The heat of neutralization for the reaction $NaOH + HCl \to NaCl + H_2O$ is $57.1 \ kJ \ mol^{-1}$. What will be the heat released when $0.25 \ mol$ of $NaOH$ is titrated against $0.25 \ mol$ of $HCl$ (in $kJ$)?

Explain the standard enthalpy of formation.

The heat of formation for the reaction $H_2 + Cl_2 \to 2HCl + 44 \ kcal$ is ..... $kcal \ mol^{-1}$.

The heat of neutralisation of one equivalent of an acid by one equivalent of a base is minimum when:

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 $........... \ kJ \ mol^{-1}$. $(Nearest \ integer)$

Calculate the enthalpy change when $12 \ g$ of carbon reacts with sufficient hydrogen to form methane. If the enthalpy of formation of methane is $-75 \ kJ \ mol^{-1}$. (in $kJ$)