$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = -94 \ kcal$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -67.7 \ kcal$
Then the heat of formation of $CO_{(g)}$ is $..... \ kcal$

The $\Delta H_f^{\circ}$ values (in $kJ \ mol^{-1}$) for graphite, diamond, and $C_{60}$ are respectively:

In the reaction $C + 2S \to CS_2 + \Delta H$,$\Delta H$ is the

The enthalpy change for the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^{-}_{(aq)}$ is $......$ $kJ \, mol^{-1}$ (Nearest integer).
Given:
$\Delta_{dis}H^{\circ}_{Cl_{2(g)}} = 240 \, kJ \, mol^{-1}$
$\Delta_{eg}H^{\circ}_{Cl_{(g)}} = -350 \, kJ \, mol^{-1}$
$\Delta_{hyd}H^{\circ}_{Cl^{-(g)}} = -380 \, kJ \, mol^{-1}$

Combustion of glucose takes place according to the equation,$C_6H_{12}O_6 + 6O_2 \to 6CO_2 + 6H_2O$,$\Delta H = -72 \ kcal$. How much energy will be required for the production of $1.6 \ g$ of glucose? (Molecular mass of glucose $= 180 \ g/mol$)

If $H^{+} + OH^{-} \to H_2O + 13.7 \ kcal$,then the heat of neutralization for complete neutralization of one mole of $H_2SO_4$ by base will be.......$kcal$