Given that bond energies of $H-H$ and $Cl-Cl$ are $430 \ kJ \ mol^{-1}$ and $240 \ kJ \ mol^{-1}$ respectively and $\Delta H_f$ for $HCl$ is $-90 \ kJ \ mol^{-1},$ the bond enthalpy of $HCl$ is ............... $kJ \ mol^{-1}$.

Calculate $\Delta H$ for the reaction: $H_{2(g)} + O_{2(g)} \rightarrow H_2O_{2(g)}$ given the bond energies: $BE_{H-H} = 436 \ kJ/mol$,$BE_{O=O} = 499 \ kJ/mol$,$BE_{O-O} = 142 \ kJ/mol$,and $BE_{O-H} = 460 \ kJ/mol$. (in $kJ$)

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:
$1) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)} ; \Delta H^{\ominus} = -286 \ kJ$
$2) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} ; \Delta H^{\ominus} = -394 \ kJ$
$3) \ CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)} ; \Delta H^{\ominus} = -890 \ kJ$

Explain the change in enthalpy related to a chemical reaction.

The enthalpy of neutralization of a strong acid by a strong base is $-57.32 \ kJ/mol$. The enthalpy of formation of water is $-285.84 \ kJ/mol$. The enthalpy of formation of hydroxyl ion is......$kJ/mol$. (Assume $\Delta H_{f}^{\circ}(H^{+}_{(aq)}) = 0 \ kJ/mol$)

How much heat in $kJ$ is produced by the combustion of $15.5 \, g$ of propane? ${C_3H_8} + 5{O_2} \to 3{CO_2} + 4{H_2O}; \Delta{H^o} = -2219 \, kJ/mol$