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 $kJ/mole$?

Given the thermochemical reactions:
$C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta H = -110.5 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta H = -283.2 \ kJ$
Calculate the heat of reaction for $C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}$ in $kJ$.

If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be:

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:

If the combustion of $1 \ g$ of graphite produces $20.7 \ kJ$ of heat,what will be the molar enthalpy change? Give the significance of the sign also.

The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$