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

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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:

$$∆G^circ = -RTln K quadtoquad ln K = -frac{∆G^circ}{RT}$$

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

$$ln K = -frac{∆H^circ}{RT} + frac{∆S^circ}{R},$$

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 33 mins ago

Karsten Theis

4,564542

asked 3 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 | 

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:

$$∆G^circ = -RTln K quadtoquad ln K = -frac{∆G^circ}{RT}$$

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

$$ln K = -frac{∆H^circ}{RT} + frac{∆S^circ}{R},$$

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 33 mins ago

Karsten Theis

4,564542

asked 3 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:

$$∆G^circ = -RTln K quadtoquad ln K = -frac{∆G^circ}{RT}$$

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

$$ln K = -frac{∆H^circ}{RT} + frac{∆S^circ}{R},$$

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 33 mins ago

Karsten Theis

4,564542

asked 3 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 33 mins ago

Karsten Theis

4,564542

asked 3 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 33 mins ago

Karsten Theis

4,564542

asked 3 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 33 mins ago

Karsten Theis

4,564542

edited 33 mins ago

Karsten Theis

4,564542

asked 3 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 3 hours ago

Mateen KasimMateen Kasim

212

1 Answer
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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 $$m = -frac{Delta H}{R} + frac{T Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-frac{Delta H}{R} times frac{1}{T}$ tends to zero and $frac{T Delta S}{R} times frac{1}{T}$ cancels to be just $frac{Delta S}{R}$.

share|improve this answer

edited 34 mins ago

answered 2 hours ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

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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 $$m = -frac{Delta H}{R} + frac{T Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-frac{Delta H}{R} times frac{1}{T}$ tends to zero and $frac{T Delta S}{R} times frac{1}{T}$ cancels to be just $frac{Delta S}{R}$.

share|improve this answer

edited 34 mins ago

answered 2 hours 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 $$m = -frac{Delta H}{R} + frac{T Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-frac{Delta H}{R} times frac{1}{T}$ tends to zero and $frac{T Delta S}{R} times frac{1}{T}$ cancels to be just $frac{Delta S}{R}$.

share|improve this answer

edited 34 mins ago

answered 2 hours 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 $$m = -frac{Delta H}{R} + frac{T Delta S}{R}$$

The $y$-intercept corresponds to an infinitely high temperature where $-frac{Delta H}{R} times frac{1}{T}$ tends to zero and $frac{T Delta S}{R} times frac{1}{T}$ cancels to be just $frac{Delta S}{R}$.

share|improve this answer

edited 34 mins ago

answered 2 hours ago

Karsten TheisKarsten Theis

4,564542

$endgroup$

share|improve this answer

edited 34 mins ago

answered 2 hours ago

Karsten TheisKarsten Theis

4,564542

answered 2 hours ago

Karsten TheisKarsten Theis

4,564542

answered 2 hours ago

Karsten TheisKarsten Theis

4,564542