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Difficulty: MediumNitrogen Gas, Nitrogen Cycle, and Oxides of Nitrogen
During atmospheric lightning strikes, nitrogen gas reacts with oxygen gas to form nitrogen(II) oxide, which further oxidizes to nitrogen(IV) oxide gas (NO2NO_2). What volume of NO2NO_2 gas, measured at 27∘C27^\circ\text{C} and 1 atm1\text{ atm}, is produced when 14.0 g14.0\text{ g} of nitrogen gas reacts completely according to the following equation?
N2(g)+2O2(g)→2NO2(g)N_2(g) + 2O_2(g) \rightarrow 2NO_2(g)
[Molar gas volume at STP = 22.4 dm3mol−122.4\text{ dm}^3\text{mol}^{-1}, atomic mass of N=14.0 g mol−1N = 14.0\text{ g mol}^{-1}, 0∘C=273 K0^\circ\text{C} = 273\text{ K}]
  1. 24.6 dm324.6\text{ dm}^3Answer
  2. B
    22.4 dm322.4\text{ dm}^3
  3. C
    2.22 dm32.22\text{ dm}^3
  4. D
    15.0 dm315.0\text{ dm}^3

Answer

The correct volume of nitrogen(IV) oxide gas produced at 27∘C27^\circ\text{C} and 1 atm1\text{ atm} is 24.6 dm324.6\text{ dm}^3.
The complete reaction converts 14.0 g14.0\text{ g} of N2N_2 (0.50 mol0.50\text{ mol}) into 1.00 mol1.00\text{ mol} of NO2NO_2. At STP (273 K273\text{ K}), 1.00 mol1.00\text{ mol} occupies 22.4 dm322.4\text{ dm}^3. Expanding to 27∘C27^\circ\text{C} (300 K300\text{ K}) using Charles's Law yields 22.4×(300/273)=24.6 dm322.4 \times (300 / 273) = 24.6\text{ dm}^3.

Step-by-Step Solution

1
Calculate the moles of nitrogen gas (N2N_2) reacted.
Moles of N2=14.0 g28.0 g mol−1=0.50 mol\text{Moles of } N_2 = \frac{14.0\text{ g}}{28.0\text{ g mol}^{-1}} = 0.50\text{ mol}.
Molar mass of N2=2×14.0=28.0 g mol−1N_2 = 2 \times 14.0 = 28.0\text{ g mol}^{-1}.
2
Determine moles of NO2NO_2 gas produced using the mole ratio from the balanced equation.
Moles of NO2=0.50 mol N2×2 mol NO21 mol N2=1.00 mol NO2\text{Moles of } NO_2 = 0.50\text{ mol } N_2 \times \frac{2\text{ mol } NO_2}{1\text{ mol } N_2} = 1.00\text{ mol } NO_2.
The balanced chemical equation shows a 1:21:2 mole ratio between N2N_2 and NO2NO_2.
3
Calculate the volume of 1.00 mol1.00\text{ mol} of NO2NO_2 at standard temperature and pressure (STP).
VSTP=1.00 mol×22.4 dm3mol−1=22.4 dm3V_{\text{STP}} = 1.00\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 22.4\text{ dm}^3.
One mole of any ideal gas occupies 22.4 dm322.4\text{ dm}^3 at STP (273 K273\text{ K} and 1 atm1\text{ atm}).
4
Convert the volume from STP (273 K273\text{ K}) to the target temperature (27∘C=300 K27^\circ\text{C} = 300\text{ K}) at constant pressure.
V2=V1×T2T1=22.4 dm3×300 K273 K=24.615 dm3≈24.6 dm3V_2 = V_1 \times \frac{T_2}{T_1} = 22.4\text{ dm}^3 \times \frac{300\text{ K}}{273\text{ K}} = 24.615\text{ dm}^3 \approx 24.6\text{ dm}^3.
According to Charles's Law, volume is directly proportional to absolute temperature when pressure is held constant.

Key Concept

Gas Stoichiometry and Temperature-Volume Relationship (Charles's Law)
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Topics

Nitrogen Gas, Nitrogen Cycle, and Oxides of Nitrogen

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