The heat of combustion of ethanol into carbon dioxide and water is $-327 \ kcal$ at constant pressure. The heat evolved (in $cal$) at constant volume and $27^{\circ} C$ (assuming all gases behave ideally) is $\left( R = 2 \ cal \ mol^{-1} \ K^{-1} \right)$

The heat of neutralization of a strong dibasic acid by a dilute solution of $NaOH$ is approximately ....... $Kcal/equivalent$.

Two reactions are given below:
$2 Fe_{(s)} + \frac{3}{2} O_{2_{(g)}} \rightarrow Fe_2 O_{3_{(s)}}, \Delta H^{o} = -822 \ kJ/mol$
$C_{(s)} + \frac{1}{2} O_{2_{(g)}} \rightarrow CO_{(g)}, \Delta H^{o} = -110 \ kJ/mol$
Then,the enthalpy change for the following reaction is:
$3 C_{(s)} + Fe_2 O_{3_{(s)}} \rightarrow 2 Fe_{(s)} + 3 CO_{(g)}$

From the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:

$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$	$\Delta H = 42.09 \ kJ \ mol^{-1}$
$H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$	$\Delta H = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)}$	$\Delta H = 436 \ kJ \ mol^{-1}$
$O_{2(g)} \rightarrow 2 O_{(g)}$	$\Delta H = 496 \ kJ \ mol^{-1}$

If the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:

What is the enthalpy change when $6.80 \, g$ of $NH_3$ is passed over hot $CuO$ (in $, kJ$)? The standard enthalpies of formation for $NH_3(g)$,$CuO(s)$,and $H_2O(l)$ are $-46.0$,$-155.0$,and $-285.0 \, kJ \, mol^{-1}$ respectively. The reaction is: $NH_3(g) + \frac{3}{2}CuO(s) \to \frac{1}{2}N_2(g) + \frac{3}{2}H_2O(l) + \frac{3}{2}Cu(s)$