Calculating Final Pressure in a Sublimation and Pressure Homework Problem

Click For Summary

Discussion Overview

The discussion revolves around a homework problem involving the sublimation of solid iodine and the reaction with fluorine gas to form iodine heptafluoride. Participants explore the calculation of final pressure in a closed container after heating, considering the ideal gas law and Dalton's Law of Partial Pressures. The scope includes theoretical reasoning and mathematical calculations related to gas behavior under changing temperatures.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant outlines the initial conditions and calculations for the moles of fluorine and iodine, concluding that iodine is the limiting reactant.
  • Another participant questions the formula for iodine heptafluoride, leading to a clarification about the correct stoichiometry.
  • A later reply suggests a new balanced equation for the reaction, indicating a shift in understanding regarding the products formed.
  • Participants discuss the need to convert pressures to atmospheres at the final temperature of 550 K, emphasizing the importance of temperature in pressure calculations.
  • There is mention of setting up a stoichiometry problem to determine the total pressure based on the pressures of the product and remaining reactants.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations, as one participant expresses uncertainty about the accuracy of their initial solution, while another proposes a different approach without resolving the discrepancies in their methods.

Contextual Notes

Participants note the importance of considering temperature changes in gas calculations, and there are unresolved aspects regarding the application of Dalton's Law and the ideal gas law under varying conditions.

Qube
Gold Member
Messages
461
Reaction score
1

Homework Statement



When fluorine and solid iodine are heated at 550. K, the iodine completely sublimes and gaseous iodine heptafluoride forms. If 350. torr of fluorine gas and 2.50 g of solid iodine are put into a 2.50 L container at 250. K and the container is heated to 550. K, what is the final pressure in the flask?

Homework Equations



5F2 (g) + I2 (s) --> 2IF5(g)

The Attempt at a Solution



We have 350/760 atm of fluorine gas. Using the ideal gas law, P = nRT/V, we find taht we have 0.056 moles of F.

We also have 0.009 moles of solid iodine (2.5 grams / twice the molecular mass of elemental iodine).

This yields 0.0197 moles of product. Iodine is limiting and fluorine is in excess. The final pressure, by Dalton's Law of Partial Pressures, is the pressure of the product and the pressure of the remaining reactants.

We have 0.0197 * (5/2) moles of fluorine used. This means we have 0.056 minus 0.0475 moles of fluorine remaining. This means there are 0.085 moles of fluorine remaining. The pressure of this remaining fluorine is:

P = nRT/V = 0.085(0.0821)(550)/2.5 = 0.15 atm.

The pressure of the product is also nRT/V = 0.0197 * 0.0821 * 550 / 2.5 = 0.355 atm.

Summing these two pressures, we get 0.50 atm of pressure. However, this isn't an answer choice. The closest choice is 0.534 atm, leading me to believe there is an error in my calculations, and not a rounding one at that. Is there any flaw in my line of reasoning?
 
Physics news on Phys.org
What is the formula of the iodine heptafluoride?
 
Oh snap. Hept ... that's prefix for 7. I confused that with pentafluoride. Thanks for the catch!
 
Okay, new balanced equation:

7F2 + I2 --> 2IF7.

New process.

We have to convert everything to atmospheres of pressure at 550 K to solve this problem. This is the most convenient way. My method doesn't work because it doesn't account for the temperature change and how that might affect the Fluorine in terms of pressure. I can't use pressure of F at the initial temperature and arrive at an amount of Fluorine; it'll change when the temperature changes.

1) Find the atmospheres of I2 at 550 K. P = nRT/V. P = 0.177 atm.

2) Find the atmospheres of F2 at 550 K. P / T = P / T. Convert torr to atm (350/760) and temperatures are in Kelvins (250 and 550 K). Solve for the new pressure of F2.

3) Now we can set up this classic problem (classic for my teacher; he often gives us atmospheres of pressure and has us to a stoichiometry problem; this just takes it a step further by withholding the proper pressure at the proper temperature). Use atmospheres of pressure for each reactant and find which one will result in the least amount (pressure) of product. Total pressure = pressure of product + pressure of left over reactants.
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 10 ·
Replies
10
Views
5K
Replies
1
Views
5K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
16
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
9K
Replies
1
Views
7K
Replies
5
Views
16K