Question

The enthalpies of combustion of $C_{\left(graphite\right)}$ and $C_{\left(diamond\right)}$ are $-393.5$ and $-395.4kJ/mol$ respectively. The enthalpy of conversion of $C_{\left(graphite\right)}$ to $C_{\left(diamond\right)}$ in $kJ/mol$ is:

• A. $-1.9$
• B. $-788.9$
• C. $1.9$
• D. $788.9$

Answer 1

The correct option is C $1.9$

The combustion reactions for $C_{\left(graphite\right)}$ and $C_{\left(diamond\right)}$ can be written as,
$C_{\left(graphite\right)}+O_2\to C{O}_2$ $\Delta H_1=-393.5\frac{KJ}{mol}$ (1)
$C_{\left(diamond\right)}+O_2\to C{O}_2$ $\Delta H_2=-395.4\frac{KJ}{mol}$ (2)
On subtracting reaction (2) from reaction (1) we will get,
$C_{\left(graphite\right)}\to C_{\left(diamond\right)}$ with $\Delta H=\Delta H_1-\Delta H_2=-393.5-(-395.4)=1.9\frac{KJ}{mol}$
Hence, answer is option C.