.atm.mol^–1.K^–1 * 273.15 K/2.0 L) = 1.25 atm

Partial pressure of NO gas = 0.00 atm

Partial pressure of O₂ gas = nRT/V = (0.020066 mol * 0.0821 L.atm.mol^–1.K^–1 * 273.15 K/2.0 L) = 0.225 atm