Problem 15 - Chemistry and Chemical Reactivity

Question

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})$$ (a) You wish to react \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio. The sample of NO has a volume of 150 mL. What volume of \(\mathrm{O}_{2}\) is required (at the same pressure and temperature)? (b) What volume of \(\mathrm{NO}_{2}\) (at the same pressure and temperature) is formed in this reaction?

Step-by-Step Solution

Answer

75 mL of \(\mathrm{O}_{2}\) is required, and 150 mL of \(\mathrm{NO}_{2}\) is formed.

Step 1: Identify Reaction Stoichiometry

The balanced chemical equation is \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\), indicating that 2 volumes of NO react with 1 volume of \(\mathrm{O}_{2}\) to produce 2 volumes of \(\mathrm{NO}_{2}\). This implies a 2:1:2 stoichiometric ratio of \(\mathrm{NO}\):\(\mathrm{O}_{2}\):\(\mathrm{NO}_{2}\).

Step 2: Calculate Volume of Oxygen Required

Given 150 mL of \(\mathrm{NO}\), and using the stoichiometric ratio 2:1, the required volume of \(\mathrm{O}_{2}\) is half the volume of \(\mathrm{NO}\). Thus, \( \frac{150 \text{ mL}}{2} = 75 \text{ mL} \) of \(\mathrm{O}_{2}\) is required.

Step 3: Determine Volume of Nitrogen Dioxide Formed

Since the molar volume ratio of \(\mathrm{NO}\):\(\mathrm{NO}_{2}\) is 1:1, the volume of \(\mathrm{NO}_{2}\) formed is equal to the initial volume of \(\mathrm{NO}\) reacted. Therefore, 150 mL of \(\mathrm{NO}_{2}\) is formed.

Key Concepts

Balanced Chemical Equation

A balanced chemical equation is fundamental to understanding reactions. It ensures that the number of atoms for each element is equal on both sides of the equation. This balance reflects the conservation of mass principle, where no atoms are lost or gained during the reaction.
For the reaction: \[2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\]

• The equation is balanced, meaning there are 2 nitrogen atoms and 4 oxygen atoms on both sides.
• This balance helps us understand the reaction’s stoichiometry, which is crucial for accurately calculating reactant and product volumes.

Understanding a balanced equation provides the groundwork for solving stoichiometry problems in chemistry.

Volume Ratio

The volume ratio in a gas reaction provides insight into how much of each gas is used or produced. When gases react, their volumes are often directly proportional to their mole ratios if conditions are constant. This means we can use the coefficients from the balanced equation.
In our reaction:

• The volume ratio of \(\mathrm{NO}\) to \(\mathrm{O}_2\) is 2:1.
• The volume ratio of \(\mathrm{NO}\) to \(\mathrm{NO}_2\) is 1:1.

These ratios are vital when determining how much of a substance is needed or produced without needing to convert to moles.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to determine the proportions of reactants and products. It is crucial for calculating quantities needed in a chemical reaction.
The stoichiometric ratio is derived from the coefficients of each substance in the balanced equation. In our example:

• 2 moles of \(\mathrm{NO}\) react with 1 mole of \(\mathrm{O}_2\) to produce 2 moles of \(\mathrm{NO}_2\).
• This ratio helps in calculating the exact amounts needed or produced.

Applying stoichiometry, 150 mL of \(\mathrm{NO}\) requires 75 mL of \(\mathrm{O}_2\) and produces 150 mL of \(\mathrm{NO}_2\). This approach ensures precision in chemical reactions.

Gas Reactions

Gas reactions behave differently due to the properties of gases. One key feature is that, under the same conditions of temperature and pressure, equal volumes of gases contain equal numbers of molecules, according to Avogadro's law. This is why gas volume ratios can be directly taken from the coefficients in a balanced equation.
This implies that in the reaction:

• 150 mL of \(\mathrm{NO}\) under the same conditions require 75 mL of \( \mathrm{O}_2 \).
• The same 150 mL of \(\mathrm{NO}\) will produce 150 mL of \(\mathrm{NO}_2\).

Gas reactions are straightforward to analyze because volume, rather than mass, can be directly used. This simplifies calculations considerably, especially when dealing with reactions at constant temperature and pressure.