From the data given below at 298 K for the reaction : CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) Calculate the enthalpy of formation of CH_(4)(g) at 298 K. Enthalpy of reaction is = -893.5 kJ Enthalpy of formation of CO_(2)(g) = 393. kJ mol^(-1) Enthalpy of formation of H_(2)O(l) = 286.0 kJ mol^(-1).

From the data given below at 298 K for the reaction : CH_(4)(g) + 2O_

1.	From the data given below at 298 K for the reaction : CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) Calculate the enthalpy of formation of CH_(4)(g) at 298 K. Enthalpy of reaction is = -893.5 kJ Enthalpy of formation of CO_(2)(g) = 393. kJ mol^(-1) Enthalpy of formation of H_(2)O(l) = 286.0 kJ mol^(-1).
Answer» Solution :`Delta H = Delta H_(f)CO_(2)(g) + 2DELTA H_(f)H_(2)O(L) - Delta H_(f)CH_(4)(g) - Delta H_(f)O_(2)(g)` `-890.5 KJ = -393.5 kJ + 2 XX -286 kJ - Delta H_(f)CH_(4)(g) - 0` `Delta H_(f)CH_(4) = -75.0 kJ`.