Problem 43 - Chemistry The Molecular Science

Question

If \(1.0 \times 10^{3} \mathrm{~g}\) uranium metal is converted to gaseous \(\mathrm{UF}_{6},\) calculate the pressure of \(\mathrm{UF}_{6}\) at \(62{ }^{\circ} \mathrm{C}\) in a chamber that has a volume of \(3.0 \times 10^{2} \mathrm{~L}\).

Step-by-Step Solution

Answer

The pressure of UF6 is approximately 0.384 atm.

Step 1: Determine Moles of Uranium Metal

First, calculate the number of moles of uranium in the sample. The molar mass of uranium (U) is approximately 238 g/mol. We have 1000 g of uranium.\[\text{Moles of } U = \frac{1000 \text{ g}}{238 \text{ g/mol}} \approx 4.20 \text{ moles}\]

Step 2: Convert Uranium to UF6

Each mole of uranium reacts to form one mole of \( \text{UF}_6 \). So, the moles of \( \text{UF}_6 \) produced is the same as that of uranium, which is 4.20 moles.

Step 3: Use Ideal Gas Law

Use the ideal gas law \( PV = nRT \) to find the pressure. Rearrange it to solve for pressure \( P \):\[P = \frac{nRT}{V}\]Where:- \( n = 4.20 \) moles- \( R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1} \) (ideal gas constant)- \( V = 300 \text{ L} \)- \( T = 62 + 273 = 335 \text{ K} \) (Temperature in Kelvin)

Step 4: Calculate the Pressure

Substitute the known values into the equation from Step 3:\[P = \frac{(4.20 \text{ moles})(0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1})(335 \text{ K})}{300 \text{ L}}\]Simplify to find \( P \):\[P \approx \frac{115.15}{300} \approx 0.384 \text{ atm}\]

Step 5: Verify Units and Solution

Ensure that all units are consistent and appropriate for the ideal gas law calculations. The units used (moles, liters, Kelvin, and the ideal gas constant in L atm K⁻¹ mol⁻¹) confirm that the pressure is correctly computed in atmospheres.

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products. In the given problem, we transform uranium (U) metal into uranium hexafluoride (UF extsubscript{6}). The chemical reaction can be expressed as:\[ \text{U} + 6\text{F}_2 \rightarrow \text{UF}_6 \]In this case, it's particularly simplified since each mole of uranium converts directly to one mole of UF extsubscript{6}. This one-to-one mole ratio makes it straightforward to determine that the moles of UF extsubscript{6} produced are equal to the moles of uranium used.
Understanding these stoichiometric relationships is crucial for calculating how much of each product will form from given quantities of reactants, which allows us to move on to other calculations, such as gas pressure under specific conditions.

Molar Mass Calculation

Calculating molar mass is a fundamental part of stoichiometry. In our exercise, we need to know how many moles of uranium are in 1000 g of uranium to proceed correctly with further calculations.

• The molar mass of uranium (U) is 238 g/mol.
• To find the moles of uranium, use the formula:

\[ \text{Moles of U} = \frac{1000 \text{ g}}{238 \text{ g/mol}} \]This calculation tells us that there are approximately 4.20 moles of uranium.
Once you have the molar mass and the weight in grams, you simply divide the mass by the molar mass. This conversion from grams to moles is critical because stoichiometry works in terms of moles, not grams, allowing us to use equations like the ideal gas law accurately.

Gas Pressure Calculation

Once we determine the amount of UF extsubscript{6} gas produced, we use the Ideal Gas Law to find the pressure it exerts within a defined volume and temperature. The formula used is:\[ PV = nRT \]

• P is the pressure (in atmospheres).
• V is the volume of the container (300 L in our scenario).
• n is the number of moles (4.20 moles as calculated).
• R is the ideal gas constant, 0.0821 L atm K^-1 mol^-1.
• T is the temperature in Kelvin (62°C + 273 = 335 K).

Rearrange for pressure:\[ P = \frac{nRT}{V} \]By substituting the known values, we find:\[ P = \frac{(4.20) (0.0821) (335)}{300} \]Simplifying this gives a pressure of 0.384 atm.
This calculation shows how changes in temperature, volume, and the amount of substance can affect the pressure of a gas, reflecting the dynamic nature of gas behavior.