Question

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$ \mathrm{I}_{2}(\mathrm{~s})+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g) $$ A 5.00-L flask containing \(10.0 \mathrm{~g} \mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g}\) \(\mathrm{F}_{2}\), and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is \(125^{\circ} \mathrm{C}\). (a) What is the partial pressure of \(\mathrm{IF}_{5}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{5}\) in the flask?

Short answer

(a) The partial pressure of \(\mathrm{IF}_{5}\) in the flask is \(2.59 \textrm{atm}\). (b) The mole fraction of \(\mathrm{IF}_{5}\) in the flask is \(0.544\).

Step-by-step solution

Determine the number of moles of reagents #from the given masses

First, let's find the number of moles of \(\mathrm{I}_{2}\) and \(\mathrm{F}_{2}\) using their molar masses: For \(\mathrm{I}_{2}\), molar mass \(= 253.8 \frac{\textrm{g}}{\textrm{mol}}\) For \(\mathrm{F}_{2}\), molar mass \(= 38.0 \frac{\textrm{g}}{\textrm{mol}}\) Moles of \(\mathrm{I}_{2} = \frac{10.0 \textrm{g}}{253.8 \frac{\textrm{g}}{\textrm{mol}}} = 0.0394 \textrm{mol}\) Moles of \(\mathrm{F}_{2} = \frac{10.0 \textrm{g}}{38.0 \frac{\textrm{g}}{\textrm{mol}}} = 0.263 \textrm{mol}\)

Identify the limiting reactant

The balanced chemical equation tells us that \(1\) mole of \(\mathrm{I}_{2}\) reacts with \(5\) moles of \(\mathrm{F}_{2}\). Let's find the mole ratio of \(\frac{\textrm{F}_{2}}{_{\textrm{I}_{2}}}\) in the flask: Mole ratio \(= \frac{0.263 \textrm{mol}}{0.0394 \textrm{mol}} = 6.68\) Since \(6.68 > 5\), the mole ratio of \(\mathrm{F}_{2}\) to \(\mathrm{I}_{2}\) is greater than required by the stoichiometry of the reaction. This means \(\mathrm{I}_{2}\) is the limiting reactant.

Find the number of moles of products and remaining reactants

Using the stoichiometry of the reaction, we can determine the number of moles of \(\mathrm{IF}_{5}\) produced: The balanced chemical equation tells us that \(1\) mole of \(\mathrm{I}_{2}\) produces \(2\) moles of \(\mathrm{IF}_{5}\): Moles of \(\mathrm{IF}_{5} = 2 \times 0.0394 \textrm{mol} = 0.0788 \textrm{mol}\) Additionally, let's calculate the remaining moles of \(\mathrm{F}_{2}\): Moles of \(\mathrm{F}_{2}\) consumed \(= 5 \times 0.0394 \textrm{mol} = 0.197 \textrm{mol}\) Remaining moles of \(\mathrm{F}_{2} = 0.263 \textrm{mol} - 0.197 \textrm{mol} = 0.066 \textrm{mol}\)

Calculate the final pressure of the system

We are given that the final temperature is \(125^{\circ} \mathrm{C}\). Converting this to Kelvin gives: \(T = 125 + 273.15 = 398.15 \mathrm{K}\) Using the ideal gas law, \(PV = nRT\), where \(n\) is the total number of moles in the system (both products and remaining reactants), and \(R\) is the ideal gas constant (\(0.0821 \frac{\textrm{L} \cdot \textrm{atm}}{\textrm{K} \cdot \textrm{mol}}\)): Total moles, \(n = 0.0788 \textrm{mol}(\mathrm{IF}_{5}) + 0.066 \textrm{mol}(\mathrm{F}_{2}) = 0.1448 \textrm{mol}\) Solving for the pressure, \(P\): \(P = \frac{nRT}{V} = \frac{(0.1448 \textrm{mol})(0.0821 \frac{\textrm{L} \cdot \textrm{atm}}{\textrm{K} \cdot \textrm{mol}})(398.15 \textrm{K})}{5.00 \textrm{L}} = 4.76 \textrm{atm}\)

Calculate the partial pressure of \(\mathrm{IF}_{5}\)

As we know the total pressure and moles of each gas in the flask, we can calculate the partial pressure of \(\mathrm{IF}_{5}\) using the mole fraction formula, \(P_{\textrm{gas}} = mole\_fraction \times P_{total}\): Mole fraction of \(\mathrm{IF}_{5} = \frac{0.0788 \textrm{mol}}{0.1448 \textrm{mol}} = 0.544\) Partial pressure of \(\mathrm{IF}_{5} = (0.544)(4.76 \textrm{atm}) = 2.59 \textrm{atm}\) The partial pressure of \(\mathrm{IF}_{5}\) in the flask is \(2.59 \textrm{atm}\).

Calculate the mole fraction of \(\mathrm{IF}_{5}\)

The mole fraction of \(\mathrm{IF}_{5}\) has already been calculated in step 5: Mole fraction of \(\mathrm{IF}_{5} = 0.544\) (a) The partial pressure of \(\mathrm{IF}_{5}\) in the flask is \(2.59 \textrm{atm}\). (b) The mole fraction of \(\mathrm{IF}_{5}\) in the flask is \(0.544\).

Key concepts

Stoichiometry

Stoichiometry is a branch of chemistry that focuses on the quantitative relationships between reactants and products in a chemical reaction. It relies on the balanced chemical equation and uses the concept of molar ratios.
For instance, in the reaction of iodine (\( \mathrm{I}_2 \)) with fluorine (\( \mathrm{F}_2 \)), stoichiometry helps determine how much \( \mathrm{IF}_5 \) can be produced.

• The coefficients in a balanced equation indicate the proportions in which reactants combine.
• This allows chemists to predict if a reaction has enough resources to proceed completely.

Engaging with stoichiometry aids in planning reactions and conserving materials.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that gets consumed first, preventing the reaction from continuing. Identifying the limiting reactant is crucial because it determines the maximum amount of product that can be formed.
In our iodine-fluorine reaction, \( \mathrm{I}_2 \) was found to be the limiting reactant.

• A comparison of the available mole ratios to those required by the balanced equation reveals the limiting material.
• Once the limiting reactant is consumed, the reaction halts, even if other reactants remain.

Understanding this concept helps in efficiently utilizing resources in laboratory or industrial settings.

Stoichiometry Calculations

Stoichiometry calculations involve determining the quantities of reactants and products in a chemical reaction. These calculations start by converting masses into moles using molar masses.
In the iodine and fluorine reaction, the given masses were translated into moles to proceed with stoichiometric ratios.

• Calculate the moles of each reactant using their respective molar mass.
• Use the balanced equation to find the mole-to-mole relationships.
• This allows you to predict the quantities of products formed.

Accurate stoichiometry calculations are vital for predicting yields and ensuring reactions go as planned.

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's given by \( PV = nRT \).
In the given reaction, this law helped calculate the pressure after the reaction was complete.

• \( P \) is the pressure, \( V \) is the volume, \( n \) is the moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.
• Converting temperature to Kelvin is essential for accuracy.
• This law assumes gases behave ideally, which is a close approximation in many conditions.

Mastery of the Ideal Gas Law is crucial for handling gases and calculating their properties under different conditions.