Problem 98 - 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 equation for the reaction between gaseous hydrazine (N2H4) and oxygen (O2) is: \(2N_2H_4 + O_2 \rightarrow 2N_2 + 4H_2O\) The enthalpy change accompanying this reaction is -584.90 kJ/mol. To react with all the oxygen in 3.0 x 10^4 L of water under the given conditions, 546 grams of hydrazine are required.

Step 1: (a) Balanced chemical equation

First, we'll find the balanced equation for the reaction between gaseous hydrazine (N2H4) and oxygen (O2). The reaction produces nitrogen (N2) and water (H2O): \(N_2H_4 + O_2 \rightarrow N_2 + 2H_2O\) To balance this equation, we can put a 2 in front of the hydrazine (N2H4) which will result: \(2N_2H_4 + O_2 \rightarrow 2N_2 + 4H_2O\) Now, the equation is balanced.

Step 2: (b) Calculate the enthalpy change

Enthalpy change can be found using the following formula: ΔH = ΣΔH(products) - ΣΔH(reactants) Where: ΔH = Enthalpy change ΣΔH(products) = Sum of standard enthalpies of formation of products ΣΔH(reactants) = Sum of standard enthalpies of formation of reactants Standard enthalpies of formation for the involved species are as follows (in kJ/mol): ΔHf(N2H4, g) = +50.63 ΔHf(O2, g) = 0 (reference state) ΔHf(N2, g) = 0 (reference state) ΔHf(H2O, g) = -241.82 Now, we can substitute the standard enthalpies of formation into the equation: ΔH = [2*(-241.82) - (2*50.63)] - [0] ΔH = -483.64 - 101.26 ΔH = -584.90 kJ/mol Thus, the enthalpy change for this reaction is -584.90 kJ/mol.

Step 3: (c) Amount of hydrazine required

Given, Oxygen in air dissolves in water to the extent of 9.1 ppm (parts per million) at 20℃. One ppm is equivalent to 1 milligram per liter (mg/L). So we have 9.1 mg of dissolved oxygen per liter of water. Volume of water = 3.0 x 10^4 L To find the total amount of oxygen dissolved, multiply the concentration by volume: Oxygen dissolved = 9.1 mg/L * 3.0 x 10^4 L = 273000 mg Convert to grams: 273000 mg (1 g / 1000 mg) = 273 g of oxygen From the balanced chemical equation, we have: \(2N_2H_4 + O_2 \rightarrow 2N_2 + 4H_2O\) According to the balanced equation, 1 mole of O2 reacts with 2 moles of N2H4. Molar mass of O2 = 32 g/mol Molar mass of N2H4 = 32 (N) + 4 (H) = 32 g/mol Using stoichiometry, we can find the amount of hydrazine required: 273 g O2 * (1 mol O2 / 32 g O2) * (2 mol N2H4 / 1 mol O2) * (32 g N2H4/1 mol N2H4) = 546 g of hydrazine (rounded to the nearest whole number) Therefore, 546 grams of hydrazine are required to react with all the oxygen in 3.0 x 10^4 L of water under the given conditions.

Key Concepts

Enthalpy Change

Enthalpy change is a key concept in chemistry, often symbolized as \( \Delta H \). It refers to the amount of heat released or absorbed during a chemical reaction at constant pressure. This change helps us understand energy flow within the reaction.

In chemical equations, \( \Delta H \) is calculated using the standard enthalpies of formation for all reactants and products. The formula for calculating enthalpy change is:

• \( \Delta H = \Sigma \Delta H_{\text{products}} - \Sigma \Delta H_{\text{reactants}} \)

Here, \( \Sigma \Delta H_{\text{products}} \) and \( \Sigma \Delta H_{\text{reactants}} \) are the sums of the standard enthalpies of formation for the products and reactants, respectively. Enthalpy of formation is usually available in thermodynamic tables for various substances.

For example, in the reaction involving hydrazine (\( N_2H_4 \)) and oxygen (\( O_2 \)), the enthalpy change measures how heat energy is conserved or released when nitrogen (\( N_2 \)) and water (\( H_2O \)) are formed. In our given solution, it was computed as \(-584.90\) kJ/mol, indicating the reaction releases energy, making it an exothermic process.

Chemical Equation Balancing

Chemical equation balancing is crucial because it ensures the law of conservation of mass is adhered to, indicating that matter is neither created nor destroyed in a chemical reaction.

An unbalanced equation demonstrates a chemical reaction without accounting for the exact number of atoms on each side. For balance, you adjust the coefficients before the chemical formulas to ensure the same number of each type of atom on both sides of the arrow.

Taking the example of hydrazine reacting with oxygen:

• Initial reaction: \( N_2H_4 + O_2 \rightarrow N_2 + 2H_2O \)
• Balanced reaction: \( 2N_2H_4 + O_2 \rightarrow 2N_2 + 4H_2O \)

Here, we place a '2' in front of \( N_2H_4 \) to balance the nitrogen and hydrogen atoms, ensuring the equation on the left matches with the right in atom count. Balancing equations not only aligns with chemical laws but also aids in conservation calculations, such as energy and mass.

Stoichiometry

Stoichiometry is a quantitative aspect of chemistry that involves the calculation of reactants and products in chemical reactions. It uses relationships from a balanced chemical equation to determine how much of a substance is required or produced.

To understand stoichiometry, you must first comprehend the mole ratio in a balanced chemical equation. This ratio indicates the proportional relationship between reactants and products. For example, in the reaction between hydrazine and oxygen, shown as:

• \( 2N_2H_4 + O_2 \rightarrow 2N_2 + 4H_2O \)

The equation indicates that 2 moles of hydrazine react with 1 mole of oxygen.

Using stoichiometry, if you know the amount of one substance, you can calculate how much of another is necessary. For instance, when calculating the hydrazine needed to fully react with dissolved oxygen in water, we used this mole relation. The molar masses and balanced equation allow conversion from grams to moles. Therefore, 546 grams of hydrazine were calculated to react with the oxygen available in the pool water, ensuring precise proportions are fulfilled.