Calculate the pressure (in atmosphere) of IF5 after the reaction has gone to completion.

Question

A 7.78 x 10^-1 g sample or IBr and a 1.53 x 10^-1 g sample of F₂ react in a closed 9.90 L container at 495 K, according to the balanced chemical equation: IBr (g) + 4F2 (g) > IF₅ (g) + BrF₃ (g) Calculate the pressure (in atmosphere) of IF₅ after the reaction has gone to completion.

Answer 1

moles IBr present = 7.78x10^-1 g x 1 mol IBr / 207 g = 3.76x10^-3 mol

moles F₂ present = 1.53x10^-1 g x 1 mol F2 / 38.0 g = 4.03x10^-3mol

F₂ is limiting as it takes 4x as much as IBr, and there isn't 4x as much. All the F₂ will be used up.

Using an ICE table to see what happens, we have...

IBr (g) + 4F2 (g) ===> IF₅ (g) + BrF₃ (g)

3.76x10^-3.......4.03x10^-3...........0....................0...............Initial

1.01x10^-3......-4.03x10^-3........+1.01x10^-3...+1.01x10^-3....Change

2.75x10^-3............0...................1.01x10^-3.....1.01x10^-3....Equilibrium

Moles of IF₅ @ equilibrium = 1.01x10^-3 and pressure can be determined using the ideal gas law:

PV = nRT

P = pressure = ?

V = volume = 9.90 L

n = moles IF₅ = 1.01x10^-3

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 495K

Solving for P, we have...

P = nRT/V = (1.01x10^-3)(0.0821)(495) / 9.90

P = 4.15x10^-3 atm (3 sig figs)

Answer 2

Hello! The key to answering this questions comes in way for utilizing the Ideal Gas Law (PV = nRT) to find the pressure exerted by your gas product. IF5. In order to solve the problem, you'll need to figure out how many moles of product (IF5) you produced, which involves stoichiometry. Afterwards, you will have four of the five variables to use the Ideal Gas Law equation. Please let me know if you have further questions!

To calculate the pressure of IF₅ (iodine pentafluoride) after the reaction has gone to completion, we need to follow these steps:

1. Convert the masses of IBR and F₂ into moles using their molar masses.
2. Determine the limiting reactant by comparing the mole ratio from the balanced chemical equation.
3. Calculate the moles of IF₅ produced.
4. Use the Ideal Gas Law to find the pressure: PV = nRT

From the equation, 1 mole of IBr reacts with 4 moles of F₂ to produce 1 mole of IF₅.

1. Mass of IBr = 7.78×10⁻¹ g
2. Mass of F₂ = 1.53×10⁻¹ g
3. Volume of the container = 9.90 L
4. Temperature = 495 K
5. R (ideal gas constant) = 0.0821 L·atm/mol·K
6. Molar mass of IBr ≈ 206.9 g/mol
7. Molar mass of F₂ ≈ 38 g/mol

Step 1: Convert Masses to moles

IBr:

o    7.78×10⁻¹ g x (1 mol/206.9 g) = 0.00376 mol IBr

F₂

o    1.53×10⁻¹ g x (1 mol/38.0 g) = 0.00403 mol F₂

Step 2: Find the Limiting Reagent

Based on the mole-to-mole ratio, we need 4 moles of F₂ to react in order to produce 1 mole of IF₅. When doing the stoichiometry, we would find that the amount of F₂ used in this reaction will produce less IF₅ vs. the amount of IBr used, making F₂ our limiting reagent!

Step 3: Calculate moles of IF5 produced (based on our limiting reagent)

Moles of IF₅ Produce:

o    0.00403 mol F₂ x (1 mol IF5/ 4 mol F2) = 0.0010075 mol IF₅

Step 4: Ideal Gas Law

PV = nRT

|Volume of the container = 9.90 L| Temperature = 495 K| R = 0.0821 L·atm/mol·K | n = 0.0010075 mol

P = (nRT)/V --> ~plug in the variables and solve for P~ --> P = 0.00413 atm