One decides the spontaneity of a reaction by considering

= ∆S_{total} = ∆S_{sys }+ ∆S_{surr}

For calculating ∆S_{surr}, we have to consider the heat absorbed by the surroundings which is equal to −∆_{r}H^{Θ}.

**At temperature T, entropy change of the surroundings is:**

∆S_{surr} = - \(\frac{\Delta_rH^\ominus}{T}\) (at constant pressure)

= \(\frac{-(-1648\times10^3\,J\,mol^{-1})}{298\,K}\)

= 5530 JK^{-1} mol^{-1}

So, total entropy change for this reaction

Δ_{r}S_{total} = 5530JK^{−1}mol^{−1} + (−549.4 JK^{−1} mol^{−1})

= 4980.6 J K^{−1} mol^{−1}

This Show that the above reaction is spontaneous.