One decides the spontaneity of a reaction by considering

`DeltaS_("total")(DeltaS_("sys")+DeltaS_(surr")).` For calculating `DeltaS_("surr")` have to consider the heat absorbed by the surroundings which is equal to `-Delta_(r)H^(Ө)` . At temperature T, entropy change of the surroundings is

`DeltaS_("surr")=-(Delta_(r)H^(Ө))/(T)` (at constant pressure)

`=-((-1648xx19^(3)J "mol"^(1)))/(298K)`

`5530 JK^(-1)"mol"^(-1)`

Thus, total entropy change for this reaction

`Delta_(r)S_("total")=5530JK^(-1)"mol"^(-1)+`

`" " (-549.4JK^(-1)"mol"^(-1))`

`=4980.6 JK^(-1)"mol"^(-1)`

This shows that the above reaction is spontaneous.