Calculate the standard enthalpy of formation of CH_(3) OH_((l)) from the following data : CH_(3) OH_((l)) + (3)/(2) O_(2(g)) to CO_(2(g)) + 2H_(2) O_((l)) , Delta_(r) H^( Theta ) = -726 "kJ mol"^(-1) C_("(graphite)") +O_(2(g)) to CO_(2(g)) , Delta_(r) H^( Theta ) = -393 "kJ mol"^(-1) H_(2(g)) + (1)/(2) O_((g)) to H_(2) O_((l)) , Delta_(f) H^( Theta ) = -286 "kJ mol"^(-1)

Calculate the standard enthalpy of formation of CH_(3) OH_((l)) from t

1.	Calculate the standard enthalpy of formation of CH_(3) OH_((l)) from the following data : CH_(3) OH_((l)) + (3)/(2) O_(2(g)) to CO_(2(g)) + 2H_(2) O_((l)) , Delta_(r) H^( Theta ) = -726 "kJ mol"^(-1) C_("(graphite)") +O_(2(g)) to CO_(2(g)) , Delta_(r) H^( Theta ) = -393 "kJ mol"^(-1) H_(2(g)) + (1)/(2) O_((g)) to H_(2) O_((l)) , Delta_(f) H^( Theta ) = -286 "kJ mol"^(-1)
Answer» Solution :`CH_(3) OH_((l)) + (3)/(2) O_(2(G)) to CO_(2(g)) + 2H_(2) O_((l)) Delta_(r) H^( Theta ) = - 726 "kJ mol"^(-1)` `Delta_(r) H^( Theta ) = [2 Delta_(r) H^( Theta ) H_(2) O + Delta_(r) H CO_(2) ] - [ Delta_(r) H^( Theta ) CH_(3) OH]` `-726 = [2 xx (-286) + (-393) ] -[X]` `-726 = [-572-393]-x` `therefore x=-239 "kJ/mol"`