The heat evolved in the combustion of benzene is given by $C_6H_6 + 7.5 O_2 \to 6CO_2(g) + 3H_2O(l); \Delta H = -3264.6 \, kJ$. Which of the following quantities of heat energy will be evolved when $39 \, g$ of $C_6H_6$ is burnt? (in $kJ$)

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:

The heat of combustion of ethanol into carbon dioxide and water is $-327 \ kcal$ at constant pressure. The heat evolved (in $cal$) at constant volume and $27^{\circ} C$ (assuming all gases behave ideally) is $\left( R = 2 \ cal \ mol^{-1} \ K^{-1} \right)$

The enthalpy change of which reaction corresponds to $\Delta H_f^o$ for $Na_2CO_{3(s)}$ at $298 \ K$?

The enthalpies of combustion of $C(s)$, $H_2(g)$, and $CH_4(g)$ are $-390 \text{ kJ/mol}$, $-285 \text{ kJ/mol}$, and $-890 \text{ kJ/mol}$ respectively. Calculate the enthalpy of formation of methane $(CH_4)$.