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 $kJ \, mol^{-1} : C-C : 347, C=C : 611, C-H : 414, H-H : 436$]

Calculate the standard enthalpy of formation of $CH_{3}OH_{(l)}$ from the following data:
$CH_{3}OH_{(l)} + \frac{3}{2} O_{2_{(g)}} \rightarrow CO_{2_{(g)}} + 2 H_{2}O_{(l)}$; $\Delta_{r} H^{\ominus} = -726 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2_{(g)}} \rightarrow CO_{2_{(g)}}$; $\Delta_{c} H^{\ominus} = -393 \ kJ \ mol^{-1}$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_{2}O_{(l)}$; $\Delta_{f} H^{\ominus} = -286 \ kJ \ mol^{-1}$

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

Calculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)

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:

Calculate the heat of reaction for the process: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$ given the following data:
$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H = -8.4 \, Kcal$
$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H = -17.3 \, Kcal$
$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H = -12.5 \, Kcal$
$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H = +3.9 \, Kcal$ (in $, Kcal$)