Question

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)$$ If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(127^{\circ} \mathrm{C}\) in an evacuated 250-L tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has 100% yield?

Short answer

The partial pressure of nitrogen gas produced in the reaction is 13.23 atm, and the total pressure in the tank is 40.01 atm.

Step-by-step solution

Find moles of dimethylhydrazine and nitrogen gas

First, we need to find the moles of dimethylhydrazine and the moles of nitrogen gas produced from the reaction. We know that 150 g of dimethylhydrazine is used and the molar mass of dimethylhydrazine is: \([(2\times12)+(6\times1)+(2\times14)+(2\times1)]=46\ \mathrm{g/mol}\) Now we can find the moles of dimethylhydrazine: \(\mathrm{moles\ of\ dimethylhydrazine}=\frac{150\ \mathrm{g}}{46\ \mathrm{g/mol}}=3.26\ \mathrm{mol}\) Using the balanced chemical equation, we can determine the moles of nitrogen gas produced for every mole of dimethylhydrazine reacted: \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow3\ \mathrm{mol\ of\ }N_2\) So, the moles of nitrogen gas produced are: \(3.26\ \mathrm{mol}\times3=9.78\ \mathrm{mol}\)

Calculate the partial pressure of nitrogen gas

Now, we can calculate the partial pressure of the nitrogen gas, using the ideal gas law: \(PV=nRT\) Where: \(P=\mathrm{Partial\ pressure\ of\ nitrogen\ gas}\) \(V=250\ \mathrm{L}\) \(n=9.78\ \mathrm{mol}\) \(R=0.08206\ \mathrm{L\ atm/mol\cdot K}\) \(T=127^\circ\mathrm{C}+273.15=400.15\ \mathrm{K}\) Now, solve for the partial pressure (P) of nitrogen gas: \(P_\mathrm{N_2}=\frac{nRT}{V}=\frac{(9.78\ \mathrm{mol})(0.08206\ \mathrm{L\ atm/mol\cdot K})(400.15\ \mathrm{K})}{250 \mathrm{L}}=13.23\ \mathrm{atm}\)

Find moles of other gases and total moles in the tank

From the balanced equation, we can find the moles of water and carbon dioxide as follows: \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow4\ \mathrm{mol\ of\ }H_2O\) \(1\ \mathrm{mol\ of\ }(CH_3)_2N_2H_2\rightarrow2\ \mathrm{mol\ of\ }CO_2\) Calculate the moles of water and carbon dioxide produced: \(3.26\ \mathrm{mol}\times4=13.04\ \mathrm{mol\ of\ }H_2O\) \(3.26\ \mathrm{mol}\times2=6.52\ \mathrm{mol\ of\ }CO_2\) Total moles in the tank: \(9.78+13.04+6.52=29.34\ \mathrm{mol}\)

Calculate total_pressure in the tank

Now, we can find the total_pressure by using the ideal gas law (PV=nRT) for total moles in the tank: \(P_\mathrm{total}=\frac{n_\mathrm{total}RT}{V}=\frac{(29.34\ \mathrm{mol})(0.08206\ \mathrm{L\ atm/mol\cdot K})(400.15\ \mathrm{K})}{250 \mathrm{L}}=40.01\ \mathrm{atm}\) Thus, the partial pressure of nitrogen gas produced is 13.23 atm, and the total pressure in the tank is 40.01 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 expressed as \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. This law assumes that the gas behaves ideally, meaning the gas particles do not interact with each other and occupy no volume themselves. This assumption is good for most gases under moderate conditions.
Using the ideal gas law, you can calculate any one of these variables if the others are known. For instance, in a chemical reaction where gas products are collected, you can determine the pressure of the gas if you know the moles of gas produced, the volume of the container, and the temperature of the gas. This makes the ideal gas law very helpful in exercises involving chemical reactions and gases.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In stoichiometry, a branch of chemistry, it's important to use balanced chemical equations to understand the proportions in which substances react and are produced. The equation given here is:
\[ (\mathrm{CH}_3)_2\mathrm{N}_2\mathrm{H}_2(l)+2\mathrm{N}_2\mathrm{O}_4(l) \rightarrow 3\mathrm{N}_2(g)+4\mathrm{H}_2\mathrm{O}(g)+2\mathrm{CO}_2(g) \]
Each molecule of dimethylhydrazine reacts with two molecules of dinitrogen tetroxide to produce three molecules of nitrogen, four of water, and two of carbon dioxide. This equation is balanced, meaning the same number of each type of atom is present on both sides of the reaction.
By balancing equations, you can use stoichiometric coefficients (like the 1, 2, 3, 4, 2 in this reaction) to convert between moles of reactants and products, which is essential for determining how much product a reaction will yield.

Partial Pressure

In a mixture of gases, each gas contributes to the total pressure independently. This contribution is called its partial pressure. Dalton's law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of individual gases.
When considering the ideal gas law, the partial pressure of a gas in a mixture can be calculated by substituting its moles in the equation \( PV = nRT \). For example, if you know the moles of nitrogen gas produced in a reaction, you can calculate its partial pressure using the volume of the container and temperature condition.
This concept is particularly useful in clearly understanding how different gases behave when they are mixed together and allows for precise calculations in situations involving multiple gases, such as in a chemical reaction that produces several gaseous products.

Moles and Molar Mass

Moles and molar mass are fundamental concepts in stoichiometry, crucial for converting between the mass of a substance and the amount of substance in moles. A mole is a unit that represents a quantity (6.022 x 10²³) of particles, be they atoms or molecules.
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It can be calculated by adding up the atomic masses of the elements in a compound. For instance, in our exercise, the molar mass of dimethylhydrazine \((\mathrm{CH}_3)_2\mathrm{N}_2\mathrm{H}_2)\) is 46 g/mol.
Using molar mass, you can convert the mass of a reactant to moles (\( moles = \frac{mass}{molar\ mass} \)). This allows for stoichiometric calculations to find out how many moles of product you can get from a given amount of reactant in a chemical reaction. Understanding this relationship is key to performing any quantitative chemical analysis.