Question

Determine the entropy change when 1.80 mol of HBr(g) condenses at atmospheric pressure?

Answer 1

That's the question, I've seriously tried everything and I just don't know how to find the Entropy change of the solution. I don't want to loose any more points on the question and I want to understand how to solve for it. ANYONE PLEASE!

Answer 2

Entropy Change is Del S = Del H / T.. Please try to solve by using this equation.

Answer 3

@nanosaravana Del H from vapourization or fusion?

Answer 4

yes

Answer 5

@nanosaravana condensation is delta H fusion, not vapourization right?

Answer 6

Yes, condensation is not vaporization. You know the Del S per mole which is tabulated. How do you calculate for 1.8 moles is the answer:

Answer 7

Delta S= 12.922/(melting point + 273.15 ((-86.96+273.15)) then multiply that by 1.7 for moles?

Answer 8

Or should I use the temperature for boiling instead of melting?

Answer 9

Melting point can be used.

Answer 10

So it would be melting point temperature rather than boiling?

Answer 11

When considered fusion, yes. It is boiling point When working on vapurization

Answer 12

So then if I solve down, I get an entropy change of .117798 J/K.... That looks a little low.....

Answer 13

condensation is the same \(process\) as vaporization, but in the opposite direction. You need to find the molar entropy of vaporization, with the opposite sign, and then multiply by 1.80 because you have 1.80 moles, rather than 1.0 moles

Answer 14

@JFraser THANK YOU SO MUCH I was using the wrong entropy because I thought it was fusion!! You've solved my confusion!

Answer 15

YW