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

Standard molar enthalpy of formation, Delta_(f)H^(c-) is just a specia

1.	Standard molar enthalpy of formation, Delta_(f)H^(c-) is just a special case of enthalpy of reaction, Delta_(r)H^(c-) . Is the Delta_(r)H^(c-) for the following reaction same as Delta_(f)H^(c-) ? Give reaction for your answer. CaO(s) +CO_(2)(g) rarr CaCO_(3)(s), Delta_(f)H^(c-)= - 178. 3 kJ mol^(-1)
Answer» Solution :No, because enthalpy of FORMATION is the enthalpy changewhen 1 MOLE of the compound is formed from the elements , i.e., for the reaction `Ca(s) + C(s) + (3)/(2) O_(2)(g) rarr CaCO_(3)(s)`. Thus, this reaction is DIFFERENT from the givien reaction. Hence, `Delta_(r) H^(@)cancel(=) Delta_(f)H^(@)`.