$\ce{CO(g) + 2H2(g) +O2(g) <=> 2CH3OH(g)}$,

$\Delta H=-128\ \mathrm{kJ/mol}$
$\Delta S=-409.2\ \mathrm{J/(mol \cdot K)}$

Determine if the reaction if thermodynamically favored.

I just wanted someone to see if my logic is right. Since there are 4 moles on the reactants side and and 2 on the products side the reaction would not be thermodynamically favored because the reaction decreases the amount of entities, which decreases the entropy.


Your reasoning is correct but there is more to it.

Look at it from the point of view of Gibbs Free Energy,

$\Delta G = \Delta H - T \Delta S$

For a given reaction if:

$\Delta G >0$, the reaction is nonspontaneous in the forward direction, not thermodynamically favourable

$\Delta G < 0 $, the reaction is spontaneous in the forward direction, thermodynamically favourable

If we substitute the values for $\Delta H$ and $\Delta S$:

$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$

Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this, the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).

  • $\begingroup$ thank you! I did not think about the reasoning using delta G or H to figure out that delta G was negative. This helped a lot! $\endgroup$ – sloth1111 Jan 5 '14 at 22:02
  • $\begingroup$ Just one more thing, in your last sentence you mention about "when $T$ becomes very negative", but that doesn't make sense, as absolute temperatures are always positive except in certain systems while using a certain definition of absolute temperature. $\endgroup$ – Nicolau Saker Neto Jan 5 '14 at 22:13
  • $\begingroup$ I meant from a mathematical perspective, but I should make that more clear $\endgroup$ – pingOfDoom Jan 5 '14 at 22:23
  • $\begingroup$ What @pingOfDoom is saying is that only way that delta G can become positive is if T is negative. Because this is physically impossible, delta G can never be positive. $\endgroup$ – Georgeos Hardo Jan 1 '17 at 3:53

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