From the following,the heat of neutralization is maximum in:

Given
$(i) \, 2Fe_2O_{3(s)} \to 4Fe_{(s)} + 3O_{2(g)}$
$\Delta _rG^o = + 1487.0 \, kJ \, mol^{-1}$
$(ii) \, 2CO_{(g)} + O_{2(g)} \to 2CO_{2(g)}$
$\Delta _rG^o = - 514.4 \, kJ \, mol^{-1}$
Free energy change,$\Delta _rG^o$ for the reaction
$2Fe_2O_{3(s)} + 6CO_{(g)} \to 4Fe_{(s)} + 6CO_{2(g)}$ will be ..... $kJ \, mol^{-1}$

The heat of formation of $SiO_2$ and $MgO$ is respectively $-48.4 \ kJ$ and $-34.7 \ kJ$. The heat of the reaction $2Mg + SiO_2 \to 2MgO + Si$ is $...... \ kJ$.

Given,
$NO_{(g)} + O_{3(g)} \longrightarrow NO_{2(g)} + O_{2(g)}; \Delta H = -198.9 \, kJ/mol$
$O_{3(g)} \longrightarrow 3/2 O_{2(g)}; \Delta H = -142.3 \, kJ/mol$
$O_{2(g)} \longrightarrow 2O_{(g)}; \Delta H = +495.0 \, kJ/mol$
The enthalpy change $(\Delta H)$ for the following reaction is $..... \, kJ/mol$
$NO_{(g)} + O_{(g)} \longrightarrow NO_{2(g)}$

The standard enthalpies of formation of $CO$ and $CO_2$ are $-110 \, kJ \, mol^{-1}$ and $-394 \, kJ \, mol^{-1}$ respectively. What will be the heat of combustion of $1 \, mol$ of graphite in $kJ$?

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.