From the standard enthalpies of formation values for some compounds (in `KJ mol^(-1)` ), predict which one of them will be most stable?

If the standard enthalpy of formation of methanol is -238.9 kJ mol^(-1) then entropy change of the surroundings will be.

The enthalpy of formation of two compounds A and B are -84 kJ and -156 kJ respectively. Which one of the following statements is correct ?

At 25^(@)C the standard enthalpy of formation of freon gas (CHClF_(2)) is -480.0 kJ*mol^(-1) . Wirte down of therrmochemical equation representing the formation reaction of the compound.

Since their discovery in 1985 , fullerenes have received the attention of many chemical researchers. In a recentrly reported data, the specific internal energy of combustion of crystalline C_(60) is found to be -36kJ g^(-1) at 298K . Compute the standard enthalpy of combustion and formation for the same. Standard enthalpy of combustion of graphite is -395kJ mol^(-1) . If the standard enthalpy of formation of diamond is +2kJ per mol of C- atom, which is more stable :C_(60) or diamond ?

Given standard enthalpy of formation of CO(-110 "KJ mol"^(-1)) and CO_(2)(-394 "KJ mol"^(-1)) . The heat of combustion when one mole of graphite burns is

Given standard enthalpy of formation of CO(-110 "KJ mol"^(-1)) and CO_(2)(-394 "KJ mol"^(-1)) . The heat of combustion when one mole of graphite burns is