If the enthalpies of combustion of benzene $(l)$,carbon $(s)$,and hydrogen $(g)$ are $Q_1$,$Q_2$,and $Q_3$ respectively,what will be the enthalpy of formation of benzene?

When it is not possible to calculate the enthalpy of a reaction experimentally,it can be calculated by .....

$H_2 + \frac{1}{2} O_2 \to H_2O; \Delta H = -68.39 \ kcal$
$K + H_2O + \text{water} \to KOH_{(aq)} + \frac{1}{2} H_2; \Delta H = -48 \ kcal$
$KOH + \text{water} \to KOH_{(aq)}; \Delta H = -14 \ kcal$
The heat of formation of $KOH$ is (in $kcal$):

With the help of the following data,find out the change in heat content for the reaction in $kJ$:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

From the given reaction,$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)} \quad \Delta H = -92.6 \ kJ$,the enthalpy of formation of $NH_3$ is (in $kJ$):

If at $298 \, K$ the bond energies of $C-H, C-C, C=C$ and $H-H$ bonds are respectively $414, 347, 615$ and $435 \, kJ \, mol^{-1}$,the value of enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ will be $.... \, kJ$.