Hess's law is based on

The enthalpy changes at $298 \ K$ in successive breaking of $O-H$ bonds of $H_2O$ are:
$H_2O_{(g)} \to H_{(g)} + OH_{(g)}, \Delta H = 498 \ kJ \ mol^{-1}$
$OH_{(g)} \to H_{(g)} + O_{(g)}, \Delta H = 428 \ kJ \ mol^{-1}$
The bond enthalpy of the $O-H$ bond is ..... $kJ \ mol^{-1}$.

At $298 \ K$,the enthalpy change (in $kJ$) for the reaction given below is: $CH_{4(g)} + O_{2(g)} \rightarrow C_{(s)} + 2H_2O_{(l)}$
Given:
$1) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H^{\ominus} = -286 \ kJ$
$2) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H^{\ominus} = -394 \ kJ$
$3) \ CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?

The enthalpies of formation of $CO_{2(g)}$ and $CaO_{(s)}$ are $-94.0 \, kJ$ and $-152 \, kJ$ respectively. The enthalpy of the reaction $CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$ is $42 \, kJ$. The enthalpy of formation of $CaCO_{3(s)}$ is ............... $kJ$.

For the reaction $Cu_{(g)}^{+} + I_{(g)}^{-} \to CuI_{(s)}$,the value of $\Delta H^o$ is $-446 \ kJ \ mol^{-1}$. If the ionization energy of $Cu_{(g)}$ is $745 \ kJ \ mol^{-1}$ and the electron affinity of $I_{(g)}$ is $-295 \ kJ \ mol^{-1}$,calculate the value of $\Delta H^o$ for the formation of $CuI_{(s)}$ from $Cu_{(g)}$ and $I_{(g)}$ in $kJ \ mol^{-1}$.