Chemistry - What is the difference between ∆G and ∆G°?
Solution 1:
Short answer
- Does it need to be at $25~^\circ\mathrm{C}$?
No. $\Delta_\mathrm{r} G^\circ$ can be defined at any temperature you wish to define it at, since the standard state does not prescribe a particular temperature. If you change the temperature, $\Delta_\mathrm{r} G^\circ$ will change.
- Does $\Delta_\mathrm{r} G^\circ = \Delta_\mathrm{r} H^\circ - T\Delta_\mathrm{r} S^\circ$ always use $T = 298~\mathrm{K}$?
No. You use whatever temperature you are running your reaction at.
- (...maths...)
Yes, at equilibrium, $\Delta_\mathrm{r} G = 0$ and $Q = K$.
However, everything after the first bullet point is wrong. You cannot conclude that $\Delta_\mathrm{r} G^\circ = 0$, nor can you conclude that $K = 1$. The equation $\Delta_\mathrm{r} G^\circ = -RT \ln K$ does not analogously translate into $\Delta_\mathrm{r} G = -RT \ln Q$! The accurate relation is:
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$
Setting $Q = K$ and $\Delta_\mathrm{r} G = 0$ in this equation does not tell you anything about the value of $K$. In fact, if you try doing it, all you will find out is that $\Delta_\mathrm{r} G^\circ = -RT \ln K$ - no surprises there!
Long answer
Any book that writes that $\Delta_\mathrm{r} G^\circ$ is the "special case" of $\Delta_\mathrm{r} G$ at $T = 298~\mathrm{K}$ is wrong.
The Gibbs free energy of a system is defined as follows:
$$G = H - TS$$
Under constant temperature and pressure (from now on, I will just assume constant $T$ and $p$ without stating it), all systems will seek to minimise their Gibbs free energy. Equilibrium is reached when $G$ is minimised. When $G$ is at a minimum, any infinitesimal change in $G$, i.e. $\mathrm{d}G$, will be $0$. Therefore, this is equivalent to saying that the condition for chemical equilibrium is $\mathrm{d}G = 0$.
Clearly, we need a way to relate this quantity $\mathrm{d}G$ to the actual reactants and products that are in the system. This can be done by using the Maxwell relation (see any physical chemistry text for details):
$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$
Under constant $T$ and $p$, $\mathrm{d}p = \mathrm{d}T = 0$ and therefore
$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i$$
where $\mu_i$ is the chemical potential of species $i$, defined as a partial derivative:
$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$
So, we now have a refined condition for equilibrium:
$$\mathrm{d}G = \sum_i \mu_i\,\mathrm{d}n_i = 0$$
We can go further by noting that the values of $\mathrm{d}n_i$ for different components $i$, $j$, etc. are not unrelated. For example, if we have a reaction $i + j \longrightarrow k$, then for each mole of $i$ that is consumed, we must also use up one mole of $j$; this means that $\mathrm{d}n_i = \mathrm{d}n_j$.
This can be formalised using the idea of a stoichiometric coefficient $\nu_i$, which is defined to be positive for products and negative for reactants. For example, in the reaction
$$\ce{3H2 + N2 -> 2NH3}$$
we have $\nu_{\ce{H2}} = -3$, $\nu_{\ce{N2}} = -1$, and $\nu_{\ce{NH3}} = 2$.
By stoichiometry, if $1.5~\mathrm{mol}$ of $\ce{H2}$ is consumed, then $1~\mathrm{mol}$ of $\ce{NH3}$ has to be produced. We could write $\Delta n_{\ce{H2}} = -1.5~\mathrm{mol}$ and $\Delta n_{\ce{NH3}} = 1~\mathrm{mol}$. These quantities are proportional to their stoichiometric coefficients:
$$\frac{\Delta n_{\ce{H2}}}{\nu_{\ce{H2}}} = \frac{-1.5~\mathrm{mol}}{-3} = 0.5~\mathrm{mol} = \frac{1~\mathrm{mol}}{2} = \frac{\Delta n_{\ce{NH3}}}{\nu_{\ce{NH3}}}$$
The quantity $0.5~\mathrm{mol}$ is a constant for all chemical species $\ce{J}$ that participate in the reaction, and it is called the "extent of reaction" and denoted $\Delta \xi$ (that is the Greek letter xi). If the reaction is going forward, then $\Delta \xi$ is positive, and if the reaction is going backwards, then $\Delta \xi$ is negative. If we generalise the above result, we can write
$$\Delta \xi = \frac{\Delta n_i}{\nu_i}$$
and if we make $\Delta n_i$ smaller and smaller until it becomes an infinitesimal, then:
$$\begin{align} \mathrm{d}\xi &= \frac{\mathrm{d}n_i}{\nu_i} \\ \mathrm{d}n_i &= \nu_i\,\mathrm{d}\xi \end{align}$$
If we go back to our condition for equilibrium, we can substitute in the above to get:
$$\mathrm{d}G = \sum_i \mu_i\nu_i\,\mathrm{d}\xi = 0$$
Now, $\mathrm{d}\xi$ is no longer dependent on $i$, since we have established already that $\Delta \xi$ (and by extension $\mathrm{d}\xi$) is a constant for all chemical species. So, we can "divide through" by it to get:
$$\Delta_\mathrm{r} G \equiv \frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$
where $\Delta_\mathrm{r} G$ is defined to be $\mathrm{d}G/\mathrm{d}\xi$.
Note that $\Delta_\mathrm{r} G$ is an intensive property and has units of $\mathrm{kJ~mol^{-1}}$, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$. This ensures that the units we use are consistent: since we know that $\Delta_\mathrm{r}G = \Delta_\mathrm{r} G^\circ + RT\ln Q$, $\Delta_\mathrm{r}G$ must have the same units as $RT$.
How do we interpret the physical significance of $\Delta_\mathrm{r} G$, or in other words, what does it even mean? There are two ways, each based on a different mathematical expression.
- We have $\Delta_\mathrm{r}G = \sum \nu_i \mu_i$. This means that $\Delta_\mathrm{r}G$ is simply the difference between the chemical potentials of the products and the reactants, weighted by their stoichiometric coefficients. For the reaction $\ce{3H2 + N2 -> 2NH3}$, we have:
$$\Delta_\mathrm{r} G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$
- We have $\Delta_\mathrm{r}G = \mathrm{d}G/\mathrm{d}\xi$. This means that it is the slope of a curve of $G$ against $\xi$:
Note that up to this point, we have not stipulated any particular temperature, pressure, amounts of species present, or any conditions whatsoever. We have only said that the temperature and pressure must be constant.
It is important to realise that $\Delta_\mathrm{r}G$ is a well-defined quantity at all $T$, all $p$, and all possible values of $n_i, n_j, \cdots$! The shape of the curve will change when you vary the conditions. However, no matter what the curve looks like, it is always possible to find its gradient ($= \Delta_\mathrm{r}G$) at a particular point.
What exactly, then, is $\Delta G^\circ$? It is just a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:
- For a gas: pure ideal gas when the pressure $p$ is equal to the standard pressure $p^\circ$.
- For a liquid or solid: pure liquid or solid at $p = p^\circ$
- For a solution: ideal solution when the concentration $c$ is equal to the standard concentration $c^\circ$.
$p^\circ$ is most commonly taken to be $\pu{1 bar}$, although older texts may use the value $\pu{1 atm} = \pu{1.01325 bar}$. Since 1982, IUPAC has recommended the value $\pu{1 bar}$ for the standard pressure (Pure Appl. Chem. 1982, 54 (6), 1239–1250; DOI: 10.1351/pac198254061239). However, depending on the context, a different value of $p^\circ$ may prove to be more convenient. Likewise, $c^\circ$ is most commonly – but not necessarily – taken to be $\pu{1 mol dm-3}$.
Note that in the above definitions, no temperature is specified. Therefore, by defining the standard Gibbs free energy, we are fixing a particular value of $p$, as well as particular values of $n_i, n_j, \cdots$. However, the value of $T$ is not fixed. Therefore, when stating a value of $\Delta_\mathrm rG^\circ$, it is also necessary to state the temperature which that value applies to.
When a reaction vessel is prepared with all its substances in the standard state, all the components of the system will have an activity of exactly $1$ by definition. Therefore, the reaction quotient $Q$ (which is a ratio of activities) will also be exactly equal to $1$. So, we could also say that $\Delta_\mathrm{r}G^\circ$ is the value of $\Delta_\mathrm{r}G$ when $Q = 1$.
Returning to the graph of $G$ against $\xi$ above, we note that at the left-most point, $Q = 0$ since there are only reactants; at the right-most point, $Q \to \infty$ as there are only products. As we move from left to right, $Q$ increases continuously, so there must be a point where $Q = 1$. (In general, the point where $Q = 1$ will not be the same as the equilibrium point.) Since $\Delta_\mathrm{r}G$ is the gradient of the graph, $\Delta_\mathrm{r}G^\circ$ is simply the gradient of the graph at that particular point where $Q = 1$:
The gradient of the graph, i.e. $\Delta_\mathrm{r}G$, will vary as you traverse the graph from left to right. At equilibrium, the gradient is zero, i.e. $\Delta_\mathrm{r}G = 0$. However, $\Delta_\mathrm{r}G^\circ$ refers to the gradient at that one specific point where $Q = 1$. In the example illustrated above, that specific gradient is negative, i.e. $\Delta_\mathrm{r}G^\circ < 0$.
Again, I reiterate that the temperature has nothing to do with this. If you were to change the temperature, you would get an entirely different graph of $G$ versus $\xi$. You can still find the point on that graph where $Q = 1$, and the gradient of that graph at the point where $Q = 1$ is simply $\Delta_\mathrm{r} G^\circ$ at that temperature.
We have established the qualitative relationship between $\Delta_\mathrm{r} G$ and $\Delta_\mathrm{r} G^\circ$, but it is often useful to have an exact mathematical relation.
$\Delta_\mathrm{r} G^\circ$ is exactly the same as $\Delta_\mathrm{r} G$ except for the imposition of the standard state. It follows that if we take the equation
$$\Delta_\mathrm{r} G = \sum_i \mu_i \nu_i$$
and impose the standard state, we get
$$\Delta_\mathrm{r} G^\circ = \sum_i \mu_i^\circ \nu_i$$
Thermodynamics tells us that
$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$
where $a_i$ is the thermodynamic activity of species $i$. Substituting this into the expressions for $\Delta G$ and $\Delta G^\circ$ above, we obtain the result:
$$\Delta_\mathrm{r} G = \Delta_\mathrm{r} G^\circ + RT\ln Q$$
where the reaction quotient $Q$ is defined as
$$Q = \prod_i a_i^{\nu_i}$$
When equilibrium is reached, we necessarily have $\Delta_\mathrm{r} G = 0$ (see the discussion above). The equilibrium constant $K$ is defined to be the value of $Q$ at equilibrium. Therefore, at equilibrium, $Q = K$. Plugging this into the equation above gives us the famous equation:
$$\Delta_\mathrm{r} G^\circ = -RT\ln K$$
Again, no temperature is specified! In general, $K$ depends on the temperature as well; the relationship is given by the van 't Hoff equation.
Solution 2:
I think the answer to these questions can be simplified considerably. My answer would start out with the part of orthocresol's answer which says:
What exactly, then, is $\Delta G^\circ$? The truth is that is is simply a special case of $\Delta G$, where all the reactants and products are prepared in a standard state. According to IUPAC, the standard state is defined as:
- For a gas: pure ideal gas at $p = 1~\mathrm{bar}$
- For a liquid or solid: pure liquid or solid at $p = 1~\mathrm{bar}$
- For a solution: ideal solution at $c = 1~\mathrm{mol~dm^{-3}}$
I would then add the following:
For the cases of pure gases, liquids, and solids (not solutions), I like to think of the pure reactants and pure products as actually being in separate containers, each at 1 bar pressure. So, in these cases, the 0 superscript refers specifically to reactants and products all at 1 bar pressure. It can be at any specified temperature. So $\Delta G^0$ refers to the change in free energy in transitioning from thermodynamic State 1 to thermodynamic State 2, where
State 1 = Pure reactants (in stoichiometric proportions) in separate containers at 1 bar and temperature T
State 2 = Pure products (in corresponding stoichiometric proportions) in separate containers at 1 bar and temperature T
Then, for $\Delta G$, it is basically the same thing, except that here the pure reactants and pure products are at arbitrary specified pressures in their separate containers, not necessarily at 1 bar. If these pressures happen to coincide with the partial pressures in an equilibrium mixture of reactants and products (assuming ideal gas behavior), then the $\Delta G$ is zero.
The trick in all this is to figure out a reversible path between pure reactants and pure products at 1 bar (such that the changes in G can be evaluated). That's where the van't Hoff equilibrium box comes in.
Solution 3:
I see it in different way. You are right in most textbooks $\Delta G^0$ means free energy change at $25~\mathrm{^\circ C}$ but in this case you have to think in a different way. I see it as segregation of pressure and temperature dependence.
$$\Delta G(T,P)=\Delta G^0 (T)+RT\ln(K_\mathrm{eq})$$
At equilibrium you must have $\Delta G(T,P)=0$
So, it helps me to find the trade-off between free energy contribution due to temperature and due to activity of the reactant-product system at equilibrium condition.
In microkinetic modeling, I often need to include lateral interaction and in that case my $\Delta G^0 (T) $ is not constant at a particular temperature but a function of coverage, that's why I have a dynamic relation between $\Delta G^0$ and $K_\mathrm{eq}$. But in most of the practical system it's just a static relation and only means one particular point on thermodynamic landscape.