Calculate ΔSfus and ΔSvap for HF.
Determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure.
I already calculated ΔSfus as 24.103 J/k*mol and ΔSvap as 86.07 J/K*mol....These are correct (I checked them already). I need help with this part of the question:
Determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure.
I got an answer of 183.18 J/K and it marked me wrong and said "The value of ΔSfus above is for 1.00 mol of HF. You need to scale that value up for 7.60 mol of HF. Also, consider the sign of your answer. Does freezing result in an increase or decrease in entropy?"
Freezing is the opposite of melting, so reverse the ΔSfus. ΔSfus is in units of (J/K*mol). Multiply that by 7.6 mol. That should give you the right answer.
24!
How did you get that? I'm stuck on getting to 183.18
Ah, the frozen world of thermodynamics! Let's break it down, shall we?
To determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure, we need to scale up the value of ΔSfus, as pointed out.
So, since the value you have (ΔSfus = 24.103 J/K*mol) corresponds to 1.00 mol of HF, we can simply multiply it by the number of moles involved.
ΔSfus (for 7.60 mol of HF) = 24.103 J/K*mol * 7.60 mol
Now, the sign of the answer is crucial! Freezing generally results in a decrease in entropy, as the molecules become more ordered.
So, the correct answer for the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure would be -183.55 J/K.
Remember, entropy can be a bit chilly, so watch out for those signs!
To determine the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure, you need to scale up the value of ΔSfus for 1.00 mol of HF.
The given value of ΔSfus is 24.103 J/K*mol, which represents the entropy change for 1.00 mol of HF(l) freezing. To calculate the entropy change for 7.60 mol of HF(l) freezing, you can multiply the value of ΔSfus by the number of moles (7.60 mol):
Entropy change (ΔSfus) for 7.60 mol of HF(l) = 24.103 J/K*mol x 7.60 mol = 183.4468 J/K
Since entropy is an extensive property, it scales linearly with the amount of substance. Therefore, the value of ΔSfus for 7.60 mol of HF(l) is approximately 183.45 J/K.
However, there is another consideration mentioned in the feedback: the sign of your answer. When a substance freezes, it transitions from a more disordered state (liquid) to a more ordered state (solid). In other words, freezing results in a decrease in entropy. Therefore, the sign of the entropy change when HF(l) freezes should be negative.
Hence, the correct answer for the entropy change when 7.60 mol of HF(l) freezes at atmospheric pressure is approximately -183.45 J/K.