Problem 35 - Chemistry and Chemical Reactivity

Question

Hydrazine reacts with \(\mathrm{O}_{2}\) according to the following equation: $$\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$ Assume the \(\mathrm{O}_{2}\) needed for the reaction is in a \(450-\mathrm{L}\) tank at \(23^{\circ} \mathrm{C} .\) What must the oxygen pressure be in the tank to have enough oxygen to consume \(1.00 \mathrm{kg}\) of hydrazine completely?

Step-by-Step Solution

Answer

The required oxygen pressure is approximately 1.68 atm.

Step 1: Write the Balanced Equation

The chemical equation is given and is balanced as follows: \[\text{N}_2\text{H}_4(g) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(\ell)\]. This shows that 1 mole of \( \text{N}_2\text{H}_4 \) reacts with 1 mole of \( \text{O}_2 \).

Step 2: Calculate Moles of Hydrazine

To find the moles of hydrazine, use its molar mass. The molar mass of \( \text{N}_2\text{H}_4 \) is \((2\times14.01) + (4\times1.01) = 32.06\ \text{g/mol}\). Convert 1.00 kg to grams (1000g), then divide by the molar mass: \[ \text{Moles of } \text{N}_2\text{H}_4 = \frac{1000\ \text{g}}{32.06\ \text{g/mol}} \approx 31.18\ \text{moles} \].

Step 3: Moles of Oxygen Needed

From the balanced equation, 1 mole of \( \text{N}_2\text{H}_4 \) reacts with 1 mole of \( \text{O}_2 \). Thus, 31.18 moles of \( \text{O}_2 \) are needed to react with 31.18 moles of \( \text{N}_2\text{H}_4 \).

Step 4: Calculate Oxygen Pressure Using Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to solve for the pressure \( P \). Here, \( n = 31.18 \) moles, \( R = 0.0821\ \text{L atm/mol K} \), \( T = 23 + 273 = 296\ \text{K} \) (temperature in Kelvin), \( V = 450\ \text{L} \). Substitute these values: \[ P \times 450\ \text{L} = 31.18 \times 0.0821\ \times 296 \].

Step 5: Final Calculation and Pressure

Calculate for pressure: \( P = \frac{31.18 \times 0.0821 \times 296}{450} \).\ Calculating gives \( P \approx 1.68\ \text{atm} \). This is the pressure of \( \text{O}_2 \) required in the tank.

Key Concepts

Understanding the Ideal Gas Law

The ideal gas law is a fundamental equation used for relating the properties of gases. It's expressed as \( PV = nRT \), where:

• \( P \) represents pressure in atmospheres (atm).
• \( V \) is volume in liters (L).
• \( n \) is the amount of substance in moles.
• \( R \) is the ideal gas constant, \(0.0821\ \text{L atm/mol K}\).
• \( T \) is temperature in Kelvin (K).

To solve for any unknown in the equation, knowing the other variables is key.
In this exercise, the focus is on determining pressure. You need the moles of gas () and temperature in Kelvin. Converting temperature from Celsius to Kelvin is simple: add 273 to the Celsius temperature. Using these values, the ideal gas law helps find the required pressure when the volume is fixed.
By rearranging the equation to solve for pressure, the relationship becomes clear: \[ P = \frac{nRT}{V} \]. This allows us to calculate the pressure needed for the gas, like the oxygen in this scenario.

Calculating Molar Mass

Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). It's essential for converting between mass and moles. To find it, add up the atomic masses of all the atoms in a molecule.
For hydrazine \( \text{N}_2\text{H}_4 \), calculate its molar mass by combining:

• 2 nitrogen atoms: \(2 \times 14.01\ \text{g/mol} = 28.02\ \text{g/mol}\).
• 4 hydrogen atoms: \(4 \times 1.01\ \text{g/mol} = 4.04\ \text{g/mol}\).

Total molar mass of \( \text{N}_2\text{H}_4 \) is \(32.06\ \text{g/mol}\).
For conversion, use the formula: \[ \text{Moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]. Here, converting 1.00 kg of hydrazine to grams gives 1000 g. Then divide by the molar mass to find moles: \( \approx 31.18 \text{ moles} \). Knowing moles helps in stoichiometric calculations, connecting mass and chemical reactions effectively.

Exploring Gas Reactions

Gas reactions often involve the exchange of moles between reactants and products. The balanced chemical equation provides the mole ratio needed to understand how substances react.
This exercise focuses on hydrazine \( \text{N}_2\text{H}_4 \) reacting with oxygen \( \text{O}_2 \) to produce nitrogen \( \text{N}_2 \) and water \( \text{H}_2\text{O} \). The equation: \[ \text{N}_2\text{H}_4(g) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(\ell) \]
shows that one mole of hydrazine reacts with one mole of oxygen. This 1:1 ratio means that for every mole of hydrazine used, one mole of oxygen is required.
Understanding this stoichiometric relationship is crucial for calculating the amount of each reactant needed. It helps in determining how much product will form and ensures that all reactants are fully consumed without excess.

Interpreting Chemical Equations

Chemical equations are shorthand notations for chemical reactions. They show how reactants transform into products. Each equation must be balanced, meaning the number of atoms for each element must be the same on both sides.
In the equation \( \text{N}_2\text{H}_4(g) + \text{O}_2(g) \rightarrow \text{N}_2(g) + 2\text{H}_2\text{O}(\ell) \), every molecule has its place:

• Reactants: 1 molecule of hydrazine \( \text{N}_2\text{H}_4 \) and 1 molecule of oxygen \( \text{O}_2 \).
• Products: 1 molecule of nitrogen \( \text{N}_2 \) and 2 molecules of water \( \text{H}_2\text{O} \).

Balancing equations verifies the law of conservation of mass, where mass is neither created nor destroyed.
Through balancing, you ensure that mole ratios are accurate. This is vital for correctly calculating reactant and product quantities in any given reaction, like in this hydrazine and oxygen scenario.