One decides the spontaneity of a reaction by considering
= ∆Stotal = ∆Ssys + ∆Ssurr
For calculating ∆Ssurr, we have to consider the heat absorbed by the surroundings which is equal to −∆rHΘ.
At temperature T, entropy change of the surroundings is:
∆Ssurr = - \(\frac{\Delta_rH^\ominus}{T}\) (at constant pressure)
= \(\frac{-(-1648\times10^3\,J\,mol^{-1})}{298\,K}\)
= 5530 JK-1 mol-1
So, total entropy change for this reaction
ΔrStotal = 5530JK−1mol−1 + (−549.4 JK−1 mol−1)
= 4980.6 J K−1 mol−1
This Show that the above reaction is spontaneous.