If the bond dissociation energies of $XY$,$X_2$,and $Y_2$ (all diatomic molecules) are in the ratio of $1 : 1 : 0.5$ and $\Delta_f H$ for the formation of $XY$ is $-200 \ kJ \ mol^{-1}$,the bond dissociation energy of $X_2$ will be in $kJ \ mol^{-1}$:

The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

Enthalpy is an extensive property. In general,if enthalpy of an overall reaction $A \to B$ along one route is $\Delta_r H$ and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ represent enthalpies of intermediate reactions leading to product $B$. What will be the relation between $\Delta_r H$ for overall reaction and $\Delta_r H_1, \Delta_r H_2, \Delta_r H_3, \dots$ etc. for intermediate reactions?

The heat of neutralization of a strong dibasic acid by a dilute solution of $NaOH$ is approximately ....... $Kcal/equivalent$.

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

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