Match the following allotropes of carbon with their standard enthalpy of formation $(\Delta_f H^{\Theta})$:

Allotrope	$\Delta_f H^{\Theta}$
$i$. Graphite	$b$. $0 \ kJ/mol$
$ii$. Diamond	$c$. $1.90 \ kJ/mol$
$iii$. Fullerene	$a$. $38.1 \ kJ/mol$

Calculate $\Delta H \ (kJ/mol)$ for the reaction:
$2FeO_{(s)} + \frac{1}{2} O_{2_{(g)}} \to Fe_2O_{3_{(s)}}$
Given $\Delta H$ values:
$(i)$ $Fe_2O_{3_{(s)}} + 3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ : $492 \ kJ/mol$
$(ii)$ $FeO_{(s)} + C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ : $156 \ kJ/mol$
$(iii)$ $C_{(graphite)} + O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-393 \ kJ/mol$
$(iv)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-283 \ kJ/mol$

When $50 \ cm^3$ of a strong acid is added to $50 \ cm^3$ of an alkali,the temperature rises by $5 \ ^oC$. If $250 \ cm^3$ of each liquid are mixed,the temperature rise would be $... \ ^oC$.

If heat of neutralization is $-13.7 \, KCal$ at $25 \, ^oC$ and $\Delta H_f^o (H_2O) = -68 \, KCal$,then the standard enthalpy of formation of $OH^{-}$ will be.....$KCal$. (in $.3$)

The enthalpies of dissolution of $BaCl_2(s)$ and $BaCl_2 \cdot 2H_2O(s)$ are $-20.6 \ kJ \ mol^{-1}$ and $8.8 \ kJ \ mol^{-1}$ respectively. Calculate the enthalpy of hydration for the given reaction: $BaCl_2(s) + 2H_2O(l) \to BaCl_2 \cdot 2H_2O(s)$

The $H_2O_{(g)}$ molecule dissociates as:
$(i)$ $H_2O_{(g)} \to H_{(g)} + OH_{(g)}; \Delta H = 490 \ kJ$
$(ii)$ $OH_{(g)} \to H_{(g)} + O_{(g)}; \Delta H = 424 \ kJ$
The average bond energy (in $kJ$) for water is