Problem 97 - Chemistry The Central Science

Question

The dissolved oxygen present in any highly pressurized, high-temperature steam boiler can be extremely corrosive to its metal parts. Hydrazine, which is completely miscible with water, can be added to remove oxygen by reacting with it to form nitrogen and water. (a) Write the balanced equation for the reaction between gaseous hydrazine and oxygen.(b) Calculate the enthalpy change accompanying this reaction. (c) Oxygen in air dissolves in water to the extent of 9.1 ppm at \(20^{\circ} \mathrm{C}\) at sea level. How many grams of hydrazine are required to react with all the oxygen in 3.0 \(\times 10^{4} \mathrm{L}\) (the volume of a small swimming pool) under these conditions?

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between hydrazine and oxygen is: \(N_2H_4 (g) + O_2 (g) \rightarrow N_2 (g) + 2 H_2O (l)\). The enthalpy change for this reaction is -622.2 kJ/mol. To react with all the oxygen dissolved in a 3.0 x 10^4 L pool, 273 grams of hydrazine are required.

Step 1: Write the unbalanced chemical equation

Identify the reactants and products involved in the reaction. In this case, the reactants are gaseous hydrazine (N2H4) and oxygen (O2), and the products are nitrogen (N2) and water (H2O). Write the unbalanced chemical equation: N2H4 (g) + O2 (g) -> N2 (g) + H2O (l)

Step 2: Balance the chemical equation

To balance the equation, we will adjust the stoichiometric coefficients (the numbers in front of each chemical species) so that the number of atoms of each element is equal on both sides of the equation: N2H4 (g) + O2 (g) -> N2 (g) + 2 H2O (l) Now, we have 2 N atoms, 4 H atoms, and 2 O atoms on each side of the equation, so the balanced chemical equation is: N2H4 (g) + O2 (g) -> N2 (g) + 2 H2O (l) Part (b)

Step 3: Use the standard enthalpies of formation to find the enthalpy change for the reaction

In order to calculate the enthalpy change (∆H) for the reaction, you need to use the standard enthalpies of formation (∆Hf) for each component. These values can be found in a standard chemistry reference book or online database. The general formula for calculating ∆H is: ∆H = ∑ [∆Hf(products)] - ∑ [∆Hf(reactants)] Using the standard enthalpies of formation, we can calculate ∆H for the given reaction: ∆H = [1 * ∆Hf(N2) + 2 * ∆Hf(H2O)] - [1 * ∆Hf(N2H4) + 1 * ∆Hf(O2)] Note that the enthalpy of formation for an element in its standard state is zero, so ∆Hf(N2) and ∆Hf(O2) are both zero: ∆H = [2 * ∆Hf(H2O)] - [∆Hf(N2H4)] Look up the standard enthalpies of formation for H2O and N2H4: ∆Hf(H2O) = -285.8 kJ/mol ∆Hf(N2H4) = 50.6 kJ/mol Substitute these values into the formula: ∆H = [2 * -285.8] - [50.6]= -571.6 - 50.6 = -622.2 kJ/mol So the enthalpy change for the reaction is -622.2 kJ/mol. Part (c)

Step 4: Calculate the amount of dissolved oxygen

Given the dissolved oxygen concentration is 9.1 ppm (parts per million) at 20°C, we will convert this concentration into moles of dissolved oxygen in the 3.0 x 10^4 L of water. The molar mass of O2 is 32 g/mol. First, convert 9.1 ppm to grams per liter: 9.1 ppm = 9.1 mg/L = 0.0091 g/L Now, find the total mass of dissolved oxygen in the given volume of water: Total mass of dissolved oxygen = (0.0091 g/L) * (3.0 x 10^4 L) = 273 g Next, convert the mass of dissolved oxygen to moles: Moles of dissolved oxygen = 273 g / 32 g/mol ≈ 8.53 mol

Step 5: Calculate the required amount of hydrazine

Based on the balanced chemical equation, one mole of hydrazine reacts with one mole of oxygen: N2H4 (g) + O2 (g) -> N2 (g) + 2 H2O (l) So, the moles of hydrazine required will be equal to the moles of dissolved oxygen: Moles of hydrazine = 8.53 mol Now, convert the moles of hydrazine into grams. The molar mass of hydrazine (N2H4) is 32 g/mol: Mass of hydrazine = 8.53 mol * 32 g/mol ≈ 273 g So, 273 grams of hydrazine are required to react with all the oxygen dissolved in the 3.0 x 10^4 L of water.

Key Concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, interact to form different substances, known as products. Reactants undergo a transformation during the reaction, often resulting in a change of energy and matter's arrangement. In the given example, hydrazine (N\(_2\)H\(_4\)) reacts with oxygen (O\(_2\)) to produce nitrogen (N\(_2\)) and water (H\(_2\)O). This transformation releases energy due to new bonds being formed and broken.
To represent this process, we use a chemical equation, which must be balanced to follow the Law of Conservation of Mass. Balancing ensures that each element has the same number of atoms on both sides of the equation. In this exercise, the balanced reaction is:
\[\text{N}_2\text{H}_4 (g) + \text{O}_2 (g) \rightarrow \text{N}_2 (g) + 2\text{H}_2\text{O} (l)\]
This balanced equation tells us that one mole of gaseous hydrazine reacts with one mole of diatomic oxygen to produce one mole of nitrogen gas and two moles of liquid water.

Stoichiometry

Stoichiometry is the quantitative relation between reactants and products in a chemical reaction. It is derived from the balanced chemical equation and helps us calculate the amount of each substance involved or produced in a reaction. In this exercise, stoichiometry allows us to determine how much hydrazine is required to react with dissolved oxygen.
For example, from the balanced equation, we see that one mole of hydrazine reacts with one mole of oxygen. By using stoichiometry, we can convert between various quantities such as moles, grams, or liters using molar masses or concentrations, ensuring we have the correct amounts for reactions.
In part (c) of the problem, the stoichiometric relationship helps us calculate the amount of hydrazine needed for the reaction with the dissolved oxygen from a swimming pool scenario. With the conversion from parts per million (ppm) to grams of oxygen dissolved, stoichiometry facilitates the computation for the mass of hydrazine needed to fully react with the available oxygen.

Enthalpy Change

Enthalpy change (ΔH) is a measure of the heat energy absorbed or released during a chemical reaction at constant pressure. It indicates whether a reaction is exothermic (releases heat, ΔH < 0) or endothermic (absorbs heat, ΔH > 0).
To calculate the enthalpy change for a reaction, we use the standard enthalpies of formation (ΔH\(_f\)) for the reactants and products. The formula is:
\[\Delta H = \sum [\Delta H_f(\text{products})] - \sum [\Delta H_f(\text{reactants})]\]
In this exercise, the enthalpy change for the transformation of hydrazine and oxygen into nitrogen and water is calculated using standard enthalpy values:
- ΔH\(_f\)(H\(_2\)O) = -285.8 kJ/mol
- ΔH\(_f\)(N\(_2\)H\(_4\)) = 50.6 kJ/mol
By substituting these into the formula, we find that the reaction has a ΔH of -622.2 kJ/mol, making it exothermic. The negative value reflects energy release, which is typical of reactions forming strong bonds, such as water from hydrogen and oxygen.

Dissolved Oxygen

Dissolved oxygen is the amount of oxygen that is mixed into a liquid such as water. In natural and industrial processes, dissolved oxygen plays a critical role by supporting aquatic life and facilitating various chemical reactions.
In this case, the dissolved oxygen is detrimental as it can cause corrosion in the metal parts of steam boilers. The exercise calculates how much hydrazine is needed to remove this oxygen by reacting with it. Oxygen's solubility in water is generally measured in parts per million (ppm), indicating how many milligrams of oxygen are present per liter of water.
For our calculation, the dissolved oxygen concentration is given as 9.1 ppm at 20°C. To find the total amount of oxygen in a volume of water (such as a pool), we convert ppm to grams and then to moles using the molar mass of oxygen (32 g/mol). This allows us to determine the reactivity with hydrazine, ensuring enough is present to neutralize the corrosive potential of the dissolved oxygen.