Is the reaction below spontaneous at 298 K?

a) 2H2(g) + O2(g) ----> 2H2O(g)


First we can calculate H for the reaction,

The S was calculated earlier and is shown below

So now we have H and S. G can now be calculated using the equation,

Gsystem = Hsystem - TSsystem

Gsystem = -484 kJ - 298 K()

Gsystem = -484 kJ + 26.2 kJ = -458 kJ

So we conclude the following, the reaction is very exothermic (-484 kJ), becomes more ordered (), but is still spontaneous. How could that be you might ask? Did you not say for a reaction to be spontaneous the entropy had to increase?

And yes I did say that, but listen carefully, for a reaction to be spontaneous the entropy of the UNIVERSE must increase! In this problem the entropy of SYSTEM decreased. The reaction is spontaneous because the reaction is VERY exothermic, releasing heat to the surroundings causing the entropy of surrounding to increase much more than the entropy of the system decreased.


We notice that the calculated value is negative. Just as we predicted on the basis of comparing the relative order/disorder of the reactants and products. In this case we concluded S was likely to be negative because there are more moles of gas on the reactants side compared to the product side.