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Why is my conclusion inconsistent with the van't Hoff equation?

4

$begingroup$

Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:

$$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$

This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.

However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that

$$ln K = frac∆H^circRT - frac∆S^circR,$$

but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?

thermodynamics free-energy

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edited 1 hour ago

andselisk

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asked 2 hours ago

Mateen KasimMateen Kasim

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Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

4

$begingroup$

Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:

$$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$

This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.

However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that

$$ln K = frac∆H^circRT - frac∆S^circR,$$

but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?

thermodynamics free-energy

share|improve this question

edited 1 hour ago

andselisk

19.6k665127

asked 2 hours ago

Mateen KasimMateen Kasim

212

New contributor

Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Let's say I hypothesize that a graph of $ln K$ vs. $1/T$ has a slope of $-∆G^circ/R$ and a $y$-intercept of $0$. I prove it simply:

$$∆G^circ = -RTln K quadtoquad ln K = -frac∆G^circRT$$

This matches the linear form $y = mx + b$. Thus, plotting $ln K$ vs. $1/T$ would have a slope $m = -∆G^circ/R$ and a $y$-intercept $b = 0$.

However, I understand that a van't Hoff plot defines a graph of $ln K$ vs. $1/T$ to have a slope of $-ΔH^circ/R$ and a $y$-intercept of $∆S^circ/R$. It is clear from the relation $∆G^circ = ∆H^circ - TΔS^circ$ that my final equation is thermodynamically equivalent to the van't Hoff equation. I do not disagree that

$$ln K = frac∆H^circRT - frac∆S^circR,$$

but if I were to experimentally measure temperature and calculate the equilibrium constant temperature, why should I expect the y-intercept to be $∆S^circ/R$ as defined by van't Hoff rather than $0$ as I defined above? Why should I expect the slope to be $-ΔH^circ/R$ instead of $-ΔG^circ/R$? What makes the van't Hoff equation match experimentally determined values over the equation $ln K = -∆G^circ/(RT)$?

thermodynamics free-energy

share|improve this question

edited 1 hour ago

andselisk

19.6k665127

asked 2 hours ago

Mateen KasimMateen Kasim

212

New contributor

Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 1 hour ago

andselisk

19.6k665127

asked 2 hours ago

Mateen KasimMateen Kasim

212

New contributor

Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 1 hour ago

andselisk

19.6k665127

asked 2 hours ago

Mateen KasimMateen Kasim

212

New contributor

Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 1 hour ago

andselisk

19.6k665127

edited 1 hour ago

andselisk

19.6k665127

asked 2 hours ago

Mateen KasimMateen Kasim

212

New contributor

Mateen Kasim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 2 hours ago

Mateen KasimMateen Kasim

212

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4

$begingroup$

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$

The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $Delta S$.

share|improve this answer

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

add a comment | 

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4

$begingroup$

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$

The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $Delta S$.

share|improve this answer

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

add a comment | 

4

$begingroup$

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$

The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $Delta S$.

share|improve this answer

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

add a comment | 

$begingroup$

In the linear form $y = mx + b$, both $m$ and $b$ are constants, i.e. they don't depend on $x$. On the other hand, $Delta G^circ$ definitely depends on the temperature (and consequently on its inverse $1/T$). So if you plot a function $$f(x) = m x$$ where $m$ is not a constant but a function dependent on $x$, you might get something unexpected. In your case, $x$ is $1/T$ and $$m = -fracDelta HR + fracT Delta SR$$

The $y$-intercept corresponds to an infinitely high temperature where $-fracDelta HR times frac1T$ tends to zero and $fracT Delta SR times frac1T$ cancels to be just $Delta S$.

share|improve this answer

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

share|improve this answer

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

edited 1 hour ago

andselisk

19.6k665127

edited 1 hour ago

andselisk

19.6k665127

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542

answered 1 hour ago

Karsten TheisKarsten Theis

4,564542