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\[\Delta G = \Delta H -\Delta S(T)\] Does this help?
am, i don't think so.. How can I use that formula?
Well when the change in Gibbs free energy is negative it denotes a spontaneous reaction, can you devise values for enthalpy and entropy that would allow for the reaction to be spontaneous at non-standard temperatures (273K)
what? i don't understand...
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This might help you to understand
okay.. i got it. :) thank you po. :)
po?
"po" is a sign of our respect.. :)
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Fair enough
:)
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