Calculate the enthalpy change of the reaction for $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, given the bond energies (in $\text{kJ mol}^{-1}$): $H-H = 436$, $Cl-Cl = 242$, $H-Cl = 431$.

Which among the following represents the reaction of formation of the product?

The $\Delta H_f^o$ for $CO_{2(g)}$,$CO_{(g)}$ and $H_2O_{(g)}$ are $-393.5$,$-110.5$ and $-241.8 \ kJ \ mol^{-1}$ respectively. The standard enthalpy change (in $kJ$) for the reaction $CO_{2(g)} + H_{2(g)} \to CO_{(g)} + H_2O_{(g)}$ is

On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.

If heat of neutralization is $-13.7 \, KCal$ at $25 \, ^oC$ and $\Delta H_f^o (H_2O) = -68 \, KCal$,then the standard enthalpy of formation of $OH^{-}$ will be.....$KCal$. (in $.3$)

The enthalpy of neutralization of any strong acid and strong base is constant because......