The heat of neutralization of a weak acid and a strong base is less than the heat of neutralization of a strong acid and a strong base due to:

From the following data:
$CH_3OH_{(l)} + \frac{3}{2}O_{2(g)} \longrightarrow CO_{2(g)} + 2H_2O_{(l)}$; $\Delta_rH^{\circ} = -726 \ kJ \ mol^{-1}$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_2O_{(l)}$; $\Delta_rH^{\circ} = -286 \ kJ \ mol^{-1}$
$C_{(graphite)} + O_{2(g)} \longrightarrow CO_{2(g)}$; $\Delta_rH^{\circ} = -393 \ kJ \ mol^{-1}$
The standard enthalpy of formation of $CH_3OH_{(l)}$ in $kJ \ mol^{-1}$ is:

Given the following thermochemical equations:
$(1) \ S + O_2 \rightarrow SO_2 ; \Delta H = -298.2 \ kJ$
$(2) \ SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H = -98.7 \ kJ$
$(3) \ SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H = -130.2 \ kJ$
$(4) \ H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H = -287.3 \ kJ$
Calculate the enthalpy of formation of $H_2SO_4$ at $298 \ K$ in $kJ$.

If $C$ (diamond) $\rightarrow C$ (graphite) $+ X \ kJ \ mol^{-1}$,$C$ (diamond) $+ O_{2(g)} \rightarrow CO_{2(g)} + Y \ kJ \ mol^{-1}$,and $C$ (graphite) $+ O_{2(g)} \rightarrow CO_{2(g)} + Z \ kJ \ mol^{-1}$,at constant temperature,then which of the following relations is correct?

$AB$,$A_2$,and $B_2$ are diatomic molecules. Enthalpies of dissociation of $AB$,$A_2$,and $B_2$ are in the ratio of $1:1:0.5$. Enthalpy of formation of $AB$,$\Delta_f H = -100 \ kJ \ mol^{-1}$. Find the dissociation enthalpy of $A_2$?
Reaction : $\frac{1}{2} A_2 + \frac{1}{2} B_2 \to AB$

Which of the following compounds has a standard molar enthalpy of formation equal to zero at $298 \ K$?