Problem 35 - Chemistry and Chemical Reactivity

Question

Hydrazine reacts with \(\mathrm{O}_{2}\) according to the following equation: $$ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ Assume the \(\mathrm{O}_{2}\) needed for the reaction is in a \(450-\mathrm{I}\), 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 must be approximately 1.68 atm.

Step 1: Calculate moles of hydrazine

First, we need to find the moles of hydrazine (\(\mathrm{N}_2\mathrm{H}_4\)) from the given mass. The molar mass of \(\mathrm{N}_2\mathrm{H}_4\) is \(32.05\, \mathrm{g/mol}\). The mass of hydrazine is \(1.00\, \mathrm{kg} = 1000\, \mathrm{g}\).\[\text{Moles of } \mathrm{N}_2\mathrm{H}_4 = \frac{1000\, \mathrm{g}}{32.05\, \mathrm{g/mol}} = 31.20\, \mathrm{mol}\]

Step 2: Use stoichiometry to find moles of oxygen required

The balanced chemical equation shows that 1 mole of \(\mathrm{N}_2\mathrm{H}_4\) reacts with 1 mole of \(\mathrm{O}_2\). Therefore, the moles of \(\mathrm{O}_2\) needed are equal to the moles of \(\mathrm{N}_2\mathrm{H}_4\).\[\text{Moles of } \mathrm{O}_2 = 31.20\, \mathrm{mol}\]

Step 3: Calculate pressure using the Ideal Gas Law

Use the Ideal Gas Law \(PV = nRT\) to find the pressure \(P\) of \(\mathrm{O}_2\) in the tank. Given:- \(V = 450\, \mathrm{L}\)- \(n = 31.20\, \mathrm{mol}\)- \(R = 0.0821\, \mathrm{L\cdot atm \cdot mol^{-1} \cdot K^{-1}}\)- \(T = 23^\circ \mathrm{C} = 296\, \mathrm{K}\)\[P = \frac{nRT}{V} = \frac{31.20 \times 0.0821 \times 296}{450}\]\[P \approx 1.68\, \mathrm{atm}\]

Step 4: Conclusion

The required pressure of \(\mathrm{O}_2\) in the tank to ensure complete reaction with 1.00 kg of hydrazine is approximately \(1.68\, \mathrm{atm}\).

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is given by the formula \( PV = nRT \), where:

• \( P \) is the pressure of the gas in atmospheres (atm),
• \( V \) is the volume of the gas in liters (L),
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant, approximately \( 0.0821 \, \mathrm{L\cdot atm \cdot mol^{-1} \cdot K^{-1}} \),
• \( T \) is the temperature in Kelvin (K).

This equation helps us calculate any one of these properties if the others are known. For example, in the provided exercise, it is used to determine the necessary oxygen pressure in a tank to react with hydrazine.
By substituting the given values (\( n = 31.20 \) moles, \( R = 0.0821 \), \( T = 296 \) K, \( V = 450 \) L), we find the pressure \( P \) to ensure the right amount of gas is there for the reaction.
Understanding this law is crucial for predicting how gases behave under different conditions, making it a powerful tool in chemical applications.

Molar Mass

Molar mass is a property that represents the mass of one mole of a substance, typically measured in grams per mole (g/mol). It is used to calculate the amount of substance present in a given sample.
For hydrazine, a molecule consisting of nitrogen and hydrogen, the molar mass is calculated from its chemical formula \( \mathrm{N}_2\mathrm{H}_4 \).

• The atomic mass of nitrogen (N) is approximately \( 14.01 \, \mathrm{g/mol} \),
• And for hydrogen (H) it is approximately \( 1.01 \, \mathrm{g/mol} \).

Thus, the molar mass of hydrazine is: \[ (2 \times 14.01) + (4 \times 1.01) = 32.05 \, \mathrm{g/mol} \]This value is vital when working in stoichiometry, as it allows conversion from grams to moles, which is necessary for applying the Ideal Gas Law.
By calculating the moles from a given mass (for instance, the 1000 grams of hydrazine in the example), scientists can better prepare and analyze chemical reactions.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products, indicated by a balanced chemical equation. In the exercise, hydrazine reacts with oxygen, producing nitrogen gas and water as products.
The equation can be expressed as:\[ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \]This equation is balanced, meaning the number of each atom is the same on both sides ensuring mass conservation. Balancing equations is essential before performing calculations in stoichiometry, as it impacts the relative quantities of reactants and products.
In the problem, the equation shows a 1:1 molar ratio between hydrazine and oxygen. Thus, to know how much oxygen is needed, we first calculate how much hydrazine we have in moles, and then we directly know the moles of oxygen required.
Understanding these reactions helps in preparing the correct amounts of substances, ensuring reactions proceed completely, and are foundational in both laboratory and industrial settings.