The heat of formation of $H_2O$ is $-286 \, kJ/mol$ and $H_2O_2$ is $-188 \, kJ/mol$. The enthalpy change for the reaction $2H_2O_2 \to 2H_2O + O_2$ is......$kJ$.

Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$

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 $:-$

Enthalpy of sublimation of a substance is equal to

$PbO$ exists in two crystalline forms: yellow and red. The standard enthalpies of formation for these two forms are $-217.3 \ kJ/mol$ and $-219.0 \ kJ/mol$,respectively. Calculate the enthalpy of transition for the process: $PbO \text{ (yellow)} \rightarrow PbO \text{ (red)}$ in $kJ/mol$.