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(D) The given reaction is $C(	ext{diamond}) 	o C(	ext{graphite})$ with $\Delta H = -1.89 \ kJ$. This means diamond is at a higher energy level than

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More than in the second case by $0.945 \ kJ$

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(D) The given reaction is $C(	ext{diamond}) 	o C(	ext{graphite})$ with $\Delta H = -1.89 \ kJ$. This means diamond is at a higher energy level than graphite by $1.89 \ kJ/mol$. When $1 \ mol$ $(12 \ g)$ of diamond burns,it releases $1.89 \ kJ$ more heat than $1 \ mol$ of graphite. For $6 \ g$ of diamond,the amount of substance is $6/12 = 0.5 \ mol$. Therefore,the extra heat liberated compared to graphite is $0.5 	imes 1.89 \ kJ = 0.945 \ kJ$. Thus,the heat liberated in the first case (diamond) is more than in the second case (graphite) by $0.945 \ kJ$.

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is:

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