The enthalpy change for the transition of carbon from diamond to graphite is $\Delta H = -453.5 \ \text{cal}$. What does this indicate?

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 $........... kJ \ mol^{-1}$ $(Nearest \ integer)$.

At $25^{\circ} C$, the enthalpy of the following processes are given:
$H_{2(g)} + O_{2(g)} \rightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = 78 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} \quad \Delta H^{\circ} = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)} \quad \Delta H^{\circ} = 436 \ kJ \ mol^{-1}$
$\frac{1}{2} O_{2(g)} \rightarrow O_{(g)} \quad \Delta H^{\circ} = 249 \ kJ \ mol^{-1}$
What would be the value of $X$ for the following reaction? (Nearest integer)
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} \quad \Delta H^{\circ} = X \ kJ \ mol^{-1}$

The value of heat generated when $36.5 \, g$ of $HCl$ and $40 \, g$ of $NaOH$ react during neutralization is.....$kcal$.

Observe the following reaction:
$2 A_{2(g)} + B_{2(g)} \xrightarrow{T(K)} 2 A_2 B_{(g)} + 600 \ kJ$
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of $A_2 B_{(g)}$ is:

If at $298 \, K$ the bond energies of $C-H, C-C, C=C$ and $H-H$ bonds are respectively $414, 347, 615$ and $435 \, kJ \, mol^{-1}$,the value of enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ will be $.... \, kJ$.