If $\Delta H_f^o$ for $H_2O_2$ and $H_2O$ are $-188 \ kJ/mole$ and $-286 \ kJ/mole$ respectively,what will be the enthalpy change of the reaction $2H_2O_{2(l)} \to 2H_2O_{(l)} + O_{2(g)}$ in $kJ/mole$?

For the reaction $H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)}; \Delta H = 12.40 \, Kcal$,the heat of formation of $HI$ is ....... $Kcal$.

The heat of neutralization of a strong acid with a strong base is constant and is equal to:

$H_2 + Cl_2 \longrightarrow 2HCl$ ; $\Delta H = -x \ kJ$
$NaCl + H_2SO_4 \longrightarrow NaHSO_4 + HCl$ ; $\Delta H = -y \ kJ$
$2H_2O + 2Cl_2 \longrightarrow 4HCl + O_2$ ; $\Delta H = -z \ kJ$
From the above equations,the value of $\Delta H_f$ of $HCl$ is:

Given:
$2C + 2O_2 \to 2CO_2 : \Delta H = -787 \text{ kJ}$
$H_2 + \frac{1}{2}O_2 \to H_2O : \Delta H = -286 \text{ kJ}$
$C_2H_2 + \frac{5}{2}O_2 \to 2CO_2 + H_2O : \Delta H = -1310 \text{ kJ}$
Calculate the heat of formation of acetylene $(C_2H_2)$ in $\text{kJ}$.

The enthalpy change at $298 \ K$ for the decomposition of water is given in the following two steps:
Step $I$: $H_2O_{(g)} \to H_{(g)} + OH_{(g)}$; $\Delta H = 498 \ kJ \ mol^{-1}$
Step $II$: $OH_{(g)} \to H_{(g)} + O_{(g)}$; $\Delta H = 428 \ kJ \ mol^{-1}$
The average bond enthalpy of the $O-H$ bond is $.... \ kJ \ mol^{-1}$