The enthalpy of the reaction,$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}$ is $\Delta H_1$ and that of $H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}$ is $\Delta H_2$. Then:

In an exothermic reaction,$\Delta H$ is

Hess's law deals with

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 $kJ$)?

Calculate the standard enthalpy change of the following reaction: $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(\ell)}$ Given that: $\Delta_{f} H^{\circ}(CH_4) = -75 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(CO_2) = -394 \ kJ \ mol^{-1}$,$\Delta_{f} H^{\circ}(H_2O) = -286 \ kJ \ mol^{-1}$

Given that the bond energies of $N \equiv N$ is $946 \ kJ \ mol^{-1}$,$H-H$ is $435 \ kJ \ mol^{-1}$,$N-N$ is $159 \ kJ \ mol^{-1}$,and $N-H$ is $389 \ kJ \ mol^{-1}$,calculate the enthalpy of formation for the gas phase reaction $N_2 + 2H_2 \rightarrow N_2H_4$ in $kJ \ mol^{-1}$.