Calculating Gibbs free energy for the reaction of burning methanol


> 
> Calculate $Δ\_\mathrm{r}G^\circ$ for the following reaction:
> 
> 
> $$\ce{2 CH3OH(l) + 3 O2(g) -> 2 CO2(g) + 4 H2O(l)}$$
> 
> 
> using $Δ\_\mathrm{r}H^\circ$ and $Δ\_\mathrm{r}S^\circ$ at $\pu{298.13 K}$ $(\pu{25 °C}).$ The values for $Δ\_\mathrm{f}H^\circ$ and $Δ\_\mathrm{f}S^\circ$ can be found in Appendix G of the textbook.
> 
> 
> Also state if this reaction is spontaneous at this temperature.
> 
> 
> 


In [Appendix G of the textbook (*Openstax Chemistry*)](https://openstax.org/books/chemistry-2e/pages/g-standard-thermodynamic-properties-for-selected-substances), I found the $Δ\_\mathrm{f}H^\circ$ to be $\pu{-239.2 kJ mol-1}$ and the $Δ\_\mathrm{f}S^\circ$ to be $\pu{126.8 kJ mol-1}.$


The values of the Gibbs free energy of formation for both the reactants and products were also listed in the Appendix G, this is where I got them. So I plugged those values into the equation, leaving out the $\ce{O2}$ because it is just an element, and multiplied the appropriate values by the numbers in front of the compounds, and that is how I got the Gibbs free energy of the reaction:


$$
\begin{align}
Δ\_\mathrm{r}G^\circ &= \sum Δ\_\mathrm{f}G^\circ(\text{products}) - \sum Δ\_\mathrm{f}G^\circ(\text{reactants})\\
 &= [2 × (\pu{-394.4 kJ mol-1}) + 4 × (\pu{-228.6 kJ mol-1})] - [2 × (\pu{-163.3 kJ mol-1})]\\
 &= \pu{-1379 kJ mol-1}
\end{align}
$$


However, I don't know how to get the Gibbs free energy by using the $Δ\_\mathrm{f}H^\circ$ and $Δ\_\mathrm{f}S^\circ,$ and I don't know if it was spontaneous or not. Can anybody point me in the right direction, please?


