Find the bond enthalpy of $N-H$ bond in ammonia by using the change in enthalpy for the reaction given below: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta H = -23 \ kcal$. Given bond energies: $N \equiv N = 226 \ kcal/mol$,$H-H = 103 \ kcal/mol$.

Considering the reaction $3O_2 \rightarrow 2O_3$; $\Delta H = +ve$,what can we conclude about the reaction?

The heat of neutralisation of one equivalent of an acid by one equivalent of a base is minimum when:

The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

$A$,$B$,$C$ and $D$ are some compounds. The enthalpy of formation of $A_{(g)}$,$B_{(g)}$,$C_{(g)}$ and $D_{(g)}$ is $9.7, -110, 81$ and $-393 \ kJ \ mol^{-1}$ respectively. What is $\Delta_r H$ (in $kJ \ mol^{-1}$) for the given reaction?
$A_{(g)} + 3B_{(g)} \longrightarrow C_{(g)} + 3D_{(g)}$

Consider the following cases of standard enthalpy of reaction $\Delta H_{r}^{\circ}$ in $kJ \ mol^{-1}$:
$C_{2}H_{6(g)} + \frac{7}{2} O_{2(g)} \rightarrow 2 CO_{2(g)} + 3 H_{2}O(\ell)$,$\Delta H_{1}^{\circ} = -1550$
$C(\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H_{2}^{\circ} = -393.5$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_{2}O(\ell)$,$\Delta H_{3}^{\circ} = -286$
The magnitude of $\Delta H_{f, C_{2}H_{6(g)}}^{\circ}$ is $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.