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Sodium carbonate `(Na_(2)CO_(3)0`can 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 temperature above 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` |
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Answer» `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 reaction will be spontaneous. |
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