# Short answer
> 1. Does it need to be at $25~^\circ\mathrm{C}$?
**No.** $\Delta G^\circ$ can be defined at any temperature you wish to define it at, since the [standard state](http://goldbook.iupac.org/S05925.html) does not prescribe a particular temperature. If you change the temperature, $\Delta G^\circ$ will change.
> 2. Does $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ always use $T = 298~\mathrm{K}$?
**No.** You use whatever temperature you are running your reaction at.
> 3. (...)
You got it correct up to a certain point. Yes, at equilibrium, $\Delta G = 0$ and $Q = K$. However, everything after that is wrong. You cannot conclude that $\Delta G^\circ = 0$. There is no constraint that $\Delta G = \Delta G^\circ$. They are related by the equation
$$\Delta G = \Delta G^\circ + RT\ln Q$$
$Q$ is not necessarily equal to $1$, so $\ln Q$ is not necessarily $0$ and $\Delta G$ is not necessarily equal to $\Delta G^\circ$.
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# Long answer
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. When $G$ is 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:
$$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T + \sum_i \mu_i\,\mathrm{d}n_i$$
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:
$$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{n_{j\neq i}}$$
A proof of this equation can be found in any physical chemistry textbook and I will not go into it. The chemical potential can be interpreted as a *partial* molar Gibbs free energy, $G_m$. Suppose we have a system which only has one component $i$. Then, the chemical potential of the component $i$ is exactly equivalent to the Gibbs free energy, since there are no other species $j$ present which can interact with $i$:
$$\mu_i = \frac{\mathrm{d}G}{\mathrm{d}n_i} = G_{m,i}$$
However, if the system has more than one component (as will be the case in a reaction), then the usage of the partial derivative becomes important. In any case, 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. At this stage, we will have to introduce further notation. If we treat a reaction as a mathematical equation, we can move all the reactants of a reaction to the right-hand side. For example,
$$\ce{3H2 + N2 -> 2NH3}$$
becomes
$$\ce{0 -> -3H2$\,$ - $\,$N2 + 2NH3}$$
This allows us to write all equations in the general form
$$0 \ce{->} \sum_j \nu_j \ce{J}$$
where $\ce{J}$ refers to a chemical species and $\nu_j$ refers to its (signed) stoichiometric coefficient. From the above example, we can see that for reactants like $\ce{H2}$ and $\ce{N2}$, $\nu_j$ is negative ($\nu_{\ce{H2}}$ and $\nu_{\ce{N2}}$ are $-3$ and $-1$ respectively). For products, $\nu_j$ is positive (here, $\nu_{\ce{NH3}}$ is $2$).
Clearly, 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](http://spikedmath.com/498.html)). 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
$$\xi = \frac{\Delta n_i}{\nu_i}$$
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:
$$\frac{\mathrm{d}G}{\mathrm{d}\xi} = \sum_i \mu_i\nu_i = 0$$
The derivative on the left, $\mathrm{d}G/\mathrm{d}\xi$, is given the special symbol $\Delta G$. This is somewhat confusing in two ways. Firstly, since $\mathrm{d}\xi$ has units of $\mathrm{mol}$, $\Delta G$ has units of $\mathrm{kJ~mol^{-1}}$, which I find a little irritating since $G$ itself has units of $\mathrm{kJ}$. However, it's just a convention that we will have to deal with. Secondly, it may be odd to see this defined as a derivative, since $\Delta$ usually represents a difference between something. In fact, it is. You can see from the above equation it 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:
$$\Delta G = \sum_i \mu_i\nu_i = 2\mu_{\ce{NH3}} - 3\mu_{\ce{H2}} -\mu_{\ce{N2}}$$
**Note that up to this point, we have not stipulated any particular temperature, pressure, or extent of reaction. We have only said that the temperature and pressure must be constant.** In general, the chemical potentials $\mu_i$ will depend on the exact temperature, pressure, and amounts of chemical species $i, j, \ldots$ present; this dependency is reflected in the very first definition of chemical potential that was given. Therefore, $\Delta G$ will depend on temperature, pressure, as well as the relative quantities of reactants and products. We could perhaps draw a graph of $G$ against $\xi$ ($\xi$ without the $\Delta$ simply refers to $\Delta \xi$ relative to the pure reactants):
$\hspace{40 mm}$ <img src="https://i.sstatic.net/Si4I6.jpg" width="320" height="320" alt="Graph">
$\Delta G = \mathrm{d}G/\mathrm{d}\xi$ is to be interpreted as the slope of the graph at any particular point in time. There is a magical value of $\xi$, which I have labelled $\xi'$, for which $\Delta G = 0$. As I have mentioned before, this is the condition for equilibrium. If you have knowledge of the starting composition, as well as the stoichiometry of the reaction, it is possible to mathematically relate $\xi$ to $Q$ and $\xi'$ to $K$. The reason why we use $Q$ and $K$ is because these numbers are independent of the initial composition of the reaction vessel (whereas $\xi$ is not - it is defined relative to a particular initial composition). The condition for equilibrium can therefore be stated in several different ways:
- $\Delta G = 0$
- $\xi = \xi'$
- $Q = K$
If you prepared a reaction vessel with an exact composition that corresponded to $\xi < \xi'$ (such as pure reactants where $\xi = 0$), then $\Delta G < 0$, as you can see from the graph. This means that the forward reaction will be spontaneous, i.e. $\xi$ will *increase* in order for the Gibbs free energy of the system to be minimised. Conversely, if you prepared a reaction vessel with $\xi > \xi'$ (such as pure products), then $\Delta G > 0$ and the reverse reaction will be spontaneous.
---------------
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](http://goldbook.iupac.org/S05925.html), 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 $c = 1~\mathrm{mol~dm^{-3}}$
**Note that no temperature is specified.** Therefore, by defining the *standard* Gibbs free energy, we are fixing a particular value of $\xi$ and $p$. However, the value of $T$ is not fixed. Since we earlier described $\Delta G$ as a derivative, you could think of $\Delta G^\circ$ as being the value of that derivative at a particular value of $\xi$, just like how the value of $\mathrm{d}y/\mathrm{d}x$ depends on $x$.
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 $1$. This will correspond to one particular value of $\xi$, which we could call $\xi^\circ$. We now have three possible cases:
- If $\xi^\circ < \xi'$, then *if we prepared a reaction vessel with all its substances in the standard state*, the forward reaction would be spontaneous, and when the system reaches equilibrium, the products will be favoured. This corresponds to the case where $K > 1$.
- If $\xi^\circ = \xi'$, then $K = 1$.
- If $\xi^\circ > \xi'$, then *if we prepared a reaction vessel with all its substances in the standard state*, the reverse reaction would be spontaneous and the reactants are favoured; $K < 1$.
Therefore, $\Delta G^\circ$ is purely **a measure of the equilibrium position**. Strictly speaking, it should *not* be used to predict whether a reaction is spontaneous or not, since **spontaneity depends on the exact composition of the reaction mixture**. If I prepared a reaction vessel with *only* reactants and no products, the forward reaction will always be spontaneous, because there's just no possible reverse reaction that can happen! In this case, $\Delta G^\circ$ is entirely irrelevant.
In practice, we will loosely say that "$\Delta G^\circ$ is negative, so the forward reaction is spontaneous". It is important to remember that what we *actually* mean is "$\Delta G^\circ$ is negative, so if we prepared a reaction vessel with all components present in their standard state, the forward reaction will be spontaneous", *or* "$\Delta G^\circ$ is negative, so at equilibrium there will be an excess of products over reactants".
Since $\Delta G^\circ$ is exactly the same as $\Delta G$ except for the imposition of the standard state, we can write the same equations:
$$\begin{align}
\Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\
\Delta G^\circ &= \sum_i \mu_i^\circ \nu_i
\end{align}$$
Again, since no temperature is specified in $\Delta G$ and no temperature is specified in the standard state, **no temperature is specified in $\Delta G^\circ$**.
----------------
A brief ending note: We have established the qualitative relationship between $\Delta G$ and $\Delta G^\circ$, but it is often useful to have an exact mathematical relation. We could substitute the equation
$$\mu_i = \mu_i^\circ + RT\ln{a_i}$$
into the expressions for $\Delta G$ and $\Delta G^\circ$:
$$\begin{align}
\Delta G &= \sum_i \mu_i\nu_i \\
\Delta G^\circ &= \sum_i \mu_i^\circ\nu_i
\end{align}$$
After some manipulation, we obtain the result:
$$\Delta G = \Delta G^\circ + RT\ln Q$$
When equilibrium is reached, we necessarily have $\Delta G = 0$ and $Q = K$. This gives us the famous equation:
$$\Delta G^\circ = -RT\ln K$$
Again, **no temperature is specified**! In general, $K$ depends on the temperature as well; the relationship is given by the [van 't Hoff equation](http://chemistry.stackexchange.com/questions/32385/temperature-dependence-of-equilibrium-constant).