Question

For the allotropic change represented by the equation $C\left(graphite\right)\to C\left(diamond\right),\delta H=1.9kJ.$ If 6 g of diamond and 6 g of graphite are separately burnt to yield $C{O}_2,$ the enthalpy liberated in first case is:

• A. less than in the second case by 1.9 kJ
• B. less than in the second case by 11.4 kJ
• C. less than in the second case by 0.95 kJ
• D. less than in the second case by 3.6 kJ

Answer 1

The correct option is 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 $C\left(graphite\right)\to C\left(diamond\right),\delta H=1.9kJ.$.
So for 6gm (0.5 mole) of graphite $C\left(graphite\right)\to C\left(diamond\right),\delta H=0.95kJ.$.
So combustion enthalpy of diamond is less than the combustion enthalpy of graphite by 0.95 kJ.