On the basis of the following thermochemical data: $(\Delta_fG^o H^{+}_{(aq)} = 0)$
$H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \,; \, \Delta H = 57.32 \, kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \,; \, \Delta H = -286.20 \, kJ$
The value of enthalpy of formation of $OH^{-}$ ion at $25 \, ^oC$ is : .............. $kJ$

Using the given reaction enthalpies,find the enthalpy of formation of $H_2O_2(l)$ in $kJ/mol$.
$(i) N_2H_4(l) + 2H_2O_2(l) \rightarrow N_2(g) + 4H_2O(l); \Delta_r H_1^\circ = -818 \, kJ/mol$
$(ii) N_2H_4(l) + O_2(g) \rightarrow N_2(g) + 2H_2O(l); \Delta_r H_2^\circ = -622 \, kJ/mol$
$(iii) H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O(l); \Delta_r H_3^\circ = -285 \, kJ/mol$

The bond dissociation enthalpy of $X_2$,$\Delta H_{\text{bond}}^{\circ}$,calculated from the given data is $...$ $kJ \ mol^{-1}$. (Nearest integer)
$M^{+}X^{-}_{(s)} \rightarrow M^{+}_{(g)} + X^{-}_{(g)} \quad \Delta H_{\text{lattice}}^{\circ} = 800 \ kJ \ mol^{-1}$
$M_{(s)} \rightarrow M_{(g)} \quad \Delta H_{\text{sub}}^{\circ} = 100 \ kJ \ mol^{-1}$
$M_{(g)} \rightarrow M^{+}_{(g)} + e^{-}_{(g)} \quad \Delta H_{i}^{\circ} = 500 \ kJ \ mol^{-1}$
$X_{(g)} + e^{-}_{(g)} \rightarrow X^{-}_{(g)} \quad \Delta H_{\text{eg}}^{\circ} = -300 \ kJ \ mol^{-1}$
$M_{(s)} + \frac{1}{2}X_{2(g)} \rightarrow M^{+}X^{-}_{(s)} \quad \Delta H_{f}^{\circ} = -400 \ kJ \ mol^{-1}$
[Given : $M^{+}X^{-}$ is a pure ionic compound and $X$ forms a diatomic molecule $X_2$ in gaseous state]

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

If the heats of combustion of $C$,$S$,and $CS_2$ are $x$,$y$,and $z \, kJ \, mol^{-1}$ respectively,what will be the heat of formation of $CS_2$?

While performing a thermodynamics experiment,a student made the following observations:
$HCl + NaOH \rightarrow NaCl + H_{2}O$ $\Delta H = -57.3 \ kJ \ mol^{-1}$
$CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ $\Delta H = -55.3 \ kJ \ mol^{-1}$
The enthalpy of ionization of $CH_{3}COOH$ as calculated by the student is $kJ \ mol^{-1}$. (nearest integer)