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

The enthalpy of atomization of $PH_3(g)$ is $228 \, kcal \, mol^{-1}$ and that of $P_2H_4(g)$ is $355 \, kcal \, mol^{-1}$. The $P-P$ bond energy (in $kcal \, mol^{-1}$) is:

The heat evolved in the combustion of benzene is given by the equation $C_6H_{6(l)} + 7.5 O_{2(g)} \to 3H_2O_{(l)} + 6CO_{2(g)}$,$\Delta H = -781.0 \ kcal \ mol^{-1}$. Which of the following quantities of heat energy will be evolved when $39 \ g$ of benzene is burnt in an open container?

At $25^{\circ} C$, the enthalpy of the following processes are given:
$H_{2(g)} + O_{2(g)} \rightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = 78 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} \quad \Delta H^{\circ} = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)} \quad \Delta H^{\circ} = 436 \ kJ \ mol^{-1}$
$\frac{1}{2} O_{2(g)} \rightarrow O_{(g)} \quad \Delta H^{\circ} = 249 \ kJ \ mol^{-1}$
What would be the value of $X$ for the following reaction? (Nearest integer)
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} \quad \Delta H^{\circ} = X \ kJ \ mol^{-1}$

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

For the reaction $N_2 + 3 X_2 \longrightarrow 2 NX_3$,where $X = F, Cl$ (the average bond energies are $F-F = 155 \ kJ \ mol^{-1}$,$N-F = 272 \ kJ \ mol^{-1}$,$Cl-Cl = 242 \ kJ \ mol^{-1}$,$N-Cl = 200 \ kJ \ mol^{-1}$ and $N \equiv N = 941 \ kJ \ mol^{-1}$),the heats of formation of $NF_3$ and $NCl_3$ in $kJ \ mol^{-1}$,respectively,are closest to