The $H-H$ bond energy is $430 \ kJ \ mol^{-1}$ and $Cl-Cl$ bond energy is $240 \ kJ \ mol^{-1}$. $\Delta H$ for the formation of $HCl$ is $-90 \ kJ \ mol^{-1}$. The $H-Cl$ bond energy is about:

Enthalpy change for a reaction does not depend upon

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

The values of bond enthalpy of $H_2$,$Cl_2$ and $HCl$ are $104$,$58$ and $103 \ kcal/mol$ respectively. Find the enthalpy of formation of $HCl_{(g)}$. Reaction: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \to HCl_{(g)}$

For the reaction $F_2 + 2HCl \rightarrow 2HF + Cl_2$,the $\Delta H^o$ is $-352.8 \ kJ$. If the $\Delta H_f^o$ for $HF$ is $-268.3 \ kJ \ mol^{-1}$,then the $\Delta H_f^o$ for $HCl$ will be . . . . . . $kJ \ mol^{-1}$.

Calculate the enthalpy of formation of ethylene $(C_2H_4)$ from the following data:
$(I)$ $C_{\text{(graphite)}} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta H = -393.5 \ kJ$
$(II)$ $H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta U = -256.2 \ kJ$
$(III)$ $C_2H_{4(g)} + 3 O_{2(g)} \longrightarrow 2 CO_{2(g)} + 2 H_2O_{(l)}$; $\Delta H = -1410.8 \ kJ$ (in $kJ$)