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

The value of enthalpy of neutralization is highest for the reaction between:

$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = -94 \ kcal$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -67.7 \ kcal$
Then the heat of formation of $CO_{(g)}$ is $..... \ kcal$

Calculate the heat of formation of $Ca(OH)_{2(s)}$ at $1.8\,^{\circ}C$ from the following data:
$CaO_{(s)} + H_2O_{(l)} \to Ca(OH)_{2(s)}$; $\Delta H_{1.8\,^{\circ}C} = -15.26\,K\,cal$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H_{1.8\,^{\circ}C} = -68.37\,K\,cal$
$Ca_{(s)} + \frac{1}{2}O_{2(g)} \to CaO_{(s)}$; $\Delta H_{1.8\,^{\circ}C} = -151.80\,K\,cal$

What is bond enthalpy? Explain bond enthalpy in polyatomic molecules like $H_2O$.

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$
Assertion $(A)$: Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \ kJ \ mol^{-1}$.
Reason $(R)$: Enthalpy of neutralisation is the amount of heat liberated when one mole of $H^{+}$ ions furnished by acid combine with one mole of $OH^{-}$ ions furnished by base to form one mole of water.
In the light of the above statements,choose the correct answer from the options given below.