The heat of neutralization of $HCl$ by $NaOH$ is $-57.3 \, kJ/mol$. If the heat of neutralization of $HCN$ by $NaOH$ is $-12.1 \, kJ/mol$,then the enthalpy of dissociation of $HCN$ is $...... \, kJ$.

Calculate the standard enthalpy change for the reaction,$C_2H_5OH_{(\ell)} + 3O_{2_{(g)}} \rightarrow 2CO_{2_{(g)}} + 3H_2O_{(\ell)}$. Given: $\Delta_{f}H^{\circ}(C_2H_5OH) = -280 \ kJ \ mol^{-1}$,$\Delta_{f}H^{\circ}(CO_2) = -390 \ kJ \ mol^{-1}$,and $\Delta_{f}H^{\circ}(H_2O) = -285 \ kJ \ mol^{-1}$.

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

The heat of formation of methane $C_{(s)} + 2H_{2(g)} \to CH_{4(g)}$ at constant pressure is $-18500 \ cal$ at $25 \ ^oC$. The heat of reaction at constant volume would be (in $cal$)

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

Consider the following data :
Heat of formation of $CO_{2(g)} = -393.5 \ kJ \ mol^{-1}$
Heat of formation of $H_2O_{(l)} = -286.0 \ kJ \ mol^{-1}$
Heat of combustion of benzene $= -3267.0 \ kJ \ mol^{-1}$
The heat of formation of benzene is $........... \ kJ \ mol^{-1}$. $(Nearest \ integer)$