Sodium carbonate, `Ba_(2)CO_(3)` can be obtained by heating sodium hydrogen carbonate `NaHCO_(3)` as
`2NaHCO_(3)(s) to Na_(2)CO_(3)(s) + H_(2)O(g) + CO_(2)(g)`
The essential data are :
`NaHCO_(3)(s) Na_(2)CO_(3)(s) + H_(2)O(g) + CO_(2)(g) Delta_(f)H^(@) -947.7 -1130.9 -393.51 -241.82 (kJ "mol"^(-1))`
Calculate the temperature above which `NaHCO_(3)` decomposes to give products at 1 bar.

Explanation

`2NaHCO(3)(s) to Na_(2)CO_(3)(s) + CO_(2)(g) + H_(2)O(g)`
`Delta_(r)H^(@) = Delta_(f)H^(@) (NA_(2)CO_(3)) + Delta_(f)H^(@)(CO_(2)) + Delta_(f)H^(@)(H_(2)O) - 2Delta_(f)H^(@)(NaHCO_(3))`
` = -1130.9 + (-393.51) + (-241.82) - 2 xx (-947.7)` = `- 1766.23 + 1895.4 = 129.17 kJ "mol"^(-1)`
`Delta_(r)S^(@) = S^(@)_(m)(Na_(2)CO_(3))+S_(m)^(@)(CO_(2)) + S_(m)^(@)(H_(2)O)- 2S_(m)^(@)(NaHCO_(3))`
` = 136.0 + 188.83 + 213.74 - 2 xx 102.1`
` = 538.57 - 204.2 = 334.37 J K^(-1) "mol"^(-1)`
Now at equilibrium, `Delta_(r)G^(@) =0` so that
`Delta_(r)G^(@) = Delta_(r)H^(@) - TDelta_(r)S^(@) = 0`
`therefore T = (Delta_(r)H^(@))/(Delta_(r)S^(@)) = (129.17)/(334.37 xx 10^(-3)) = 386 K`
`therefore` Reaction will be spontaneous above 386 K.