Problem 61 - General Chemistry: Principles and Modern Applications

Question

Use information from this chapter and previous chapters to write chemical equations to represent the following: (a) equilibrium between nitrogen dioxide and dinitrogen tetroxide in the gaseous state (b) the reduction of nitrous acid by \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}\) forming hydrazoic acid, followed by the reduction of additional nitrous acid by the hydrazoic acid, yielding nitrogen and dinitrogen monoxide (c) the neutralization of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})\) to the second equivalence point by \(\mathrm{NH}_{3}(\mathrm{aq})\).

Step-by-Step Solution

Answer

The chemical reactions are: (a) \(2NO2(g) \leftrightarrows N2O4(g)\), (b) \(2HNO2(aq) + N2H5+(aq) \rightarrow 2HN3(aq) + H2O(l)\), and \(HN3(aq) + HNO2(aq) \rightarrow N2(g) + N2O(g) + H2O(l)\), (c) \(H3PO4(aq) + 2NH3(aq) \rightarrow (NH4)2HPO4(aq)\)

Step 1: Nitrogen Equilibrium

To represent the equilibrium between nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4), the reactants and products should be written on separate sides of the equation, with an equilibrium arrow in between. Both compounds are in the gaseous state: \(2NO2(g) \leftrightarrows N2O4(g)\)

Step 2: Nitrous Acid Reduction

The reduction of nitrous acid (HNO2) by N2H5+ forms hydrazoic acid (HN3). Another HNO2 then reacts with the HN3 to yield nitrogen (N2) and dinitrogen monoxide(N2O): \(2HNO2(aq) + N2H5+(aq) \rightarrow 2HN3(aq) + H2O(l)\), \(HN3(aq) + HNO2(aq) \rightarrow N2(g) + N2O(g) + H2O(l)\)

Step 3: PHosphoric Acid Neutralization

The neutralization of phosphoric acid (H3PO4) to the second equivalence point by ammonia (NH3). This means that two equivalence of ammonia react with one of phosphoric acid: \(H3PO4(aq) + 2NH3(aq) \rightarrow (NH4)2HPO4(aq)\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction in a chemical process. Imagine a seesaw perfectly balanced in the middle. In the context of gaseous substances, such as nitrogen dioxide (\( \text{NO}_2 \)) and dinitrogen tetroxide (\( \text{N}_2\text{O}_4 \)), equilibrium becomes a visual demonstration of balance.
In the equation \( 2 \text{NO}_2(g) \leftrightarrows \text{N}_2\text{O}_4(g) \), you'll see a special arrow. This equilibrium arrow signals the reaction can go both ways.
This does not mean nothing is happening; it's like a busy highway that maintains balance in how many cars enter and exit.

• When nitrogen dioxide gases collide, they can form dinitrogen tetroxide.
• Conversely, dinitrogen tetroxide can split apart back into nitrogen dioxide.

At equilibrium, both these processes occur at the same rate, keeping the concentration of \( \text{NO}_2 \) and \( \text{N}_2\text{O}_4 \) constant, though not necessarily equal.

Nitrous Acid Reduction

The reduction of nitrous acid involves transforming it into something simpler. When nitrous acid (\( \text{HNO}_2 \)) meets \( \text{N}_2\text{H}_5^+ \), a change takes place that eventually leads to the creation of nitrogen (\( \text{N}_2 \)) and dinitrogen monoxide (\( \text{N}_2\text{O} \)). Crunching through the steps:
The first reaction: \( 2\text{HNO}_2(aq) + \text{N}_2\text{H}_5^+(aq) \rightarrow 2\text{HN}_3(aq) + \text{H}_2\text{O}(l) \).
In this dance, \( \text{N}_2\text{H}_5^+ \) steps in and fosters the cytogenesis of hydrazoic acid (\( \text{HN}_3 \)). As they bow out, the cycle ignites the next reaction.

• Hydrazoic acid,\( \text{HN}_3 \), extends its hand to another \( \text{HNO}_2 \) molecule, continuing its task.
• This welcomed engagement: \( \text{HN}_3(aq) + \text{HNO}_2(aq) \rightarrow \text{N}_2(g) + \text{N}_2\text{O}(g) + \text{H}_2\text{O}(l) \).

The end result is the production of nitrogen and dinitrogen monoxide, with water (H2O) gracefully returning to the wings.

Acid-Base Neutralization

Neutralization reactions are like polite exchanges where acids and bases trade pleasantries to form a salt and water. When a strong acid such as phosphoric acid (\( \text{H}_3\text{PO}_4 \)) meets ammonia (\( \text{NH}_3 \)), an exciting neutralization journey begins.
Here, we focus until the second equivalence point, meaning each phosphoric acid molecule pairs with two ammonia molecules:
\( \text{H}_3\text{PO}_4(aq) + 2\text{NH}_3(aq) \rightarrow (\text{NH}_4)_2\text{HPO}_4(aq) \).

• Each \( \text{NH}_3 \) contributes a hydrogen ion exchange, eventually forming \( \text{NH}_4^+ \), which pairs with phosphate ions.
• This creates ammonium phosphate, a soluble salt.

The result of this neutralization is not just salt formation, but also the delightful completion of balancing acidic and basic components. Through these processes, the chemical world showcases its knack for creating order and stability.