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?

Short answer

The balanced chemical equation for the reaction between hydrazine (N₂H₄) and oxygen (O₂) is: $N_2{H}_4(g)+O_2(g)\to N_2(g)+2H_{2}O(g)$ The enthalpy change (ΔH) for this reaction is -484.64 kJ/mol. To remove all the dissolved oxygen in 3.0 x 10^4 L of water at 20°C, 2.754 grams of hydrazine are required.

Step-by-step solution

Write the balanced chemical equation for the reaction between hydrazine and oxygen

The reaction between gaseous hydrazine (N₂H₄) and oxygen (O₂) forms nitrogen (N₂) and water (H₂O). To balance this chemical equation, it is required: $N_2{H}_4(g)+O_2(g)\to N_2(g)+2H_{2}O(g)$ Now, the equation is balanced.

Calculate the enthalpy change for the reaction

The balanced chemical equation given in the problem states that 1 mole of hydrazine reacts with 1 mole of oxygen to produce 1 mole of nitrogen and 2 moles of water. We need to find the enthalpy change for this reaction. To calculate the enthalpy change, we need to subtract the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products: $\Delta H_{reaction}=\Sigma [\Delta H_f(products)]-\Sigma [\Delta H_f(reactants)]$ Using the following standard enthalpies of formation ($\Delta H_f^{\circ }$): $\Delta H_f^{\circ }(N_2{H}_4(g))=+95.39kJmo{l}^{-1}$ , $\Delta H_f^{\circ }(O_2(g))=0kJmo{l}^{-1}$, $\Delta H_f^{\circ }(N_2(g))=0kJmo{l}^{-1}$, $\Delta H_f^{\circ }(H_{2}O(g))=-241.82kJmo{l}^{-1}$ Now, we can calculate the enthalpy change of the reaction: $\Delta H_{reaction}=[(1)(0)+(2)(-241.82)]-[(1)(95.39)+(1)(0)]$ $\Delta H_{reaction}=-484.64kJmo{l}^{-1}$

Calculate the grams of hydrazine needed to remove all dissolved oxygen

In this step, we need to compute how much hydrazine is needed to react with all dissolved oxygen present in 3.0 x 10^4 L of water. We know that the amount of dissolved oxygen in water is 9.1 ppm (parts per million) at 20°C. First, let's convert ppm to grams: 9.1 ppm = (9.1 g of O₂ / 10^6 g of water) Now, we can calculate the grams of dissolved oxygen in the water: $gram{s}_{O_2}=(3.0\times {10}^{4}L\times 1gm{L}^{-1})\times \frac{9.1g_{O_2}}{{10}^6{g}_{H_{2}O}}$ $gram{s}_{O_2}=2.754g$ Since the ratio of hydrazine to oxygen in the reaction is 1:1, the same amount of moles of hydrazine is required to react with oxygen. Now, we need to convert the mass of oxygen in moles: Molar mass of O₂: 32 g/mol moles of O₂ = $\frac{2.754g}{32gmo{l}^{-1}}=0.0860625mol$ Using the stoichiometry of the reaction (1:1 ratio), we find the grams of hydrazine required: Molar mass of N₂H₄: 32 g/mol grams of N₂H₄ = (0.0860625 mol) * (32 g/mol) = 2.754 g Therefore, 2.754 grams of hydrazine are needed to remove all the dissolved oxygen under the given conditions.

Key concepts

Chemical Equation Balancing

Understanding how to balance a chemical equation is essential for studying chemical reactions. When hydrazine, a simple diamine compound, reacts with oxygen, it's important to ensure the number of atoms for each element is the same on both sides of the reaction. Balancing becomes straightforward when you recall that conservation of mass is a fundamental principle of chemistry; atoms are neither created nor destroyed in a chemical reaction.

For the hydrazine-oxygen reaction, we can start by listing the number of atoms for each element on both sides of the equation and adjust the coefficients accordingly to balance the atoms. Always begin with elements that appear in the least number of compounds, and save balancing hydrogen and oxygen for last, as they are often found in multiple compounds in the reaction.

Enthalpy Change Calculation

The change in enthalpy, or $\mathrm{\Delta }H$, of a reaction represents the heat absorbed or released. Calculating this value involves using the concept of standard enthalpies of formation. These values are determined under standard conditions and indicate how much energy is absorbed or released when one mole of a compound is formed from its elements.

To find the enthalpy change for the hydrazine-oxygen reaction, you subtract the sum of the standard enthalpies of formation of the reactants from the products, as seen in Step 2 of the solution. It's crucial to ensure units are consistent and consider whether the enthalpy values are for the gases or liquids since these phases have different enthalpies of formation.

Stoichiometry

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction. It tells us how much of each substance is needed or produced. The stoichiometric coefficients from a balanced equation provide the ratios necessary to perform these calculations.

For example, with the balanced equation for the reaction of hydrazine and oxygen, we have a 1:1 molar ratio, meaning 1 mole of hydrazine will react with 1 mole of oxygen to produce products. Using stoichiometry, you can determine how many grams of hydrazine are needed by calculating the moles of oxygen present and then applying the molar ratio.

Dissolved Oxygen

Dissolved oxygen refers to the oxygen gas present in a liquid, such as water. It's crucial for aquatic life and influences corrosion rates in metal structures. The amount present is often expressed in ppm (parts per million), reflecting the mass of oxygen relative to the mass of water.

When considering reactions involving dissolved oxygen, like the removal of oxygen in steam boilers using hydrazine, you must first convert the ppm value to a more usable mass-based value, such as grams, to apply stoichiometric calculations. This conversion is an important step in solving the problem dealing with the amount of hydrazine needed to react with a certain volume of water containing dissolved oxygen.