$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

The combustion of carbon produces two oxides,$CO$ and $CO_2$,respectively. Their enthalpies of formation are $26 \ kcal$ and $94.3 \ kcal$ respectively. What is the enthalpy of combustion of carbon in $kcal$?

$Fe_2O_{3(s)} + \frac{3}{2} C_{(s)} \to \frac{3}{2} CO_{2(g)} + 2Fe_{(s)}$
$\Delta H^o = +234.1 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}$
$\Delta H^o = -393.5 \ kJ$
Use these equations and $\Delta H^o$ values to calculate $\Delta H^o$ for this reaction:
$4Fe_{(s)} + 3O_{2(g)} \to 2Fe_2O_{3(s)}$
..... $kJ$

The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.

Use the data from the table to estimate the enthalpy of formation of $CH_3CHO$.

Bond	Bond Enthalpy $(kJ \ mol^{-1})$	Enthalpy of formation $(kJ \ mol^{-1})$
$C-H$	$400$	$C(g): 700$
$C-C$	$350$	$H(g): 200$
$C=O$	$700$	$O(g): 250$

The energy absorbed by each molecule $(A_2)$ of a substance is $4.4 \times 10^{-19} \ J$ and bond energy per molecule is $4.0 \times 10^{-19} \ J.$ The kinetic energy of the molecule per atom will be $...... \times 10^{-20} \ J$