The standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus the amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction: $H_{2(g)} + Br_{2(g)} \to 2HBr_{(g)}$? Given that the bond energy of $H_2$,$Br_2$,and $HBr$ is $435 \ kJ \ mol^{-1}$,$192 \ kJ \ mol^{-1}$,and $368 \ kJ \ mol^{-1}$ respectively.

Consider the following processes:

Process	$\Delta H \ (kJ/mol)$
$I. \ \frac{1}{2} A \rightarrow B$	$+150$
$II. \ 3B \rightarrow 2C + D$	$-125$
$III. \ E + A \rightarrow 2D$	$+350$

For $B + D \rightarrow E + 2C, \Delta H$ will be ............. $kJ/mol$

Calculate the standard enthalpy of formation of $ICl_{(g)}$ based on the following reactions. The standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$ respectively.
$(i)$ $Cl_{2(g)} = 2Cl_{(g)}$,$\Delta H = 242.3 \text{ kJ mol}^{-1}$
$(ii)$ $I_{2(g)} = 2I_{(g)}$,$\Delta H = 151.0 \text{ kJ mol}^{-1}$
$(iii)$ $ICl_{(g)} = I_{(g)} + Cl_{(g)}$,$\Delta H = 211.3 \text{ kJ mol}^{-1}$
$(iv)$ $I_{2(s)} = I_{2(g)}$,$\Delta H = 62.76 \text{ kJ mol}^{-1}$
Result in $\text{kJ mol}^{-1}$:

The enthalpy change for a reaction does not depend on:

The energy required to break one mole of $Cl-Cl$ bonds in $Cl_2$ is $242 \, kJ \, mol^{-1}.$ The longest wavelength of light capable of breaking a single $Cl-Cl$ bond is $............ \, nm.$
$(c = 3 \times 10^8 \, m \, s^{-1}$ and $N_A = 6.02 \times 10^{23} \, mol^{-1}).$