If the heat of neutralization of an acid-base reaction is $56 \ kJ \ mol^{-1}$,then the substances could be:

The enthalpies of combustion of $C$(graphite) and $C$(diamond) are $-393.8 \ kJ \ mol^{-1}$ and $-395.3 \ kJ \ mol^{-1}$ respectively. The enthalpy of conversion of $C$(graphite) to $C$(diamond) is

The correct order of $\Delta_{f} H^{\circ}$ values of Diamond $(I)$, Graphite $(II)$, and Fullerene $(III)$ is:

Molar enthalpy of combustion of $C_2H_{2(g)}$,$C_{(s)}$,and $H_{2(g)}$ are $-1300$,$-394$,and $-286 \ kJ/mol$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is $:$

From the following data,the enthalpy of vaporization of liquid water in $KJ \, mol^{-1}$ will be:
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(l); \Delta H = -285.77 \, KJ \, mol^{-1}$
$H_2(g) + 1/2 O_2(g) \rightarrow H_2O(g); \Delta H = -241.84 \, KJ \, mol^{-1}$

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 ..... $kJ \, mol^{-1}$