Question

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If 25.0 \(\mathrm{mL}\) \(\mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Short answer

The volume of N₂O₄ gas formed when 25.0 mL of NO₂ gas is completely converted under the same conditions is 12.5 mL.

Step-by-step solution

Find moles of NO₂ gas

First, we need to find the number of moles of NO₂ using the volume given. Since the gas conditions are the same, we can assume equal molar volumes for the two gases. We will use the formula: Number of moles (NO₂) = \(\frac{Volume\: of\: NO₂}{Molar\: Volume}\) However, we don't have the molar volume, so let's substitute it with a variable m: Number of moles (NO₂) = \(\frac{25.0}{m}\)

Find moles of N₂O₄ gas

Now, we need to find the number of moles of N₂O₄ using stoichiometry, which is the mole ratio between NO₂ and N₂O₄ in the balanced equation. The balanced equation is: $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ Based on the stoichiometry, 2 moles of NO₂ produce 1 mole of N₂O₄. So the mole ratio is 2:1. Number of moles (N₂O₄) = \(\frac{Number\: of\: moles\: (NO₂)}{2}\) Now, substitute the value we found in Step 1 for moles of NO₂: Number of moles (N₂O₄) = \(\frac{\frac{25.0}{m}}{2}\)

Find the volume of N₂O₄ gas

To find the volume of N₂O₄ gas, we will use the same formula that we used to find the moles of NO₂: Number of moles (N₂O₄) = \(\frac{Volume\: of\: N₂O₄}{Molar\: Volume}\) Now, we can set up the equation as follows: \(\frac{\frac{25.0}{m}}{2} = \frac{Volume\: of\: N₂O₄}{m}\) We can solve for the volume of N₂O₄ by multiplying both sides by the molar volume (m): \(\frac{25.0}{2} = Volume\: of\: N₂O₄\)

Calculate the volume of N₂O₄ gas

Now we can simply calculate the volume of N₂O₄ gas: Volume of N₂O₄ = \(\frac{25.0}{2} = 12.5 \mathrm{mL}\) The resulting volume of N₂O₄ gas is 12.5 mL.

Key concepts

Chemical Equations

Chemical equations are like the language of chemistry. They represent chemical reactions where reactants are transformed into products. The equation consists of chemical formulas that detail the substances involved. In our example of converting NO₂ to N₂O₄, the equation shows us that two molecules of nitrogen dioxide gas (NO₂) react together to form a single molecule of dinitrogen tetroxide gas (N₂O₄).

Balancing the chemical equation is crucial because it provides the exact proportions of the reactants and products and obeys the Law of Conservation of Mass, meaning mass is neither created nor destroyed in a chemical reaction. So, the number of atoms for each element in the reactants must equal the number in the products. In this example, the balanced chemical equation is: \[ 2 \text{NO}_{2}(g) \rightarrow \text{N}_{2} \text{O}_{4}(g) \] This tells us that two moles of NO₂ will produce one mole of N₂O₄ under the conditions defined by the reaction.

Gas Laws

Gas laws are mathematical relationships that describe the behavior of gases. They allow us to predict changes to the properties of gases, like volume and pressure, under different conditions. Although not explicitly used in this exercise, understanding gas laws such as Boyle's Law, Charles's Law, and Avogadro's Law can be very helpful.

For instance, Avogadro's Law states that equal volumes of gases, at the same temperature and pressure, contain the same number of molecules or moles. In our exercise, the conditions are constant, so we can assume that the volumes given are directly proportional to the number of moles involved. This principle underpins our calculations, allowing us to convert moles directly to volume.

Moles Conversion

The concept of moles is fundamental in chemistry. A mole is a unit that measures an amount of substance. It allows chemists to count particles by weighing them, making dealing with atoms and molecules much more practical.

Moles conversion is the process of using stoichiometry to convert from one substance's moles to another's during a chemical reaction. This uses the coefficients from the balanced chemical equation.

In our example, the balanced equation \( 2 \text{NO}_{2} \rightarrow \text{N}_{2} \text{O}_{4} \) provides a 2:1 mole ratio. Thus, if you know the number of moles of NO₂, the moles of N₂O₄ can be calculated by dividing by two. This step allows us to bridge the gap from what we have (NO₂) to what we want to find (N₂O₄). Here, moles of NO₂ are derived from the volume, using the assumed constant conditions.

Volume Calculation

Volume calculation in gas reactions can sometimes be straightforward using stoichiometry when the conditions are constant. It relies heavily on the relationships and ratios established in the balanced chemical equation.

In this exercise, we start with the 25.0 mL of NO₂ and use the mole ratio from the balanced chemical equation to find the corresponding volume of N₂O₄. Since 2 moles of NO₂ form 1 mole of N₂O₄, the volume of N₂O₄ occupies half the volume of NO₂ if all other conditions remain constant.

Thus, by applying the stoichiometric ratio, the volume of N₂O₄ is calculated as 12.5 mL, showing how closely volume ties to mole calculations when conditions do not change. This illustration is a simplification that highlights stoichiometric principles without bringing pressure and temperature variables into play.