According to $\ce{\Delta G^\mathrm{o}= \Delta H^\mathrm{o} -T\Delta S^\mathrm{o}}$, if a reaction

a) is exothermic,

b) reduces system entropy, and

c) the temperature is low,

then it is spontaneous.

Consider the combustion of hydrogen gas, with $$2\ce{H2}+\ce{O2}\to 2\ce{H2O}.$$ This reaction is exothermic, and the system's entropy decreases. So does it mean that:

1) this reaction is spontaneous at low temperatures?

2) this reaction is not spontaneous at high temperatures?

If so, it seems counter-intuitive that placing the reagents in a high-temperature environment will not cause spontaneous combustion. Can you explain why this should be the case?

I really appreciate your help!

  • $\begingroup$ I no expert in thermochemistry, but from the equation, low temperature means a neglictible impact of −TΔS on the free energy. So if the reaction is exothermmic, the reaction would be spontaneous. At high temperature, the kinetics are higher, but the entropy loss of the conbustion would be a lot more important considering the second part of the equation. $\endgroup$ – Gwendal Grelier Aug 27 '18 at 8:45
  • $\begingroup$ Yes, this leads to my 3rd question: seems unintuitive that under High Temperature, the loss of entropy is amplified by High Temperature so much that the reaction is not spontaneous (because it seems that in High Temperature, this combustion should occur with even more ease). Any idea? $\endgroup$ – Jorge Mercent Aug 27 '18 at 8:52
  • $\begingroup$ Why don't you just use the van't Hopf equation to calculate the equilibrium constant for the reaction of hydrogen and oxygen to produce water as a function of temperature to see quantitatively how this plays out? $\endgroup$ – Chet Miller Aug 27 '18 at 14:54
  • $\begingroup$ Even at 1000 K, the equilibrium constant is still huge. $\endgroup$ – Chet Miller Aug 27 '18 at 15:15

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