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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), 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?

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    $\begingroup$ I'm guessing that, in the same way that you've calculated the Gibbs Free Energy based off of the energy of formation for each component, you can also calculate the Entropy & Enthalpy of the reaction. Since the Gibbs Free Energy is defined as: dG = dH - TdS you would need to plug in the result of the Entropy / Enthalpy of reaction, and the temperature of interest (which happens to be 298.15K). $\endgroup$ – samp May 12 '20 at 2:58
  • $\begingroup$ I took a liberty to correct formatting, notations and fix missing links and units (please check and correct, if necessary). Do note that omitting units is never permitted, and that "k" is not a unit, it's a decimal prefix — never shorten $\pu{kJ mol-1}$ to $\pu{k},$ this is plain wrong. As for the formatting, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. $\endgroup$ – andselisk May 12 '20 at 6:09
  • $\begingroup$ I also cleaned up comments section and added your essential comment as an edit to the question. $\endgroup$ – andselisk May 12 '20 at 6:13
  • $\begingroup$ Like @samp stated, you can calculate the enthalpy and entropy of the reaction using the individual enthalpies and entropies of formation for each compound in the reaction. Then you can use the derived definition of Gibbs Free Energy as above. An equation is spontaneous if Gibbs is negative. $\endgroup$ – TheGodlyBeast May 12 '20 at 6:18
  • $\begingroup$ You need to look up $\Delta S_f$ for the elements as well because they are non-zero at that temperature. $\endgroup$ – Karsten Theis May 12 '20 at 11:23

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