Q. For the change, $C_{d i a m o n d} \rightarrow C_{g r a p h i t e};\Delta H=-1.89kJ,$ if $6g$ of diamond and $6g$ of graphite are separately burnt to yield $CO_{2}$ , the heat liberated in first case is
NTA AbhyasNTA Abhyas 2022
Solution:
$C_{diamond} \rightarrow C_{graphite};ΔH=-1.89kJ$ …(i)
Enthalpy of combustion of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.
$C_{d i a m o n d}+O_{2} \rightarrow CO_{2};\Delta H=-q_{1}$ …(ii)
$C_{g r a p h i t e}+O_{2} \rightarrow CO_{2};\Delta H=-q_{2}$ …(iii)
By equations [(i) – (ii)]
$C_{d i a m o n d} \rightarrow C_{g r a p h i t e};\Delta H=-q_{1}+q_{2}$
So $-q_{1}+q_{2}=-1.89$
or $q_{2}-q_{1}=-1.89$ for $12g$ $C_{d i a m o n d \rightarrow g r a p h i t e}$
Thus, for $6g$ $C_{d i a m o n d \rightarrow g r a p h i t e}$
$=\frac{- 1.89}{2}=0.945 kJ$ .
Enthalpy of combustion of a substance is the change in enthalpy produced when one mole of the substance is completely burnt in air or oxygen at a given temperature.
$C_{d i a m o n d}+O_{2} \rightarrow CO_{2};\Delta H=-q_{1}$ …(ii)
$C_{g r a p h i t e}+O_{2} \rightarrow CO_{2};\Delta H=-q_{2}$ …(iii)
By equations [(i) – (ii)]
$C_{d i a m o n d} \rightarrow C_{g r a p h i t e};\Delta H=-q_{1}+q_{2}$
So $-q_{1}+q_{2}=-1.89$
or $q_{2}-q_{1}=-1.89$ for $12g$ $C_{d i a m o n d \rightarrow g r a p h i t e}$
Thus, for $6g$ $C_{d i a m o n d \rightarrow g r a p h i t e}$
$=\frac{- 1.89}{2}=0.945 kJ$ .