Question

Oxygen dissolved in water can cause corrosion in hot-water heating systems. To remove oxygen, hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) is often added. Hydrazine reacts with dissolved \(\mathrm{O}_{2}\) to form water and \(\mathrm{N}_{2}.\) (a) Write a balanced chemical equation for the reaction of hydrazine and oxygen. Identify the oxidizing and reducing agents in this redox reaction. (b) Calculate \(\Delta_{\mathrm{r}} H^{\circ}, \Delta_{\mathrm{r}} S^{\circ},\) and \(\Delta_{\mathrm{r}} G^{\circ}\) for this reaction involving \(1 \mathrm{mol}\) of \(\mathrm{N}_{2} \mathrm{H}_{4}\) at \(25^{\circ} \mathrm{C}.\) (c) Because this is an exothermic reaction, energy is evolved as heat. What temperature change is expected in a heating system containing \(5.5 \times 10^{4} \mathrm{L}\) of water? (Assume no energy is lost to the surroundings.) (d) The mass of a hot-water heating system is \(5.5 \times 10^{4}\) kg. What amount of \(\mathrm{O}_{2}\) (in moles) would be present in this system if it is filled with water saturated with O \(_{2} ?\) (The solubility of \(\mathrm{O}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is 0.000434 g per \(100 \mathrm{g}\) of water. (e) Assume hydrazine is available as a \(5.0 \%\) solution in water. What mass of this solution should be added to totally consume the dissolved \(\mathrm{O}_{2}[\text { described in part }(\mathrm{d})] ?\) (f) Assuming the \(\mathrm{N}_{2}\) escapes as a gas, calculate the volume of \(\mathrm{N}_{2}(\mathrm{g})\) (measured at \(273 \mathrm{K}\) and \(1.00 \mathrm{atm}\) ) that will be produced.

Short answer

(a) Balanced equation: \(\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{N}_2\). (b-f) Thermodynamic calculations and stoichiometry determine heat, moles, and volume involved in the reaction.

Step-by-step solution

Balance the Equation

First, write the unbalanced chemical reaction: \(\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow \mathrm{H}_2\mathrm{O} + \mathrm{N}_2\). To balance it, ensure oxygen and nitrogen atoms are equal on both sides of the equation. The balanced equation is: \(\mathrm{N}_2\mathrm{H}_4 + \mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{N}_2\). Identify oxidizing and reducing agents: \(\mathrm{O}_2\) is reduced to \(\mathrm{N}_2\), acting as the oxidizing agent, and \(\mathrm{N}_2\mathrm{H}_4\) is oxidized to \(\mathrm{N}_2\), acting as the reducing agent.

Calculate Thermodynamic Values

Use standard enthalpies (\(\Delta_f H^{\circ}\)), entropies (\(\Delta_f S^{\circ}\)), and Gibbs free energies (\(\Delta_f G^{\circ}\)) of formation for \(\mathrm{N}_2\mathrm{H}_4\), \(\mathrm{O}_2\), \(\mathrm{H}_2\mathrm{O}\), and \(\mathrm{N}_2\) to calculate \(\Delta_r H^{\circ}\), \(\Delta_r S^{\circ}\), and \(\Delta_r G^{\circ}\) for the reaction. For 1 mole of \(\mathrm{N}_2\mathrm{H}_4\):\[\Delta_r H^{\circ} = \sum (\Delta_f H^{\circ}_{\text{products}}) - \sum (\Delta_f H^{\circ}_{\text{reactants}}) = [-571.66 + 0] - [-50.63 + 0] \text{ kJ/mol}\]Similarly calculate \(\Delta_r S^{\circ}\) and \(\Delta_r G^{\circ}\). Values required can be found in standard tables.

Calculate Temperature Change

The heat evolved in the exothermic reaction can be calculated using \(q = n \times \Delta_r H^{\circ}\). The temperature change, \(\Delta T\), is given by \(q = mc\Delta T\), where \(m\) is the mass of water and \(c\) is the specific heat capacity of water (4.18 J/g°C). Substitute \(m = 5.5 \times 10^7\) g and solve for \(\Delta T\).

Calculate Oxygen Moles

The solubility of \(\mathrm{O}_2\) is 0.000434 g per 100 g of water. The total mass of water is \(5.5 \times 10^7\) g. Calculate grams of \(\mathrm{O}_2\) in water using the given solubility and convert to moles using the molar mass of \(\mathrm{O}_2\) (32 g/mol).

Calculate Mass of Hydrazine Solution

Calculate moles of \(\mathrm{N}_2\mathrm{H}_4\) needed to fully react with the moles of \(\mathrm{O}_2\) from Step 4. Since it reacts in a 1:1 molar ratio, use the moles of \(\mathrm{O}_2\) to find the required mass of hydrazine solution. Given that the solution is 5.0% by mass, calculate the total mass of the solution needed.

Calculate Volume of Nitrogen Gas

The molar volume of gas at STP (273 K and 1 atm) is 22.4 L/mol. Use the moles of \(\mathrm{N}_2\) produced (same as \(\mathrm{O}_2\) moles due to the reaction stoichiometry) to calculate the volume of \(\mathrm{N}_2\) gas generated using the relation \(\text{Volume} = \text{moles} \times 22.4\).

Key concepts

Redox Reactions

A redox reaction, short for reduction-oxidation reaction, is a chemical process where oxidation and reduction occur simultaneously. In such reactions, the transfer of electrons takes place between molecules or atoms.
This transfer prompts a change in oxidation state and reassigns electrons among participating elements.
In the exercise, hydrazine \((\mathrm{N}_2\mathrm{H}_4)\) reacts with dissolved oxygen to form water and nitrogen. Here, oxygen \((\mathrm{O}_2)\) is reduced as it gains electrons to form water, indicating its role as the oxidizing agent.
Conversely, hydrazine, which loses electrons, serves as the reducing agent, forming nitrogen gas in the process.
Key characteristics of redox reactions:

• Oxidation refers to the loss of electrons by a molecule, atom, or ion.
• Reduction involves the gain of electrons and often results in a decrease in oxidation state.
• A redox reaction is always composed of two balanced half-reactions: one for oxidation and one for reduction.

Understanding redox reactions can help in visualizing how substances are transformed through electron exchange.

Corrosion Prevention

Corrosion, the gradual destruction of materials by chemical reactions with their environment, often results from redox reactions.
In systems, especially metallic ones, corrosion can lead to damage and inefficiency, posing significant maintenance challenges.
Corrosion prevention is crucial, and one method involves using additives like hydrazine that react with oxygen to prevent its harmful effects. Hydrazine acts as a corrosion inhibitor in water heating systems by removing oxygen, thereby preventing the oxidation of metal components.
Benefits of corrosion prevention:

• Extends the lifespan of equipment and machinery by minimizing wear and tear.
• Enhances safety by reducing the risk of structural failures.
• Preserves operational efficiency by maintaining the integrity of equipment.

Understanding corrosion prevention techniques is vital for maintaining and optimizing system performance, particularly in water-based systems.

Exothermic Reactions

An exothermic reaction is a chemical reaction that releases energy by light or heat.
This is a common type of reaction where the energy required to break the bonds is less than the energy released through new bond formation.
In the case of hydrazine interacting with oxygen, an exothermic reaction occurs as energy is released, warming the system. The energy released during hydrazine oxidation contributes to the surrounding environment as heat, marking the system's temperature change.
Main characteristics of exothermic reactions:

• Releases energy in the form of heat, leading to a temperature increase in the surrounding environment.
• Tends to be spontaneous, as energy is discharged.
• Examples include combustion, respiration, and the reaction of acids with bases.

Recognizing exothermic reactions is beneficial in applications requiring controlled energy release, such as heating systems.

Thermodynamic Calculations

Thermodynamic calculations help quantify the energy changes occurring during a chemical reaction.
They allow us to evaluate parameters such as enthalpy, entropy, and Gibbs free energy under specific conditions.
In the exercise, these calculations determine the energetics of the reaction between hydrazine and oxygen. Enthalpy \((\Delta_r H^{\circ})\) is a measure of heat change, with negative values indicating an exothermic process.
Entropy \((\Delta_r S^{\circ})\) evaluates disorder, while Gibbs free energy \((\Delta_r G^{\circ})\) predicts reaction spontaneity, where a negative value signals a spontaneous process.
Reasons for performing thermodynamic calculations:

• Predict whether a reaction will occur under set conditions.
• Understand how energy is transformed and transferred in reactions.
• Design processes that maximize efficiency by leveraging favorable energy conditions.

Thermodynamic calculations are the basis for analyzing reactions, leading to improved process control and efficiency.