Enthalpy change for a reaction does not depend upon

At $25^{\circ} C$, the enthalpy of the following processes are given:
$H_{2(g)} + O_{2(g)} \rightarrow 2 OH_{(g)} \quad \Delta H^{\circ} = 78 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(g)} \quad \Delta H^{\circ} = -242 \ kJ \ mol^{-1}$
$H_{2(g)} \rightarrow 2 H_{(g)} \quad \Delta H^{\circ} = 436 \ kJ \ mol^{-1}$
$\frac{1}{2} O_{2(g)} \rightarrow O_{(g)} \quad \Delta H^{\circ} = 249 \ kJ \ mol^{-1}$
What would be the value of $X$ for the following reaction? (Nearest integer)
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} \quad \Delta H^{\circ} = X \ kJ \ mol^{-1}$

Which is the best definition of “heat of neutralization”?

Given that $:$
$2 C_{(s)} + 2 O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} ; \Delta H = -787 \ kJ$
$H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \rightarrow H_2 O_{(l)} ; \Delta H = -286 \ kJ$
$C_2 H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \rightarrow 2 CO_{2_{(g)}} + H_2 O_{(l)} ; \Delta H = -1301 \ kJ$
The heat of formation of acetylene will be $:-$

The heat of neutralization for the reaction $NaOH + HCl \to NaCl + H_2O$ is $57.1 \ kJ \ mol^{-1}$. What will be the heat released when $0.25 \ mol$ of $NaOH$ is titrated against $0.25 \ mol$ of $HCl$ (in $kJ$)?

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 $...... \ kcal$.