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 : .............. $kJ$

If $C + O_2 \to CO_2 + 94.2 \ kcal$,$H_2 + \frac{1}{2} O_2 \to H_2O + 68.3 \ kcal$,and $CH_4 + 2O_2 \to CO_2 + 2H_2O + 210.8 \ kcal$,then the possible heat of formation of methane will be $...... \ kcal$.

The heat of complete neutralization of $1 \ mol$ of $H_2SO_4$ with a strong base is ....... $Kcal$.

Based on the following thermochemical equations,find the value of $x$ in $kJ$.
$(i) \ H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} ; \Delta H = 131 \ kJ$
$(ii) \ CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} ; \Delta H = -282 \ kJ$
$(iii) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} ; \Delta H = -242 \ kJ$
$(iv) \ C_{(s)} + O_{2(g)} \to CO_{2(g)} ; \Delta H = -x \ kJ$

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

Calculate $\Delta H \ (kJ/mol)$ for the reaction:
$2FeO_{(s)} + \frac{1}{2} O_{2_{(g)}} \to Fe_2O_{3_{(s)}}$
Given $\Delta H$ values:
$(i)$ $Fe_2O_{3_{(s)}} + 3C_{(graphite)} \to 2Fe_{(s)} + 3CO_{(g)}$ : $492 \ kJ/mol$
$(ii)$ $FeO_{(s)} + C_{(graphite)} \to Fe_{(s)} + CO_{(g)}$ : $156 \ kJ/mol$
$(iii)$ $C_{(graphite)} + O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-393 \ kJ/mol$
$(iv)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}$ : $-283 \ kJ/mol$