Identify the factor from the following on which the heat of reaction does not depend.

For the reaction $H_{2} + I_{2} \rightarrow 2 HI$,provide the potential energy profile diagram and prove that it is an exothermic reaction.

Diborane is formed from the elements as shown in equation $(i)$:
$2 B_{(s)} + 3 H_{2(g)} \longrightarrow B_2H_{6(g)} \dots (i)$
Given that:
$H_2O_{(l)} \longrightarrow H_2O_{(g)}, \quad \Delta H_1^{\circ} = 44 \, kJ$
$2 B_{(s)} + \frac{3}{2} O_{2(g)} \longrightarrow B_2O_{3(s)}, \quad \Delta H_2^{\circ} = -1273 \, kJ$
$B_2H_{6(g)} + 3 O_{2(g)} \longrightarrow B_2O_{3(s)} + 3 H_2O_{(g)}, \quad \Delta H_3^{\circ} = -2035 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(l)}, \quad \Delta H_4^{\circ} = -286 \, kJ$
The $\Delta H^{\circ}$ for the reaction $(i)$ is $..... \, kJ$.

If the enthalpy of combustion of carbon is $-393 \, kJ \, mol^{-1}$,calculate the heat produced by the combustion of $1 \, g$ of carbon in $kJ$.

$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$; $\Delta H$ at $298 \ K = -285.8 \ kJ$. The molar enthalpy of vaporization of water at $1 \ atm$ and $25^{\circ}C$ is $44 \ kJ$. The standard enthalpy of formation of $1 \ mole$ of water vapor at $25^{\circ}C$ is $...... \ kJ$. (in $.8$)

The heat of neutralization for $2 \ moles$ of $\text{LiOH}$ and $1 \ mole$ of $H_2SO_4$ at $25^{\circ} C$ is $-69.6 \ kJ$. The heat of ionisation of $\text{LiOH}$ will be nearly $:-$