Problem 29 - Chemistry The Science in Context

Question

Automobile air bags produce nitrogen gas from the reaction: $$2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ a. If \(2.25 \mathrm{g}\) of \(\mathrm{NaN}_{3}\) reacts to fill an air bag, how much \(P-V\) work will the \(\mathrm{N}_{2}\) do against an external pressure of 1.00 atm given that the density of nitrogen is \(1.165 \mathrm{g} / \mathrm{L}\) at \(20^{\circ} \mathrm{C} ?\) b. If the process releases \(2.34 \mathrm{kJ}\) of heat, what is \(\Delta E\) for the system?

Step-by-Step Solution

Answer

Question: Calculate the change in internal energy (ΔE) when 2.25 g of sodium azide (NaN₃) breaks down to produce nitrogen gas with the help of the given data: External pressure is at 1.00 atm, the density of nitrogen gas is 1.165 g/L at 20°C, and the heat released is 2.34 kJ. Answer: The change in internal energy (ΔE) is 2214 Joules.

Step 1: Find the number of moles of \(\mathrm{NaN}_3\)

To find the number of moles of \(\mathrm{NaN}_3\), we can use the formula: Number of moles = \(\frac{\text{mass of the compound}}{\text{molar mass of the compound}}\) The molar mass of \(\mathrm{NaN}_3\) is \(22.99 (\mathrm{Na}) + 3(14.01 (\mathrm{N})) = 65.02 \,\text{g/mol}\). Now, we can find the moles of given mass as: Number of moles = \(\frac{2.25 \,\text{g}}{65.02 \,\text{g/mol}} = 0.0346 \,\text{mol}\)

Step 2: Calculate the moles of nitrogen gas produced

Now, let's find the moles of nitrogen gas produced using the stoichiometry of the reaction: $$2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ From the balanced equation, we can see that for every 2 moles of \(\mathrm{NaN}_3\), 3 moles of \(\mathrm{N}_2\) are produced. Therefore, the moles of \(\mathrm{N}_2\) gas formed are: Moles of \(\mathrm{N}_2 = 0.0346 \,\text{mol} \times \frac{3 \,\text{mol} \, \mathrm{N}_2}{2 \,\text{mol} \, \mathrm{NaN}_3} = 0.0519 \,\text{mol}\)

Step 3: Find the volume of the nitrogen gas

We are given the density of nitrogen gas as \(1.165 \,\mathrm{g}/\mathrm{L}\) at \(20^{\circ}\mathrm{C}\). We can derive the volume from the density using the formula: Volume = \(\frac{\text{mass of gas}}{\text{density of gas}}\) First, we need to find the mass of nitrogen gas produced: Mass of \(\mathrm{N}_2 = 0.0519 \,\text{mol} \times 28.02 \,\text{g/mol} = 1.45 \,\text{g}\) Now we can find the volume: Volume = \(\frac{1.45 \,\text{g}}{1.165 \,\mathrm{g}/\mathrm{L}} = 1.244 \,\mathrm{L}\)

Step 4: Calculate the \(P-V\) work done

We can now calculate the \(P-V\) work done by the nitrogen gas using the formula: \(P-V\) Work = \(-P \times \Delta V\) The negative sign indicates that the system is doing the work against external pressure. We are given the external pressure as 1.00 atm. However, we need to convert the volume to liters and pressure to the same unit. We'll use the conversion factor: \(1 \,\mathrm{L \, atm} = 101.3 \,\mathrm{J}\). \(P-V\) Work = \(-1.00 \,\text{atm} \times 1.244 \,\text{L} \times \frac{101.3 \,\mathrm{J}}{1 \,\mathrm{L \, atm}} = -126 \,\mathrm{J}\) (rounded to 3 significant figures)

Step 5: Calculate \(\Delta E\)

Now, we can compute the change in internal energy (\(\Delta E\)) using the first law of thermodynamics: \(\Delta E = q + w\) We are given that the heat released (q) is \(2.34 \,\mathrm{kJ} = 2340 \,\mathrm{J}\). The work done by the system (w) is \(-126 \,\mathrm{J}\). Therefore, the change in internal energy is: \(\Delta E = 2340 \,\mathrm{J} - 126 \,\mathrm{J} = 2214 \,\mathrm{J}\) So, the change in internal energy (\(\Delta E\)) is 2214 Joules.

Key Concepts

Chemical Reaction Equations

Chemical reaction equations, like the one used in the air bag deployment exercise, are crucial because they inform us how reactants transform into products during a chemical reaction. The equation for the reaction in automobile air bags is: \[2 \mathrm{NaN}_3(s) \rightarrow 2 \mathrm{Na}(s) + 3 \mathrm{N}_2(g)\].
This balanced equation tells us that two moles of sodium azide (\(\mathrm{NaN}_3\)) decompose to form two moles of sodium metal and three moles of nitrogen gas. Having a balanced reaction is essential, as it obeys the Law of Conservation of Mass, which states that matter cannot be created or destroyed. This balance allows us to predict how much of each substance is involved in the reaction and is the key to solving stoichiometry problems.

Molar Mass Calculations

Molar mass calculations are a foundational concept in stoichiometry because they allow us to convert between grams and moles of a substance. In the air bag reaction, knowing the molar mass of sodium azide (\(\mathrm{NaN}_3\)) is essential for calculating the moles of reactants and products. The molar mass of a compound is the sum of the masses of all atoms within a single molecule of that compound. For instance, the molar mass of \(\mathrm{NaN}_3\) is calculated by summing the atomic mass of sodium (Na) and three times the atomic mass of nitrogen (N), leading to:\[65.02 \, \text{g/mol}\].
We use this value to determine that \(2.25 \, \text{g}\) of \(\mathrm{NaN}_3\) equates to \(0.0346 \, \text{mol}\). This step is critical for figuring out how much gas is produced to inflate the air bag.

Gas Laws and PV Work

Understanding gas laws and PV work is essential when dealing with gases, as seen in the air bag example. PV work refers to the product of pressure and volume change (\(P\Delta V\)) associated with a gas, which is work done by or against the system during a process. As per the ideal gas law, gases exert pressure when they are confined in a volume and subject to temperature. This principle allows us to calculate the volume of gas produced using the density provided and to then change it into work (energy) using the relationship where \(1 \text{L atm} = 101.3 \text{J}\).
For our nitrogen gas (\(\mathrm{N}_2\)) expanding in the air bag, \(-P \times \Delta V\) gives us the energy expended against the atmospheric pressure. It is an important concept that connects the chemical reaction to the physical work performed in creating a force, such as inflating an air bag.

First Law of Thermodynamics

The first law of thermodynamics, often summarized as \(\Delta E = q + w\), describes the relationship between the internal energy change (\(\Delta E\)) of a system and the heat (\(q\)) exchanged with the surroundings plus the work (\(w\)) done by or on the system. In the context of the air bag deployment, heat is released during the reaction, and the nitrogen gas does work on the surroundings to inflate the air bag. The internal energy (\(\Delta E\)) provides a comprehensive view of the energy changes within a reaction, which includes both heat and the expansion work:\[\Delta E = 2340 \, \mathrm{J} - 126 \, \mathrm{J} = 2214 \, \mathrm{J}\].
This formula helps us understand the energy dynamics of the chemical reaction and their implications on physical processes such as inflation of an air bag, which directly relate to safety applications in automotive design.