Question

The enthalpy of combustion of four allotropic forms of element 'X' are given as
Allotropic forms comb Hº(KJ/mol)
A -270.3
B -189.1
c -390.5
D -465.0
The most stable allotropic form of element 'X' is ​

Answer 1

Answer:

C

less than in the second case by 0.95 kJ

As graphite to diamond change is endothermic and for one mole energy change is C(graphite)→C(diamond),δH=1.9kJ..

So for 6gm (0.5 mole) of graphite C(graphite)→C(diamond),δH=0.95kJ..

So combustion enthalpy of diamond is less than the combustion enthalpy of graphite by 0.95 kJ.