For the reaction $H_{2(g)} + C_{2}H_{4(g)} \rightarrow C_{2}H_{6(g)}$,the enthalpy change is ....... $Kcal \, mol^{-1}$. Given bond energies: $H-H = 103$,$C-H = 99$,$C-C = 80$,and $C=C = 145 \, Kcal \, mol^{-1}$.

Find the value of enthalpy of formation of $PCl_5(s)$ given the following thermochemical equations:
$1) \frac{1}{2} P_{4(s)} + 3Cl_{2(g)} \to 2PCl_3(\ell) ; \Delta H = -635 \ kJ$
$2) PCl_3(\ell) + Cl_{2(g)} \to PCl_{5(s)} ; \Delta H = -137 \ kJ$

Enthalpy change for the reaction,$\frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \to HCl(g)$,is called:

$H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}$; $\Delta H_{298K} = -68.32 \, kcal$. The enthalpy of vaporization of water at $25 \, ^\circ C$ and $1 \, atm$ pressure is $10.52 \, kcal$. Calculate the standard enthalpy of formation of $1 \, mole$ of water vapor at $25 \, ^\circ C$ (in $kcal$).

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

The heats of solution of anhydrous $CuSO_4$ and $CuSO_4 \cdot 5H_2O$ are $-15.89 \, kcal \, mol^{-1}$ and $2.80 \, kcal \, mol^{-1}$ respectively. What is the heat of hydration of anhydrous $CuSO_4$ in $kcal \, mol^{-1}$?