Question

Clausius-Clapeyron equation derivation

Answer 1

\[\ln(\frac{ p }{ p_{o} }) = \frac{ \Delta H }{ R }(\frac{ 1 }{ T ^{0} }-\frac{ 1 }{ T })\]

Answer 2

Can any good math people explain how you can derive this equation?

Answer 3

@Woodward

Answer 4

@Kainui

Answer 5

@Empty

Answer 6

Alright, so we know that\[\Delta G^{\circ} = -RT \ln K_{eq}\]So we can make these two equations:\[\Delta H^{\circ} - T_1 \Delta S^{\circ} =-RT_1 \ln K_1\]\[\Delta H^{\circ} - T_2 \Delta S^{\circ} = -RT_2 \ln K_2\]Equating \(\Delta S^{\circ}\) we've got\[\frac{\Delta H^{\circ}}{T_1}+R \ln K_1 =\frac{\Delta H^{\circ}}{T_2} +R\ln K_2 \]\[\Rightarrow \ln \frac{K_2}{K_1} = \frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\]

Answer 7

Thank you

Answer 8

@parthkohli

So essentially, this is a way to express the change in entropy. in terms by relating it to K the reaction quotient.

This is very interesting because our reaction quotient K is actually temperature dependent.

\[k = \frac{ [C]^{X}[D]^{Y} }{ [A]^{Z}[B]^{D} }\]

so it seems this formula is built around the change in entropy delta S.

where Delta H and R are constants.

Answer 9

so it seems that a change in entropy is dependent on a change in temperature, which would also affect k since that's temperature dependent.