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 oxygen pressure should be around 1.69 atm.

Step 1: Determine Moles of Hydrazine

First, calculate the number of moles of hydrazine (\( \mathrm{N}_2\mathrm{H}_4 \)) present. The molecular weight of hydrazine is approximately \(32.05\, \text{g/mol}\). Given the mass of hydrazine is \(1.00\, \text{kg}\), or \(1000\, \text{g}\), the number of moles is: \[ \text{moles of } \mathrm{N}_2\mathrm{H}_4 = \frac{1000\, \text{g}}{32.05\, \text{g/mol}} \approx 31.2\, \text{mol} \]

Step 2: Calculate Oxygen Moles Needed

Using the stoichiometry of the balanced chemical equation, we know that 1 mole of \( \mathrm{N}_2\mathrm{H}_4 \) reacts with 1 mole of \( \mathrm{O}_2 \). Thus, 31.2 moles of \( \mathrm{N}_2\mathrm{H}_4 \) would require 31.2 moles of \( \mathrm{O}_2 \).

Step 3: Convert Volume and Temperature

Convert the given volume and temperature to appropriate units for use in the ideal gas law. The tank volume is \( 450\, \text{L} \) which is already in liters. Convert temperature from Celsius to Kelvin: \[ T = 23 + 273.15 = 296.15 \, \text{K} \]

Step 4: Apply Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to find the pressure \( P \) of \( \mathrm{O}_2 \). The ideal gas constant \( R \) is \( 0.0821\, \text{L⋅atm/mol⋅K} \). Rearrange the equation to solve for \( P \): \[ P = \frac{nRT}{V} = \frac{31.2 \, \text{mol} \times 0.0821 \, \text{L⋅atm/mol⋅K}\times 296.15 \, \text{K}}{450 \, \text{L}} \] Calculating gives: \[ P \approx 1.69\, \text{atm} \]

Step 5: Conclusion

Thus, to consume 1.00 kg of hydrazine completely, the oxygen pressure in the tank must be approximately 1.69 atm.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that helps us understand the behavior of gases under different conditions. It is represented by the equation: \[ PV = nRT \]

• P stands for the pressure of the gas.
• V is the volume of the gas.
• n is the number of moles of the gas.
• R is the ideal gas constant, approximately 0.0821 L⋅atm/mol⋅K.
• T is the temperature of the gas in Kelvin.

This equation allows us to calculate any one of these properties if the other three are known.
In the exercise, we used the Ideal Gas Law to determine the pressure that the oxygen in the tank must have under given conditions. First, it was essential to convert the temperature from Celsius to Kelvin, as this is the standard unit used in the Ideal Gas Law. By rearranging the formula to solve for pressure \( P \), we could determine how much the pressure of the oxygen needed to be to facilitate the complete consumption of hydrazine in the reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products as per a balanced chemical equation. In our given exercise, hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) reacts with oxygen (\(\mathrm{O}_2\)) to produce nitrogen (\(\mathrm{N}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). This particular reaction is a combustion reaction, a type that typically involves a substance reacting with oxygen.
Balancing chemical equations ensure that the mass and number of atoms are conserved throughout the reaction. In this case, one mole of hydrazine reacts with one mole of oxygen to produce one mole of nitrogen and two moles of water. By understanding the stoichiometry of the reaction, we can predict the amounts of products formed from given quantities of reactants, and vice versa. This concept was crucial in the problem, as it enabled us to calculate the moles of oxygen needed to completely react with a specific amount of hydrazine.

Mole Calculations

Mole calculations allow us to shift from thinking about substances in terms of grams to thinking about them in terms of particles. The mole is a fundamental unit in chemistry, representing \(6.022 \times 10^{23}\) particles of a substance.
In the context of the exercise, we started by calculating the number of moles of hydrazine. By employing its molar mass (32.05 g/mol), we converted the mass of hydrazine (1.00 kg or 1000 g) into moles:\[ \text{Moles of } \mathrm{N}_2\mathrm{H}_4 = \frac{1000\, \text{g}}{32.05\, \text{g/mol}} \approx 31.2\, \text{mol} \]This result was then used in conjunction with the balanced chemical equation to determine the moles of oxygen needed. Mole calculations are essential in stoichiometry, providing a bridge between the macroscopic world we can observe and measure and the microscopic world of molecules and atoms.