Problem 96 - Chemistry The Central Science In Si Units

Question

The commercial production of nitric acid involves the following chemical reactions: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$(a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. (c) How many grams of ammonia must you start with to make \(1000.0 \mathrm{~L}\) of a \(0.150 \mathrm{M}\) aqueous solution of nitric acid? Assume all the reactions give \(100 \%\) yield.

Step-by-Step Solution

Answer

(a) The first and second reactions are redox reactions. (b) Nitrogen is oxidized and oxygen is reduced in both reactions. (c) You need 5109.0 grams of ammonia.

Step 1: Identify Redox Reactions

A redox reaction involves the transfer of electrons between substances. For each given reaction, identify if there is a change in oxidation states of any elements involved. The first reaction, \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\), shows nitrogen in \(\mathrm{NH}_3\) changing from \(-3\) to \(+2\) in \(\mathrm{NO}\) and oxygen going from \(0\) in \(\mathrm{O}_2\) to \(-2\) in \(\mathrm{H_2O}\). This is a redox reaction. The second reaction, \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\), also involves changes in oxidation states (nitrogen from \(+2\) to \(+4\) and oxygen from \(0\) to \(-2\)), thus it is a redox reaction. The third reaction, \(3 \mathrm{NO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q) + \mathrm{NO}(g)\), only shows the physical combination of elements without changes in oxidation states for nitrogen and oxygen, so it is not a redox reaction.

Step 2: Identify Elements Undergoing Oxidation and Reduction

For the redox reactions identified, determine which elements lose electrons (oxidized) and which gain electrons (reduced). In the first reaction, nitrogen in \(\mathrm{NH}_3\) increases from \(-3\) to \(+2\), indicating oxidation, while oxygen decreases from \(0\) to \(-2\), indicating reduction. In the second reaction, nitrogen in \(\mathrm{NO}\) increases from \(+2\) to \(+4\), showing oxidation, and oxygen decreases from \(0\) to \(-2\), indicating reduction.

Step 3: Calculate Moles of Nitric Acid Needed

Use the formula for molarity, \(M = \frac{n}{V}\), where \(n\) is the number of moles and \(V\) is the volume in liters. The problem states a desired volume of \(1000.0 \mathrm{~L}\) and molarity of \(0.150 \mathrm{M}\). Calculate the moles of \(\mathrm{HNO}_3\) required: \[n = M \times V = 0.150 \times 1000.0 = 150.0 \text{ moles of } \mathrm{HNO}_3\].

Step 4: Calculate Moles of Ammonia Needed

From the stoichiometry of the reactions: 4 moles of \(\mathrm{NH}_3\) produce 4 moles of \(\mathrm{NO}\), which then converts ultimately to 2 moles of \(\mathrm{HNO}_3\) via the reactions. Therefore, the moles of ammonia needed for 150.0 moles of \(\mathrm{HNO}_3\) is: \[\frac{4 \text{ moles of } \mathrm{NH}_3}{2 \text{ moles of } \mathrm{HNO}_3} \times 150.0 \text{ moles of } \mathrm{HNO}_3 = 300.0 \text{ moles of } \mathrm{NH}_3\].

Step 5: Convert Moles of Ammonia to Grams

Use the molar mass of ammonia (\(\mathrm{NH}_3\)), which is \(17.03 \text{ g/mol}\), to find the mass in grams. \[300.0 \text{ moles of } \mathrm{NH}_3 \times 17.03 \text{ g/mol} = 5109.0 \text{ grams of } \mathrm{NH}_3\].

Key Concepts

Oxidation States in Redox Reactions

Redox reactions depend heavily on the changes in oxidation states. Oxidation states, also known as oxidation numbers, are a way to keep track of electrons in a chemical reaction. They indicate the degree of oxidation a substance has undergone. In essence, an oxidation state is the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic. This helps chemists understand the electron transfer during the reaction.

To determine if a reaction is redox, first assign oxidation states to each element in the reaction. For example, in ammonia (\(\text{NH}_3\)), nitrogen has an oxidation state of \(-3\). When ammonia reacts to form nitrogen monoxide (\(\text{NO}\)), nitrogen changes to \(+2\). This shift from \(-3\) to \(+2\) shows that nitrogen has lost electrons, indicating oxidation.

Similarly, oxygen in diatomic form (\(\text{O}_2\)) has an oxidation state of \(0\), which changes to \(-2\) in water (\(\text{H}_2\text{O}\)). This indicates a gain of electrons, signifying reduction. Whenever you see a change in oxidation states, especially in terms of losing (oxidation) or gaining (reduction) electrons, you are witnessing a redox reaction.

Understanding Chemical Stoichiometry

Chemical stoichiometry involves calculations that relate amounts of reactants and products in a chemical reaction. It's like a recipe, showing how much of one substance you'll need to react with another.

In our reaction series, stoichiometry helps us balance the amount of each substance. For example, starting with ammonia (\(\text{NH}_3\)), each mole can produce a corresponding product like nitric acid (\(\text{HNO}_3\)). The balanced equations show that 4 moles of ammonia produce 4 moles of nitrogen monoxide, and through further reactions, these ultimately relate to 2 moles of nitric acid.

Stoichiometry helps determine how much product you'll end up with, based on your starting substances. Knowing this relationship is essential for practical applications, such as industrial production, where maximizing yield is important while minimizing waste.

Molar Calculations and Their Importance

Molar calculations are central to converting between different units in chemistry. They are what allow chemists to go from moles, a unit related to the number of particles, to grams, a more tangible and measurable quantity.

In our exercise, we calculated that to make \(1000.0 \text{ L}\) of a \(0.150 \text{ M}\) nitric acid solution, we need 150.0 moles of \(\text{HNO}_3\). This requires using the relationship between moles and molarity, where molarity (\(M\)) equals the number of moles (\(n\)) per volume (\(V\)) in liters.

Once we know how many moles of ammonia are needed, molar mass comes into play. The molar mass of \(\text{NH}_3\) is \(17.03 \text{ g/mol}\). By multiplying the number of moles by the molar mass, we find out how many grams of ammonia are needed, resulting in a practical number: \(5109.0 \text{ grams}\). This bridge from moles to grams is crucial in lab settings, as it translates theoretical chemistry into practical, real-world applications.