2011-07-11, 06:52 AM
The negative enthalpy change of formation of water is due to the fact that bond formation releases energy. Therefore, the reaction
H2 + 0.5O2 -> H2O
is exothermic, and thus has a negative dH.
According to the formula
dG = dH - TdS
which Spaghetti posted above, we can see that a small dS and T will result in dH being approximately equal to dG and thus, a good indicator of spontaneity. However, as gases, H2 and O2 have a LARGE entropy value (S). Water, on the other hand, is a liquid and as such has a much, much smaller entropy. Therefore, the dS of the aforementioned reaction is LARGE and NEGATIVE. At room temperature, T is small and positive, so therefore, TdS is also large and negative, just like dS. dH is negative, yes, but is also fairly small (-285.83 kJ/mol). Therefore,
dG = negative - LARGE negative = negative + LARGE positive = positive
As dG is positive, the reaction is non-spontaneous. Therefore, the above reaction does not occur spontaneously.
p.s. I'm still a student myself, so if I made a mistake anywhere, feel free to correct me.
H2 + 0.5O2 -> H2O
is exothermic, and thus has a negative dH.
According to the formula
dG = dH - TdS
which Spaghetti posted above, we can see that a small dS and T will result in dH being approximately equal to dG and thus, a good indicator of spontaneity. However, as gases, H2 and O2 have a LARGE entropy value (S). Water, on the other hand, is a liquid and as such has a much, much smaller entropy. Therefore, the dS of the aforementioned reaction is LARGE and NEGATIVE. At room temperature, T is small and positive, so therefore, TdS is also large and negative, just like dS. dH is negative, yes, but is also fairly small (-285.83 kJ/mol). Therefore,
dG = negative - LARGE negative = negative + LARGE positive = positive
As dG is positive, the reaction is non-spontaneous. Therefore, the above reaction does not occur spontaneously.
p.s. I'm still a student myself, so if I made a mistake anywhere, feel free to correct me.