When the aqueous solution of $0.5 \, mole$ of $HNO_3$ is mixed with $0.3 \, mole$ of $OH^{-}$ solution,what will be the liberated heat in $kJ$? (Enthalpy of neutralization is $= 57.1 \, kJ \, mol^{-1}$)

If the enthalpy of atomisation for $Br_{2(l)}$ is $x \ kJ/mol$ and the bond enthalpy for $Br_{2(g)}$ is $y \ kJ/mol$,what is the relation between them?

Calculate the standard enthalpy change for the synthesis of ammonia gas from the following data:
$i$. $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
$ii$. $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$ (in $kJ$)

The bond dissociation enthalpy of $H_{2(g)}$ and $N_{2(g)}$ are $436 \ kJ \ mol^{-1}$ and $940 \ kJ \ mol^{-1}$ respectively,and the enthalpy of formation of $NH_{3(g)}$ is $-45 \ kJ \ mol^{-1}$. The enthalpy of atomisation of $NH_{3(g)}$ is ..... $kJ \ mol^{-1}$.

Identify the law for the statement "Overall the enthalpy change for a reaction is equal to the sum of enthalpy changes of individual steps in the reaction".

Based on Hess's law calculations,what is the average $S-O$ bond energy in $SO_3$ if $\Delta H_f^o$ of $SO_3$ is $-270 \ kJ \ mol^{-1}$. Given: Bond energy of $O=O$ is $495 \ kJ \ mol^{-1}$,heat of sublimation for $S_{(s)}$ is $277 \ kJ \ mol^{-1}$,and bond energy of $S=O$ is not provided,but we assume the formation reaction: $S_{(s)} + \frac{3}{2} O_{2(g)} \rightarrow SO_{3(g)}$. Use the atomization energy of $S_{(s)} = 277 \ kJ \ mol^{-1}$ and $O=O = 495 \ kJ \ mol^{-1}$. Calculate the average $S-O$ bond energy in $SO_3$.