Question

Calculate the enthalpy change accompaying the conversion of 10 grams of graphite into diamond if the standard enthalpy of combustion of $C_{graphite}$ and $C_{diamond}$ are -94.05 and -94.50 kcal respectively.

• A. 0.45 kcal
• B. -0.45 kcal
• C. -0.375 kcal
• D. 0.375 kcal

Answer 1

The correct option is D 0.375 kcal

Given that
(i) $C_{graphite}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)};\Delta H=-94.05kcal$
(ii) $C_{diamond}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)};\Delta H=-94.50kcal$
Thus applying the inspection method [Eq. (i) - Eq. (ii)], we get

$\Rightarrow C_{graphite}\to C_{diamond};\Delta H=0.45kcal$
Since this enthalpy change is only for conversion of 1 mol i.e. 12 g of $C_{graphite}$ to $C_{diamond}$, therefore, for the converion of 10 g of $C_{graphite}$ to $C_{diamond}$
$\Delta H=0.45\times \frac{10}{12}$
$=0.375kcal$