5 added 8 characters in body
source | link

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

4 mathjax
source | link

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$A+B\ce{<=>}C+D$$$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$.)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$A+B \rightarrow C+D$$$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$A+B\ce{<=>}C+D$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$.)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$A+B \rightarrow C+D$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$\ce{A + B <=> C + D}$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$\ce{A + B -> C + D}$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

3 added 12 characters in body
source | link

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$A+B\ce{<=>}C+D$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G$$\Delta G^0$:

$$\Delta G=-RT\ln K$$$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G=\Delta H-T\Delta S$$\Delta G^0=\Delta H-T\Delta S$.)

Evaluate this Nernst function to conclude that the more negative $\Delta G$$\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G}{RT}$$$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G$$\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$A+B \rightarrow C+D$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$A+B\ce{<=>}C+D$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G$:

$$\Delta G=-RT\ln K$$

(Of course, as you know $\Delta G=\Delta H-T\Delta S$.)

Evaluate this Nernst function to conclude that the more negative $\Delta G$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G}{RT}$$

For very negative values of $\Delta G$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$A+B \rightarrow C+D$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:

$$A+B\ce{<=>}C+D$$

We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$

Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:

$$\Delta G^0=-RT\ln K$$

(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$.)

Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:

$$K=e^\frac{-\Delta G^0}{RT}$$

For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:

$$A+B \rightarrow C+D$$

Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

2 added 47 characters in body
source | link
1
source | link