The enthalpy of formation $(\Delta H_f)$ of methanol,formaldehyde and water are $-239, -116$ and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the oxidation of methanol to formaldehyde and water in $kJ$ is

Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

From Kirchhoff's equation,which factor affects the heat of reaction?

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......$kJ$.

$A_2 + B_2 \rightarrow 2AB$; $\Delta H_{r}^0 = -400\,kJ\,mol^{-1}$. $AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$B_2$ and $AB$ are in the ratio $1:0.5:1$,then the bond enthalpy of $A_2$ is $......\,kJ\,mol^{-1}$ (Nearest integer).

Combustion of $1$ $mol$ of benzene is expressed as:
$C_6H_{6(l)} + \frac{15}{2} O_{2(g)} \rightarrow 6CO_{2(g)} + 3H_2O_{(l)}$
The standard enthalpy of combustion of $2$ $mol$ of benzene is $-x$ $kJ$.
$x = . . . . . . . . . .$
$(1)$ Standard enthalpy of formation of $1$ $mol$ of $C_6H_{6(l)}$ is $48.5$ $kJ \ mol^{-1}$.
$(2)$ Standard enthalpy of formation of $1$ $mol$ of $CO_{2(g)}$ is $-393.5$ $kJ \ mol^{-1}$.
$(3)$ Standard enthalpy of formation of $1$ $mol$ of $H_2O_{(l)}$ is $-286$ $kJ \ mol^{-1}$.