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

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

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.

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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).

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  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Jan 1 '17 at 3:53

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