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'?
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.
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')?
If you want delta So, use the difference to do the same you did with delta Ho. But use delta G = delta H - T*delta S if you want it at some T other than 25 degrees C.
To calculate the change in enthalpy, ΔH', of the reaction 2N2O <-> 4NO2 + O2, you need to know the standard enthalpy of formation for the reactants and products involved. Let's say you are given the standard enthalpy of formation for N2O (ΔH'f(N2O)), the standard enthalpy of formation for NO2 (ΔH'f(NO2)), but not the standard enthalpy of formation for O2 (ΔH'f(O2)).
To determine whether the standard enthalpy of formation for O2 (ΔH'f(O2)) is zero, you can refer to tables or databases that provide thermodynamic data. These sources will typically have values for standard enthalpies of formation at standard conditions for various compounds.
For O2, the standard enthalpy of formation (ΔH'f(O2)) is indeed zero. This is because the reference state for standard enthalpy of formation calculations is the most stable form of the element in its standard state. Since O2 is the most stable form of oxygen under standard conditions (25°C and 1 atm pressure), its standard enthalpy of formation is considered to be zero.
Now, let's address your second question regarding when to use ΔS' or ΔG' to calculate the spontaneity of a reaction.
To determine whether a reaction is thermodynamically favorable or not, you can consider two factors: the change in enthalpy (ΔH') and the change in entropy (ΔS'). These factors are related through the Gibbs free energy (ΔG') by the equation:
ΔG' = ΔH' - TΔS',
where T represents temperature in Kelvin.
If you are only given the enthalpy values and want to evaluate the spontaneity of the reaction at a specific temperature, you can use the equation ΔG' = ΔH' - TΔS'. By comparing the sign of ΔG', you can determine whether the reaction is spontaneous or not. If ΔG' is negative, the reaction is spontaneous (favorable), and if ΔG' is positive, the reaction is non-spontaneous (unfavorable).
On the other hand, if you are given both the enthalpy and entropy values in standard conditions, you can evaluate the spontaneity of the reaction directly using the ΔS' value. The equation for this is:
ΔG' = ΔH' - TΔS',
where ΔG' is the change in Gibbs free energy, ΔH' is the change in enthalpy, ΔS' is the change in entropy, and T is the temperature in Kelvin.
In summary, you can use the ΔG' equation when provided with both ΔH' and ΔS' values, and you can use the ΔH' and ΔS' equation when only ΔH' values are given to evaluate the spontaneity of a reaction.