The enthalpy change $(\Delta H)$ for the neutralisation of $1 \ M \ HCl$ by caustic potash in dilute solution at $298 \ K$ is ..... $kJ$.

One mole of ethanol $(l)$ was completely burnt in oxygen to form $CO_{2(g)}$ and $H_2O_{(l)}$. What is the $\Delta_r H^{\ominus}$ (in $kJ \ mol^{-1}$) for this reaction?
(The standard enthalpy of formation $(\Delta_f H^{\ominus})$ of $C_2H_5OH_{(l)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ is respectively $-277, -393$ and $-286 \ kJ \ mol^{-1}$.)

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

Heat of transition is the heat evolved or absorbed when a substance is converted from

The atomization enthalpies of $NH_{3(g)}$ and $N_2H_{4(g)}$ are $+150 \ kJ \ mol^{-1}$ and $+310 \ kJ \ mol^{-1}$ respectively. The $\Delta H(N-N)$ bond enthalpy in $kJ \ mol^{-1}$ is:

Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by: