Question

Gaseous iodine pentafluoride, \(\mathrm{IF}_{5}\), can be prepared by the reaction of solid iodine and gaseous fluorine: $$ \mathrm{I}_{2}(s)+5 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{IF}_{5}(g) $$ A \(5.00-\mathrm{L}\) flask containing \(10.0 \mathrm{~g}\) of \(\mathrm{I}_{2}\) is charged with \(10.0 \mathrm{~g}\) of \(\mathrm{F}_{2}\), 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 \(\mathrm{IF}_{s}\) in the flask? (b) What is the mole fraction of \(\mathrm{IF}_{s}\) in the flask (c) Draw the Lewis structure of \(\mathrm{IF}_{\mathrm{s}}\). (d) What is the total mass of reactants and products in the flask?

Short answer

a) The partial pressure of \(\mathrm{IF}_5\) in the flask is 5.13 atm. b) The mole fraction of \(\mathrm{IF}_5\) in the flask is 0.544. c) The Lewis structure of \(\mathrm{IF}_5\) has an iodine atom in the center, bonded to five fluorine atoms and with a lone pair of electrons on iodine. d) The total mass of reactants and products in the flask is 19.99 g.

Step-by-step solution

Determine the limiting reactant

To find the limiting reactant, we will first calculate the number of moles of each reactant and then compare their mole ratios to the stoichiometry of the reaction. Moles of \(\mathrm{I}_2 = \frac{10.0 \mathrm{~g}}{253.8 \mathrm{~g/mol}} = 0.0394 \mathrm{~mol}\) Moles of \(\mathrm{F}_2 = \frac{10.0 \mathrm{~g}}{38.0 \mathrm{~g/mol}} = 0.263 \mathrm{~mol}\) The stoichiometric ratio of \(\mathrm{I}_2\) to \(\mathrm{F}_2\) is 1 : 5. Now let's check the actual ratio. Actual ratio of \(\mathrm{I}_2\):\(\mathrm{F}_2\) = \(\frac{0.0394}{0.263} = 0.15\) Since the stoichiometric ratio is smaller than the actual ratio, \(\mathrm{I}_2\) is the limiting reactant.

Calculate the number of moles of \(\mathrm{IF}_5\) formed and remaining moles of excess reactant

According to the stoichiometry of the reaction, 1 mole of \(\mathrm{I}_2\) reacts with 5 moles of \(\mathrm{F}_2\) to form 2 moles of \(\mathrm{IF}_5\). So, let's find the moles of \(\mathrm{IF}_5\) formed. Moles of \(\mathrm{IF}_5\) = \(\frac{2 \times 0.0394}{1} = 0.0788 \mathrm{~mol}\) Now, let's calculate the remaining moles of \(\mathrm{F}_2\). Remaining moles of \(\mathrm{F}_2\) = \(0.263 \mathrm{~mol} - 5 \times 0.0394 \mathrm{~mol} = 0.066 \mathrm{~mol}\)

Calculate the partial pressure of \(\mathrm{IF}_5\)

The flask has a total volume of 5.00 L and the temperature is given as 125°C (which is equal to 398.15 K). We can now use the ideal gas law to calculate the partial pressure of \(\mathrm{IF}_5\). \(P_{\mathrm{IF}_5}V = n_{\mathrm{IF}_5}RT\) \(P_{\mathrm{IF}_5} = \frac{n_{\mathrm{IF}_5}RT}{V} = \frac{0.0788 \mathrm{~mol} \times 0.0821 \mathrm{~\frac{L \cdot atm}{mol \cdot K}} \times 398.15 \mathrm{~K}}{5.00 \mathrm{~L}} = 5.13 \mathrm{~atm}\)

Calculate the mole fraction of \(\mathrm{IF}_5\)

To find the mole fraction, we first need to calculate the total number of moles in the flask after the reaction. Total moles = moles of \(\mathrm{IF}_5\) + remaining moles of \(\mathrm{F}_2\) Total moles = \(0.0788 \mathrm{~mol} + 0.066 \mathrm{~mol}\) = \(0.1448 \mathrm{~mol}\) Now, calculate the mole fraction of \(\mathrm{IF}_5\). Mole fraction of \(\mathrm{IF}_5\) = \(\frac{\text{moles of }\mathrm{IF}_5}{\text{total moles}} = \frac{0.0788}{0.1448} = 0.544\)

Draw the Lewis structure of \(\mathrm{IF}_5\)

To draw the Lewis structure of \(\mathrm{IF}_5\), we will count the total number of valence electrons and distribute them accordingly. Iodine has 7 valence electrons and each fluorine atom has 7 valence electrons. Total valence electrons = 7 (iodine) + 5 × 7 (fluorine) = 42 Now, let's connect iodine to each fluorine atom with single bonds. We have used 10 electrons so far. Next, assign the remaining 32 valence electrons as lone pairs to the fluorine atoms, with each fluorine atom getting 6 electrons. Thus, completing their octet. Finally, the central iodine atom has 2 extra electrons which form a lone pair. The result is the Lewis structure with 5 single bonds between I and F and a lone pair on iodine.

Calculate the total mass of reactants and products in the flask

The total mass will be the mass of \(\mathrm{IF}_5\) and remaining \(\mathrm{F}_2\). Calculate the mass of each and sum them Mass of \(\mathrm{IF}_5 = 0.0788 \mathrm{~mol} \times \frac{221.9\mathrm{~g}}{\mathrm{mol}} = 17.48\mathrm{~g}\) Mass of remaining \(\mathrm{F}_2 = 0.066 \mathrm{~mol} \times \frac{38\mathrm{~g}}{\mathrm{mol}} = 2.51\mathrm{~g}\) Total mass of reactant and products in the flask = Mass of \(\mathrm{IF}_5\) + Mass of remaining \(\mathrm{F}_2\) Total mass = \(17.48\mathrm{~g} + 2.51\mathrm{~g} = 19.99\mathrm{~g}\) To summarize: a) The partial pressure of \(\mathrm{IF}_5\) in the flask is 5.13 atm. b) The mole fraction of \(\mathrm{IF}_5\) in the flask is 0.544. c) The Lewis structure of \(\mathrm{IF}_5\) has an iodine atom in the center, bonded to five fluorine atoms and with a lone pair of electrons on iodine. d) The total mass of reactants and products in the flask is 19.99 g.

Key concepts

Limiting Reactant

In any chemical reaction, the limiting reactant (or limiting reagent) is the substance that is completely consumed first, stopping the reaction from continuing. It's crucial to identify the limiting reactant because it determines the maximum amount of product that can be generated.
To ascertain the limiting reactant, compare the moles of each reactant you have with the moles required by the reaction's stoichiometry. In this case, iodine (\( \mathrm{I}_2 \)) and fluorine (\( \mathrm{F}_2 \)) react in a 1:5 ratio to produce iodine pentafluoride (\( \mathrm{IF}_5 \)).

• Find moles of each reactant by using their mass and molecular weight
• Compare it to the mole ratio from the balanced equation

In the provided example, iodine is the limiting reactant, meaning the reaction will stop when iodine is exhausted, and some fluorine will remain unreacted.

Ideal Gas Law

The Ideal Gas Law is a crucial formula in chemistry that links pressure, volume, temperature, and moles of a gas. The formula is stated as:\[ PV = nRT \]Here,

• \(P\) is pressure
• \(V\) is volume
• \(n\) is number of moles
• \(R\) is the ideal gas constant (0.0821 L·atm/mol·K)
• \(T\) is temperature in Kelvin

Understanding how to use this law allows you to deduce any one of these variables when the others are known. In the given exercise, it helps calculate the partial pressure of \( \mathrm{IF}_5 \) after the reaction. By inserting the number of moles of \( \mathrm{IF}_5 \), volume of the container, temperature, and using \( R \), one can compute the pressure exerted by \( \mathrm{IF}_5 \). It's a practical tool whenever dealing with gases under typical conditions.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It's especially useful in mixtures where gases are involved. The mole fraction (\( X_A \)) is calculated as the ratio of the number of moles of that component to the total number of moles in the mixture:\[ X_A = \frac{n_A}{n_{\text{total}}} \]It's a dimensionless number, ranging between 0 and 1, representing the proportion of a particular molecule's presence compared to the total mixture.
For example, in our previous reaction, once we found the moles of \( \mathrm{IF}_5 \) formed and the leftover fluorine, we can calculate the mole fraction of \( \mathrm{IF}_5 \) in the solution. This figure helps in understanding the composition of the gas mixture, useful in many chemical calculations and scenarios.

Lewis Structure

The Lewis structure is a graphical illustration used to represent the electron configuration of molecules. It showcases the arrangement of atoms, bonds, and electrons, helping determine how atoms bond and interact within the molecule.
In constructing a Lewis structure:

• Start by calculating the total number of valence electrons available
• Use pairs of these electrons to form bonds between atoms
• Assign remaining electrons to fulfill the octet rule, mainly for outer atoms

For \( \mathrm{IF}_5 \), begin with iodine (7 valence electrons) in the center, bonded to five fluorine atoms (each having 7 valence electrons). After creating single bonds, assign the remaining electrons as lone pairs on the fluorine atoms. As a result, a lone pair remains on the iodine.
This visual representation aids in predicting molecule shape, bond angles, and potential reactivity. It simplifies understanding molecular structure significantly, a valuable tool for studying chemistry.