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Explain standard enthalpy of formation |
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Answer» Solution :The standard enthalpy CHANGE for the formation of one mole of a compound from its elements in their most stable states of aggregation is called standard molar enthalpy of formation. Its symbol is `Delta_(f) H^(Theta)`. Elements are in their most stable states of aggregation. e.g., `H_(2) and O_(2)` are in gaseous state at 298 K temperature and 1 bar PRESSURE. `H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O((l))` `Delta_(f) H^( Theta) = - 285.8 "kJ mol"^(-1)` `C_("(graphite.s)") + 2H_(2(g)) to CH_(4(g)), Delta_(f) H^( Theta) = - 74.8 "kj mol"^(-1)`. Where one mole of a compound is formed from its constituent elements such as water, methane is formed. In contrast, the enthalpy change for an exothermic reaction. `CaO_((s)) + CO_(2(g)) to CaCO_(3 (s))` `Delta_(r) H^( Theta) = -178.3 "kj mol"^(-1)` It is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from its constituent elements. Enthalpy change is not standard enthalpy of formation, `Delta_(f) H^( Theta)` for `HBr_((g))`. `H_(2(g)) + Br_(2(g)) to 2HBr_((g)), Delta_(r) H^(Theta) =-72.8 "kj mol"^(-1)` Standard enthalpy of any element is taken as zero. CALCULATION of HEAT needed in DECOMPOSITION of `CaCO_(3)` is as under. `CaCO_(3 (s)) to CaO_((s)) + CO_(2 (g)),Delta_(r) H^( Theta) = (?)` `Delta_(f) H^( Theta) = sum_(i) a_(i) Delta_(f) H_(("product"))^(Theta) - sum_(i) b_(i) Delta_(f) H_(("reactions"))^( Theta)` `Delta_(f) H^( Theta) = Delta_(f) H^( Theta) [CaO_((s)) ]+ Delta_(f) H^( Theta) [ CO_(2 (g))] - Delta_(f) H^( Theta) [CaCO_(3 (s)) ]` `=[1(-635.1)+ 1(-393.5)] - [(-1206.9)]` `=178.3 kj//mol` Thus, the. decomposition of `CaCO_(3 (s))` is an endothermic process. Standard Molar Enthalpies of Formation `(Delta_(f) H^( Theta) )` at 298 K of a Few Selected Substances 
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