Question

If the combustion of $1g$ of graphite produces $20.7kJ$ of heat, what will be molar enthalpy change? Given the significance of sign also.

Answer 1

The heat of combustion $\left(\Delta H_c\right)$ of graphite (i.e. carbon) is given as $=20.7kJ$ for $1g$ of graphite $\left(C\right)$.

$1$ mol of carbon $=12g$

Hence the molar enthalpy change i.e. $\Delta H_m$ (enthalpy for $1$ mole) would be,

$\Delta H_m=(20.7\times 12)=248.4kJmo{l}^{-1}$

Since the heat is released, the actual molar enthalpy change involving sign should be,

$\Delta H_m=-248.4kJmo{l}^{-1}$

Here, the negative sign of $\Delta H_m$ indicates that heat is released in the reaction. Hence, it is a exothermic reaction.