Problem 129 - Chemistry The Central Science

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-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 (c) Draw the Lewis structure of \(\mathrm{IF}_{5}\). (d) What is the total mass of reactants and products in the flask?

Step-by-Step Solution

Answer

The partial pressure of IF5 in the flask is 5.20 atm, the mole fraction of IF5 is 0.194, the Lewis structure of IF5 is a square pyramid with a central I atom single-bonded to five surrounding F atoms and two lone pairs on the I atom, and the total mass of reactants and products in the flask is 29.91 g.

Step 1: Calculate the initial moles of I2 and F2

Utilize the molar masses of iodine (I) and fluorine (F) to determine the moles of I2 and F2 we begin with. Iodine: molar mass = 253.8 g/mol Fluorine: molar mass = 19 g/mol For I2: moles = mass (10 g) / molar mass (253.8 g/mol) = 0.0394 mol For F2: moles = mass (10 g) / molar mass (19 g/mol) = 0.5263 mol

Step 2: Identify the limiting reactant

To determine the limiting reactant, compare the mole ratio of reactants based on the balanced chemical equation. Stoichiometry: I2 : F2 = 1 : 5 For I2: given moles (0.0394 mol) / stoichiometry (1) = 0.0394 For F2: given moles (0.5263 mol) / stoichiometry (5) = 0.1053 I2 has the smallest ratio (0.0394), so it is the limiting reactant.

Step 3: Calculate moles of IF5 produced

Use the stoichiometry from the balanced equation to determine the moles of IF5 produced. Stoichiometry: I2 : IF5 = 1 : 2 Moles of IF5 = moles of limiting reactant (0.0394 mol) * stoichiometry (2) = 0.0788 mol

Step 4: Use the ideal gas law to find the partial pressure of IF5

To calculate the partial pressure of IF5, utilize the ideal gas law: PV = nRT. Volume (V) = 5.00 L Temperature (T) = 125°C + 273.15 = 398.15 K Ideal gas constant (R) = 0.0821 L*atm/(K*mol) Rearrange for pressure (P): \(P = \dfrac{nRT}{V}\) Calculate the pressure: P(IF5) = (0.0788 mol * 0.0821 L*atm/(K*mol) * 398.15 K) / 5.00 L = 5.20 atm

Step 5: Calculate the mole fraction of IF5

Determine the total moles of substances in the flask after the reaction is complete. Total moles = moles of IF5 + remaining moles of F2 Remaining F2 = initial F2 - moles reacted, which is 0.5263 mol - (0.0394 * 5) = 0.3273 mol Mole fraction of IF5 = moles of IF5 / (moles of IF5 + remaining moles of F2) Mole fraction of IF5 = 0.0788 / (0.0788 + 0.3273) = 0.194

Step 6: Draw the Lewis structure of IF5

Attached is a Lewis structure for IF5: I atom will be the central atom, with five F atoms surrounding and single bonded to it, as iodine has 7 valence electrons and each fluorine atom has 7 valence electrons. F | F-I-F | F | F Additionally, two lone pairs are on the I atom, giving IF5 a square pyramid molecular geometry.

Step 7: Calculate the total mass of reactants and products in the flask

Although the limiting reactant has been consumed, there is still remaining F2. Total mass = mass of IF5 + mass of remaining F2 Mass of IF5 = moles of IF5 * molar mass of IF5 = 0.0788 mol * 221.9 g/mol (IF5 molar mass) = 17.49 g Mass of remaining F2 = 0.3273 mol * 38 g/mol (F2 molar mass) = 12.42 g Total mass = 17.49 g + 12.42 g = 29.91 g

Key Concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that is completely used up first and thus determines the amount of product that can be formed. It acts as a bottleneck for the reaction as once it is consumed, the reaction can no longer proceed.

To identify the limiting reactant, you compare the mole ratio of reactants based on the balanced chemical equation. For instance, when reacting iodine (I₂) and fluorine (F₂), their molar ratio according to the balanced equation is 1:5. By calculating the available moles of_each and applying this ratio, we find that I₂ is the limiting reactant because it has the smallest mole ratio when divided by the stoichiometric coefficient. This crucial step dictates the course of the rest of the calculations in stoichiometry.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that relates the pressure (P), volume (V), number of moles (n), and temperature (T) of an ideal gas through the equation PV = nRT, where R is the ideal gas constant. This law allows us to calculate any one of these variables if the others are known.

For example, after determining that I₂ is the limiting reactant in the formation of iodine pentafluoride (IF₅), the ideal gas law can be applied to find the partial pressure of IF₅ inside a flask. By rearranging the ideal gas law to solve for pressure and plugging in the known values for the number of moles, temperature, and volume, you can calculate the pressure.

Mole Fraction

The mole fraction represents the ratio of the number of moles of a component to the total number of moles of all components in a mixture. It is a dimensionless quantity that can be used to calculate partial pressures, mole ratios in reactions, and as a conversion factor in various chemical calculations.

To calculate the mole fraction of IF₅, we add together the moles of IF₅ produced and the remaining moles of F₂ after the reaction. The mole fraction of IF₅ is then the quotient of the moles of IF₅ by the total moles in the flask. This number is crucial when dealing with mixtures of gases, as it can be used in conjunction with the total pressure of a system to find the partial pressure of individual gases.

Lewis Structure

The Lewis structure is a graphical representation of the valence electrons of atoms within a molecule. It illustrates how atoms are bonded together and shows lone pairs of electrons. When creating a Lewis structure, we look at the total number of valence electrons and arrange them to satisfy the octet rule for each atom, if possible.

For IF₅, iodine is the central atom. It is surrounded by five fluorine atoms, which are bonded to it through single bonds. Since iodine is in period 5 of the periodic table, it can hold more than eight electrons in its outer shell, which allows it to accommodate the five bonds with fluorine. Completing the Lewis structure for IF₅, we place two lone pairs on the iodine atom, giving it a square pyramidal geometry.