- #1
_Greg_
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Just done an experiment where you dissolve benzoic acid in water at different temperatures and different volumes of water.
From the van't hoff equation you then plot a graph of ln(solubility) against 1/Temp which produces a straight line with gradient -ΔH/RT from which you get the enthalpy change.
(vant hoff equation): lnk = -ΔH/RT + constant
The first question was:
why isn't it necessary to express solubility in mol/l despite ΔH beight evaluated in J/mol?
I answered with:
It isn’t necessary to convert solubility into mol/l because it’s the difference in solubility that’s important. Therefore plotting solubility with either units will always produce a line of the same gradient but with different Y intercept (the intercept being insignificant)
Is that ok?
The second question I am stuck on is:
Why is it that solubility, S, may be used instead of the equilibrium constant, K, in the van't hoff equation?
I would guess they are proportional to each other or something?
would really appreciate any help
From the van't hoff equation you then plot a graph of ln(solubility) against 1/Temp which produces a straight line with gradient -ΔH/RT from which you get the enthalpy change.
(vant hoff equation): lnk = -ΔH/RT + constant
The first question was:
why isn't it necessary to express solubility in mol/l despite ΔH beight evaluated in J/mol?
I answered with:
It isn’t necessary to convert solubility into mol/l because it’s the difference in solubility that’s important. Therefore plotting solubility with either units will always produce a line of the same gradient but with different Y intercept (the intercept being insignificant)
Is that ok?
The second question I am stuck on is:
Why is it that solubility, S, may be used instead of the equilibrium constant, K, in the van't hoff equation?
I would guess they are proportional to each other or something?
would really appreciate any help