Determine the entropy change when 7.40 mol of HCl(l) boils at atmospheric pressure. I already have the delta Sfus(12.5 J/Kmol) and delta Svap(93.2 J/Kmol), I just cant seem to figure out the last part.
Is this part of a problem and there is more to post? It appears to me you are have delta S vap and that x 7.40 will do it. But I may be confusing it with something else. If you have more to the problem please post it.
To determine the entropy change when HCl(l) boils at atmospheric pressure, you will need to use the equations:
ΔS_boil = ΔS_vap + ΔS_fus
ΔS represents the change in entropy.
ΔS_vap is the entropy change during vaporization.
ΔS_fus is the entropy change during fusion or melting.
Given:
ΔS_vap = 93.2 J/(K mol)
ΔS_fus = 12.5 J/(K mol)
You are looking for ΔS_boil.
Since boiling involves both vaporization and fusion, you can simply add the two entropy changes together.
ΔS_boil = ΔS_vap + ΔS_fus
= 93.2 J/(K mol) + 12.5 J/(K mol)
= 105.7 J/(K mol)
Therefore, the entropy change when 7.40 mol of HCl(l) boils at atmospheric pressure is 105.7 J/(K mol).
To determine the entropy change when HCl(l) boils at atmospheric pressure, you can use the equation:
ΔS = ΔSfus + ΔSvap
where ΔS is the total entropy change, ΔSfus is the entropy change during fusion (melting), and ΔSvap is the entropy change during vaporization (boiling).
You mentioned that you already have the values for ΔSfus and ΔSvap, which are 12.5 J/(K*mol) and 93.2 J/(K*mol), respectively.
Since boiling involves both fusion and vaporization, we can sum up the two entropy changes to find the total entropy change.
ΔS = ΔSfus + ΔSvap
= 12.5 J/(K*mol) + 93.2 J/(K*mol)
= 105.7 J/(K*mol)
Therefore, the entropy change when 7.40 mol of HCl(l) boils at atmospheric pressure is 105.7 J/(K*mol).