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.