During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is

During complete combustion of one mole of butane, 2658 kJ of heat is r

1.	During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is
Answer» `2C_(4) H_(10(g))+ 13O_(2(g)) to 8CO_(2(g)) + 10H_(2) O_((l)) , Delta_(c) H= - 2658.0 "kJ mol"^(-1)` `C_(4) H_(10(g))+ (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= - 1329.0 "kJ mol"^(-1)` `C_(4) H_(10(g))+ (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= - 2658.0 "kJ mol"^(-1)` `C_(4) H_(10(g))+ (13)/(2)O_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l)) , Delta_(c) H= + 2658.0 "kJ mol"^(-1)` SOLUTION :GIVEN that, the complete combustion reaction of one MOLE of butane is represented thermodynamically as : `C_(4) H_(10(g)) + (13)/(2) O_(2(g)) to 4CO_(2(g)) to 4CO_(2(g)) + 5H_(2) O_((l))` We have to take the combustion of one mole of `C_(4) H_(10) and Delta_( C) H` should be negative and have a value of 2658 `"kJ mol"^(-1)`