The reaction $CH_{4(g)} + Cl_{2(g)} \to CH_3Cl_{(g)} + HCl_{(g)}$ has $\Delta H = -25 \, kcal$. Given bond energies $BE(C-H) = 84 \, kcal$,$BE(H-Cl) = 103 \, kcal$,$BE(C-Cl) = x$,and $BE(Cl-Cl) = y$. If $\frac{x}{y} = \frac{9}{5}$,then find the value of $y$. (in $, kcal$)

Given $:$
$\Delta H^{\ominus}_{sub}[C(graphite)] = 710 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C-H} = 414 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{H-H} = 436 \ kJ \ mol^{-1}$
$\Delta H^{\ominus}_{C=C} = 611 \ kJ \ mol^{-1}$
The $\Delta H^{\ominus}_{f}$ for $CH_2=CH_2$ is $............ \ kJ \ mol^{-1}$ $(nearest \ integer \ value)$

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$I. C_{(graphite)} + O_{2(g)} \rightarrow CO_{2(g)}; \Delta H = -393.5 \ kJ$
$II. H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -286.2 \ kJ$
$III. C_2H_{4(g)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 2H_2O_{(l)}; \Delta H = -1410.8 \ kJ$ (in $kJ$)

For the allotropic change represented by the equation $C(\text{diamond}) \to C(\text{graphite})$,the enthalpy change is $\Delta H = -1.89 \ kJ$. If $6 \ g$ of diamond and $6 \ g$ of graphite are separately burnt to yield carbon dioxide,the heat liberated in the first case is:

The heat of neutralization for the reaction $NaOH + HCl \to NaCl + H_2O$ is $57.1 \ kJ \ mol^{-1}$. What will be the heat released when $0.25 \ mol$ of $NaOH$ is titrated against $0.25 \ mol$ of $HCl$ (in $kJ$)?

For the following reaction,
$C (diamond) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -97.6 \ kcal$
$C (graphite) + O_2 \rightarrow CO_{2(g)}$; $\Delta H = -94.3 \ kcal$
The heat change for the conversion of $1 \ g$ of $C (diamond) \rightarrow C (graphite)$ is (in $kcal$)