The Student Room Group

Entropy and Enthalpy

Hey Guys,

I need some help with a couple of questions

1.
Calculate the ΔSo\Delta S^o (Entropy change) of the following reaction at 298K:

CH4(s)CH_{4(s)}--->CH4(l)CH_{4(l)}


given that ΔHo=0.94kjmol1\Delta H^o = 0.94 kjmol^{-1}

They don't give you the standard entropy for each of the substances.
so I guess I have to use:
ΔG=ΔH(TΔS)\Delta G = \Delta H - (T\Delta S)

But they is no value given for the ΔG\Delta G

so how do I find ΔS\Delta S for this reaction.

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b.
Here we are given a slightly diff equation (note change in states)

CH4(l)CH_{4(l)}--->CH4(g)CH_{4(g)}


They give us:
ΔH=8.2Kjmol1\Delta H = 8.2Kjmol^{-1}
ΔS=73.2Jmol1K1\Delta S = 73.2Jmol^{-1}K^{-1}

We are asked to find the boiling point for this do I just set the ΔG=0\Delta G = 0 and find the value for T??

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2.
The following enthalpy changes are at 298 K:
ΔHhyd(kjmol1)\Delta H_{hyd} (kjmol^{-1}) :
K+=322K^{+} = -322 and
Cl=364Cl^{-}=-364

Lattice Enthalpy of KCl = -701kjmol(-1)

I worked out that standard enthalpy of solution= 15kjmol-1

using:
Lattice Enthalpy + Enthalpy of solution = \sum of Hydration enthalpies



"Calculate the smallest possible entropy change there must be when KCl dissolves in water at 298K, given that it's a spontaneous change"

wtf does that mean :eek: I have no clue where to being

Any help will be highly rewarded :smile:

Thanks so much in advance
Reply 1
Q1 - I think you've answered yourself in part b - the phase change is an equilibrium process so yes, delta G = 0. Yes to part b too.
Reply 2
Q2 Spontaneous change means delta G <0

it's been a long time since I've done thermodynamics & I can't remember the definitions but are you saying:

delta H (solution) = delta H (hydration) + delta H (lattice)?

In which case, once u find delta H (solution), plug it in to delta G = delta H - T delta S

where delta H- T delta S < 0 because it's spontaneous

and you know delta H (solution) and T, so you find delta S
you probably posted this not realising that you will be helping out someone else out there, 10 years into the future. thanks for your help!!
Reply 4
Original post by reinainoue
you probably posted this not realising that you will be helping out someone else out there, 10 years into the future. thanks for your help!!

...Or 10 years and one week later, haha.