The heat of neutralization of $HCl$ by $NaOH$ under certain conditions is $-55.9 \, kJ \, mol^{-1}$ and that of $HCN$ by $NaOH$ is $-12.1 \, kJ \, mol^{-1}$. The heat of ionization of $HCN$ is .............. $kJ \, mol^{-1}$.

What is enthalpy of reaction?

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 $.... \ kJ$.

Consider the following processes:

Process	$\Delta H \ (kJ/mol)$
$I. \ \frac{1}{2} A \rightarrow B$	$+150$
$II. \ 3B \rightarrow 2C + D$	$-125$
$III. \ E + A \rightarrow 2D$	$+350$

For $B + D \rightarrow E + 2C, \Delta H$ will be ............. $kJ/mol$

Equal volumes of molar hydrochloric acid and sulphuric acid are neutralized by dilute $NaOH$ solution and $x \ kcal$ and $y \ kcal$ of heat are liberated respectively. Which of the following is true?

What is the enthalpy change when $6.80 \, g$ of $NH_3$ is passed over hot $CuO$ (in $, kJ$)? The standard enthalpies of formation for $NH_3(g)$,$CuO(s)$,and $H_2O(l)$ are $-46.0$,$-155.0$,and $-285.0 \, kJ \, mol^{-1}$ respectively. The reaction is: $NH_3(g) + \frac{3}{2}CuO(s) \to \frac{1}{2}N_2(g) + \frac{3}{2}H_2O(l) + \frac{3}{2}Cu(s)$