$AB$,$A_2$ and $B_2$ are diatomic molecules. If the bond enthalpies of $A_2$,$AB$ and $B_2$ are in the ratio $1:1:0.5$ and enthalpy of formation of $AB$ from $A_2$ and $B_2$ is $-100 \, kJ \, mol^{-1}$,what is the bond energy of $A_2$ in $kJ \, mol^{-1}$?

What is the enthalpy change for the reaction $NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$ called?

Consider the reaction $4NO_2(g) + O_2(g) \rightarrow 2N_2O_5(g)$, $\Delta_rH = -111 \ kJ$. If $N_2O_5(s)$ is formed instead of $N_2O_5(g)$, what will be the value of $\Delta_rH$ in $kJ$? (Given: $\Delta H_{sub} = 54 \ kJ \ mol^{-1}$ for $N_2O_5$)

The enthalpy change for the conversion of $\frac{1}{2} Cl_{2(g)}$ to $Cl^{-}_{(aq)}$ is $......$ $kJ \, mol^{-1}$ (Nearest integer).
Given:
$\Delta_{dis}H^{\circ}_{Cl_{2(g)}} = 240 \, kJ \, mol^{-1}$
$\Delta_{eg}H^{\circ}_{Cl_{(g)}} = -350 \, kJ \, mol^{-1}$
$\Delta_{hyd}H^{\circ}_{Cl^{-(g)}} = -380 \, kJ \, mol^{-1}$

Given:
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?

Which of the following equations is used to calculate the standard enthalpy change of a reaction $(\Delta_r H^\circ)$?