The standard enthalpies of combustion of $C_6H_{6(l)}$,$C(graphite)$ and $H_{2(g)}$ are respectively $-3270 \ kJ \ mol^{-1}$,$-394 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of $C_6H_{6(l)}$ in $kJ \ mol^{-1}$?

The heat of neutralization of $HCl$ by $NaOH$ under certain conditions is $-55.9 \, kJ \, mol^{-1}$ and that of $HCN$ by $NaOH$ is $-12.1 \, kJ \, mol^{-1}$. The heat of ionization of $HCN$ is .............. $kJ \, mol^{-1}$.

What will be the amount of heat evolved by burning $0.4 \ mol$ of methane? (Given heats of formation of $CH_4$,$CO_2$,and $H_2O$ are $-75$,$-400$,and $-240 \ kJ \ mol^{-1}$ respectively) ..... $kJ$

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

Under similar conditions,the enthalpy of freezing is exactly opposite to:

Given the thermochemical reactions: $C(\text{graphite}) + \frac{1}{2}O_2 \rightarrow CO; \Delta H = -110.5 \, kJ$ and $CO + \frac{1}{2}O_2 \rightarrow CO_2; \Delta H = -283.2 \, kJ$. Calculate the heat of reaction for $C(\text{graphite}) + O_2 \rightarrow CO_2$ in $kJ$.