Avogadro's Law: Equal Volumes, Equal Molecules
Avogadro's law
Avogadro's law states that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain equal numbers of molecules.
- This principle is grounded in the idea that the volume of a gas depends on the number of particles it contains, regardless of their type or size.
- Whether the gas consists of single atoms (e.g., helium, He), diatomic molecules (e.g., oxygen, $O_2$, or larger molecules (e.g., carbon dioxide, $CO_2$, Avogadro's Law holds true.
Mathematical Expression of Avogadro's Law
- The relationship described by Avogadro's Law can be expressed mathematically as: $$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$ where:
- $V_1$ and $V_2$ are the volumes of two gas samples.
- $n_1$ and $n_2$ are the amounts of gas (in moles) in the respective samples.
- This equation shows that the volume of a gas is directly proportional to the number of moles, provided the temperature and pressure remain constant.
When solving gas stoichiometry problems, you can use Avogadro's Law to relate volumes directly, without needing to calculate the number of moles.
Applications of Avogadro's Law in Stoichiometry
- Avogadro's Law is especially useful in gas-phase reactions, where the volume ratios of reacting gases mirror their mole ratios in a balanced chemical equation.
- This simplifies calculations and eliminates the need for intermediate mole conversions.
Combustion of Hydrogen Sulfide
The combustion of hydrogen sulfide $H_2$ is represented by the following equation:
$$2 \text{H}_2\text{S (g)} + 3 \text{O}_2\text{(g)} \rightarrow 2 \text{H}_2\text{O (l)} + 2 \text{SO}_2\text{(g)}$$
If $0.908 \, dm^3$ of $H_2S$ gas is combusted, calculate:
- The volume of oxygen $O_2$ gas consumed.
- The volume of sulfur dioxide $SO_2$ gas produced.
Solution
- Volume of $O_2$:
- From the balanced equation, the stoichiometric ratio of $H_2S$ to $O_2$ is (2:3).
- Using Avogadro's Law: $$V(\text{O}_2) = \frac{3}{2} \times V(\text{H}_2\text{S}) = \frac{3}{2} \times 0.908 \, \text{dm}^3 = 1.36 \, \text{dm}^3$$
- Volume of $SO_2$:
- The stoichiometric ratio of $H_2S$ to $SO_2$ is (1:1).
- Therefore, the volume of $SO_2$ produced is equal to the volume of $H_2S$ combusted: $$V(\text{SO}_2) = V(\text{H}_2\text{S}) = 0.908 \, \text{dm}^3$$
- Avogadro's Law applies only to gases.
- Do not use it to calculate the volume of liquid water $H_2O$ produced in this reaction.
Gas Volume Ratios and Reaction Stoichiometry
- Avogadro's Law simplifies gas stoichiometry by allowing you to work directly with gas volumes instead of converting to moles.
- This is particularly helpful when all reactants and products are gases.
Incomplete Combustion of Hydrogen Sulfide
The incomplete combustion of hydrogen sulfide produces elemental sulfur instead of sulfur dioxide:
$$2 \text{H}_2\text{S (g)} + \text{O}_2\text{(g)} \rightarrow 2 \text{H}_2\text{O (l)} + 2 \text{S (s)}$$
If $1.25 \, dm^3$ of oxygen gas is consumed, calculate the volume of hydrogen sulfide gas combusted.
Solution
From the balanced equation, the stoichiometric ratio of $H_2S$ to $O_2$ is (2:1). Using Avogadro's Law:
$$V(\text{H}_2\text{S}) = 2 \times V(\text{O}_2) = 2 \times 1.25 \, \text{dm}^3 = 2.50 \, \text{dm}^3$$
What volume of nitrogen gas $N_2$ is required to produce 3.00 $dm^3$ of ammonia $NH_3$ in the Haber process? The balanced equation is:
$$N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$$
