[Unit 3] Gases and Intermolecular Forces

[wes-winter]

The following reaction occurs in a 2.00 L vessel at standard temperature and pressure.

a) What mass of fluorine gas occupies the 2.00 L vessel at standard temperature and pressure?
b) Which reactant, sulfur difluoride or fluorine gas, has a higher boiling point? Justify your answer with a description of all intermolecular forces in each molecule.
c) As the reaction proceeds, does the pressure inside the vessel increase, decrease, or stay the same? Justify your answer.

[GG1]

a) PV=nRT, so (1)(2)=x(0.0821)(298)

x=0.0817 so there are 0.0817 moles of F2. The molar mass of F2 is 38 grams, so there are 3.10 grams of F2.

b) SF2 is polar so it has london dispersion and dipole-dipole forces. F2 is polar so it only has london dispersion forces. Because dipole-dipole interactions have a greater relative strength than ldfs, it is harder to break the bonds in SF2 and the boiling point is higher.

c) The pressure inside the vessel stays the same because the temperature and volume are not altered.

[perlaalvarez431]

Honestly, I got pretty lost on part a and c. Please help

[wes-winter]

Hello,
I want to start by saying you’re on the right track and really close with a lot of these answers, I just want to work on some wording and make sure you understand why some of the language and math you used wasn’t the best.
First I want to point out that STP, standard temperature and pressure is 1 atm and 273K, not 298. I love that you used PV=nRT. you could have also used 22.4L of gas/mol (only true at STP). This threw off your calculation slightly.
Second, I think it was a typo since your statement for 1b) was right, but F₂ is nonpolar.
Third, bonds do not break when phase changes occur. They get enough energy to overcome the intermolecular forces holding them together.
Lastly, while temperature and volume are not altered, notice the number of particles does. How does the number of particles change and how do pressure and particles relate to each other?

[wes-winter]

1a) great, you could have also used PV=nRT with 1 atm and 273 K.

For your justification of 1b) I might have added that dipole-dipole forces are generally a stronger IMF. A helpful statement but maybe not necessary.

1c) make sure to consider what happens to the number of particles as the reaction occurs. Does it increase, decrease, or stay the same? Your reasoning works if the number of particles was constant. Unfortunately, that’s not the case.

[wes-winter]

Solutions
a)

You may also use PV = nRT where P=1 atm, V=2.00 L, R=0.0821, and T=273 K. This will determine the amount of moles in the substance. Multiply this value by the molar mass and you will get the same answer as above.
b) SF₂ has dipole-dipole force and London dispersion forces. F₂ has only London dispersion forces. The intermolecular forces in sulfur difluoride are stronger than the intermolecular forces in fluorine gas. This means the boiling point is greater in sulfur difluoride.
c) As the reaction proceeds the pressure inside the vessel will decrease. There are less moles on the product side thus with less particles colliding, the pressure will be less.