The heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be:

Consider the following data :
Heat of formation of $CO_{2(g)} = -393.5 \ kJ \ mol^{-1}$
Heat of formation of $H_2O_{(l)} = -286.0 \ kJ \ mol^{-1}$
Heat of combustion of benzene $= -3267.0 \ kJ \ mol^{-1}$
The heat of formation of benzene is $........... \ kJ \ mol^{-1}$. $(Nearest \ integer)$

The heat of neutralisation of a strong acid and a strong alkali is $57.0 \ kJ \ mol^{-1}$. The heat released when $0.5 \ mol$ of $HNO_3$ solution is mixed with $0.2 \ mol$ of $KOH$ is: (in $kJ$)

For the reaction $C_{2}H_{6} \rightarrow C_{2}H_{4} + H_{2}$,the reaction enthalpy $\Delta_{r}H = \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \, kJ \, mol^{-1}$. (Round off to the Nearest Integer). [Given: Bond enthalpies in $kJ \, mol^{-1} : C-C : 347, C=C : 611, C-H : 414, H-H : 436$]

The heat of neutralisation of a strong acid and a strong alkali is $57.0 \, kJ \, mol^{-1}$. The heat released when $0.5 \, mole$ of $HNO_3$ solution is mixed with $0.2 \, mole$ of $KOH$ is $...... \, kJ$.

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 $.... \ kJ$.