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From Atkins' Elements of Physical Chemistry (6th ed.):

$$\Delta G_\text{r}^\circ = -RT \ln K \tag{8}$$ This fundamentally important equation links thermodynamic quantities – which are widely available from tables of thermodynamic data – and the composition of a system at equilibrium. Note that:

  • If $\Delta G_\text{r}^\circ$ is negative, then $\ln K$ must be positive and therefore $K > 1$; products are favored at equilibrium.

  • If $\Delta G_\text{r}^\circ$ is positive, then $\ln K$ must be negative and therefore $K < 1$; reactants are favored at equilibrium.

It says products/reactants will be favored at equilibrium, but I thought at equilibrium neither are favored?

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At equilibrium the rates of the forward and reverse reactions are equal. This means that you can have more product than reactant, or more reactant than product. It just means that the rates of formation of each are equal and so there is no net change in the amount of each. The amount of products and reactants at equilibrium can be favored, their amounts just won't change once they reach equilibrium.

Depending on the temperature and pressure, the composition of the system at equilibrium can change.

Recall that in the reaction $$\ce{aA + bB<=>cC + dD}$$

The equilibrium constant is: $$K=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$$ These concentrations refer to the concentrations of $\ce{A}$ and $\ce{B}$ at equilibrium.

Therefore in a situation where $\ce{A<=>B}$ and $\Delta G_\mathrm{r}^\circ =-ve$, then $K>1$, meaning that there is more of $\ce{B}$ than $\ce{A}$ at equilibrium (and vice versa).


If you are interested to know why $\Delta G_\mathrm{r}^\circ = -RT \ln K$ gives the standard change in Gibbs energy for a reaction, then you should note that the general form of the equation at any point in the reaction is given by $$\Delta G_\mathrm{r} = \Delta G_\mathrm{r}^\circ + RT \ln Q$$

Where $Q=\frac{[\ce{C}]^c[\ce{D}]^d}{[\ce{A}]^a[\ce{B}]^b}$ at any point in the reaction. At equilibrium, there is no more Gibbs free energy, as the system no longer tends to the formation of products or reactants. Therefore at this point $\Delta G_\mathrm{r} = 0$, and the original expression given in Atkins is what you get. Also we define this special case of $Q$, saying that at equlibrium $Q=K$, the equilibrium constant.


enter image description here Source

You might find the above image helpful. The y-axis shows the amount of Gibbs free energy, $G$. The reaction will want to tend towards the state where it has the lowest amount of Gibbs free energy (the local minimum). The slope of the curve can be therefore seen as $\Delta G_\mathrm{r}$. Therefore the point where the system has the least Gibbs free energy will be when $\Delta G_\mathrm{r} = 0$ could occur anywhere in the extent of reaction. This shows that the equilibrium concentrations need not be equal, but shows that the reaction will not progress any further when $\Delta G_\mathrm{r} = 0$.

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It's not at equillibrium ,it occurs when we are going to achieve equillibrium.More the value of K ,the reaction will get driven towards the product side and more amount of product is formed Easily .(as more -ve value of ∆G suggests) .If K<1 then ∆G becomes -ve and the reaction will drive towards reactants side ).Opposite to what happens for K>1 case.

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  • $\begingroup$ That doesn't make any sense and I'm not sure if it's a language problem or a problem in your understanding in thermodynamics. $\endgroup$ – DSVA Sep 13 '17 at 14:15

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