The standard enthalpies of formation for $NH_3$,$CO_2$,$HI$,and $SO_2$ are $-46.19$,$-393.4$,$+24.94$,and $-296.9 \, kJ/mol$ respectively. What is the increasing order of their stability?

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

The heat change $\Delta H$ for the reaction $2CO + O_2 \to 2CO_2; \Delta H = -135 \ kcal$ is called

The reaction of methanol $(\Delta H_f^o = -238.7 \ kJ \ mol^{-1})$ with $2$-methylpropene produces methyl tert-butyl ether $(\Delta H_f^o = -313.6 \ kJ \ mol^{-1})$. Given the reaction: $(CH_3)_2C = CH_2 + CH_3OH \rightarrow (CH_3)_3C - OCH_3; \Delta H^o = -57.8 \ kJ \ mol^{-1}$,calculate the $\Delta H_f^o$ for $2$-methylpropene.

Calculate $\Delta H^{\circ}$ for the reaction,$Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$ given the following:
$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$
$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$
$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$

If the ratio of bond dissociation energies of $XY$,$X_2$,and $Y_2$ is $1 : 1 : 0.5$ and the enthalpy of formation of $XY$ is $\Delta_f H = -200 \ kJ \ mol^{-1}$,then the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$ is: