Sodium carbonate (Na_(2)CO_(3)0can be obtained by heating sodium hydrogen carbonate,NaHCO_(3) 2NaHCO_(3)(s) rarr Na_(2)CO_(3)(s)+H_(2)O(g) + CO_(2)(g) Calculate the temperatureabove which NaHCO_(3) decomposes to form products at 1 bar . Given Delta_(f)H^(@) ( kJ mol^(-1)):NaHCO_(3)(s) = =- 947.7 , Na_(2)CO_(3)(s) = - 1130 .9, H_(2)O(g)= - 241.8 , CO_(2)(g) = - 393.5 S^(@) (JK^(-1) mol^(-1)) : NaHCO_(3)(s) = 102.1 , Na_(2)CO_(3) (s)=136.0, H_(2)O(g) = 188.8, CO_(2)(g) 213.7

Sodium carbonate (Na_(2)CO_(3)0can be obtained by heating sodium hydro

1.	Sodium carbonate (Na_(2)CO_(3)0can be obtained by heating sodium hydrogen carbonate,NaHCO_(3) 2NaHCO_(3)(s) rarr Na_(2)CO_(3)(s)+H_(2)O(g) + CO_(2)(g) Calculate the temperatureabove which NaHCO_(3) decomposes to form products at 1 bar . Given Delta_(f)H^(@) ( kJ mol^(-1)):NaHCO_(3)(s) = =- 947.7 , Na_(2)CO_(3)(s) = - 1130 .9, H_(2)O(g)= - 241.8 , CO_(2)(g) = - 393.5 S^(@) (JK^(-1) mol^(-1)) : NaHCO_(3)(s) = 102.1 , Na_(2)CO_(3) (s)=136.0, H_(2)O(g) = 188.8, CO_(2)(g) 213.7
Answer» Solution : `Delta_(r)H^(@) = ( -1130.9) + ( -241.8) + ( -393.5)- 2 xx ( -947.7 )= 129.2 KJ mol^(-1)` `Delta_(r)S^(@) = ( 136.0 +188.8 +213.7) - 2( 102.1) = 334.3 JK^(-1) mol^(-1)` `Delta_(r)G^(@)= Delta_(r)H^(@)-TDelta_(r)S^(@)`, At equilibrium , `DELTAG^(@)= 0, T - ( Delta_(r)H^(@))/( Delta_(r)S^(@))= ( 129200)/( 334.3) = 386.5K` Thus, above 386.5 K , `Delta_(r)G^(@)` will be negative and the reactionwill be spontaneous.