The enthalpies of combustion of $C(s)$, $H_2(g)$, and $CH_4(g)$ are $-390 \text{ kJ/mol}$, $-285 \text{ kJ/mol}$, and $-890 \text{ kJ/mol}$ respectively. Calculate the enthalpy of formation of methane $(CH_4)$.

The molar neutralization heat for $KOH$ and $HNO_3$ as compared to the molar neutralization heat of $NaOH$ and $HCl$ is:

Based on the following thermochemical reactions:
$H_2O_{(g)} + C_{(s)} \rightarrow CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H = x \ kJ$
The value of $x$ will be:

The standard enthalpies of formation for $NH_3$,$CO_2$,$HI$,and $SO_2$ are $-46.19$,$-393.4$,$+24.94$,and $-296.9 \, kJ/mol$ respectively. What is the increasing order of their stability?

The standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]

The atomisation enthalpy of $CH_4$ is $1660 \ kJ \ mol^{-1}$. The $C-H$ bond enthalpy of each successive step in $CH_4$ $\rightarrow CH_3$ $\rightarrow CH_2$ $\rightarrow CH$ are $+15, +30$ and $+45 \ kJ \ mol^{-1}$ higher than the mean bond enthalpy of $C-H$ bonds,respectively. The bond enthalpy of the last $C-H$ unit is