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))

Sodium carbonate, Ba_(2)CO_(3) can be obtained by heating sodium hydro

1.	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))
Answer» Solution :`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.