Standard molar enthalpy of formation, Delta_(f) H^( Theta ) is just a special case of enthalpy of reaction, Delta_(r) H^( Theta ) . Is the Delta_(r) H^( Theta ) for the following reaction same as Delta_(f) H^( Theta ) ?Give reason for your answer. CaO_((s)) + CO_(2(g)) to CaCO_(3(s)) , Delta_(f) H^( Theta ) = - 178 "kJ mol"^(-1)

Standard molar enthalpy of formation, Delta_(f) H^( Theta ) is just a

1.	Standard molar enthalpy of formation, Delta_(f) H^( Theta ) is just a special case of enthalpy of reaction, Delta_(r) H^( Theta ) . Is the Delta_(r) H^( Theta ) for the following reaction same as Delta_(f) H^( Theta ) ?Give reason for your answer. CaO_((s)) + CO_(2(g)) to CaCO_(3(s)) , Delta_(f) H^( Theta ) = - 178 "kJ mol"^(-1)
Answer» Solution :No, the `Delta_(r) H^( Theta )` for the given reaction is not same as `Delta_(r) H^( Theta )`. "The standard enthalpy change for the formation of one MOLE of a compound from its ELEMENTS in their most stable states (reference states) is called standard molar enthalpy of formation," `Delta_(r) H^( Theta )`. `Ca_((s)) + C_((s)) + (1)/(2) O_(2(g)) to CaCO_(3(s)) , Delta_(F) H^( Theta )` This reaction is different from the given reaction. Hence, `Delta_(f) H^(0) ne Delta_(f) H^( 0) `