how would you go about calculating the change in H' of the reaction 2N2O 4NO2+O2 if you are given the H'of formation for N2O and NO2 but not the H' of formation for O2? is H' of O2 zero? if so why? and also how can you tell to use change in S'=sum(npS'p) - sum(nrS'r) or to use change G'=change H- TchangeS'?

Question

how would you go about calculating the change in H' of the reaction 2N2O <-> 4NO2+O2 if you are given the H'of formation for N2O and NO2 but not the H' of formation for O2? is H' of O2 zero? if so why? and also how can you tell to use change in S'=sum(npS'p) - sum(nrS'r) or to use change G'=change H- TchangeS'?

DrBob222 · Answer

I'm not sure I will get all of the questions but I'll start.
First, O2 is zero (by definition, I suppose) since anything in the standard state is zero. That's how delta H is defined. Look in your table and you will find O2, H2, Cl2, etc (in the free state) is zero in the tables. You get
delta Hrxn = (delta H products) - (delta H reactants). And it's that simple. Any coefficients are multiplied, for example, in the reaction you cite, it is
delta Hrxn = (4*deltaH NO2 + 1*delta H O2) -(2*delta H N2O) = etc.
As I said earlier, delta H O2 is zero.

john · Answer

the delta S' is entropy. which equation should you use to calculate delta S', S'=(delta S products)-(delta S products) or should you use the delta G'=delta H' - (T*delta S')?

DrBob222 · Answer

If you want delta S^o, use the difference to do the same you did with delta H^o. But use delta G = delta H - T*delta S if you want it at some T other than 25 degrees C.