The enthalpies of combustion of diamond and graphite are $-395.4 \, kJ$ and $-393.5 \, kJ$ respectively. The enthalpy of transformation of diamond to graphite is ..... $kJ$.

The standard enthalpy of combustion of $C$ (graphite), $H_{2(g)}$ and $CH_3OH_{(l)}$ respectively are $-393 \ kJ \ mol^{-1}$, $-286 \ kJ \ mol^{-1}$ and $-726 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of methanol?

The standard enthalpies of formation of $CO_{2(g)}$,$H_2O_{(\ell)}$ and glucose$_{(s)}$ at $25^{\circ} C$ are $-400 \ kJ/mol$,$-300 \ kJ/mol$ and $-1300 \ kJ/mol$,respectively. The standard enthalpy of combustion per gram of glucose at $25^{\circ} C$ is

The bond dissociation enthalpy of $X_2$,$\Delta H_{\text{bond}}^{\circ}$,calculated from the given data is $...$ $kJ \ mol^{-1}$. (Nearest integer)
$M^{+}X^{-}_{(s)} \rightarrow M^{+}_{(g)} + X^{-}_{(g)} \quad \Delta H_{\text{lattice}}^{\circ} = 800 \ kJ \ mol^{-1}$
$M_{(s)} \rightarrow M_{(g)} \quad \Delta H_{\text{sub}}^{\circ} = 100 \ kJ \ mol^{-1}$
$M_{(g)} \rightarrow M^{+}_{(g)} + e^{-}_{(g)} \quad \Delta H_{i}^{\circ} = 500 \ kJ \ mol^{-1}$
$X_{(g)} + e^{-}_{(g)} \rightarrow X^{-}_{(g)} \quad \Delta H_{\text{eg}}^{\circ} = -300 \ kJ \ mol^{-1}$
$M_{(s)} + \frac{1}{2}X_{2(g)} \rightarrow M^{+}X^{-}_{(s)} \quad \Delta H_{f}^{\circ} = -400 \ kJ \ mol^{-1}$
[Given : $M^{+}X^{-}$ is a pure ionic compound and $X$ forms a diatomic molecule $X_2$ in gaseous state]

Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:

The heat of combustion of $CH_{4(g)}$,$C_{(graphite)}$ and $H_{2(g)}$ are $-20 \ kcal$,$-40 \ kcal$ and $-10 \ kcal$ respectively. The heat of formation of methane is.......$kcal$.