Which among the following represents the reaction of formation of the product?

Which factor affects the heat of reaction in the Kirchhoff equation?

Given below are two statements:
Assertion $(A)$: The enthalpy of formation of graphite is taken as zero.
Reason $(R)$: Graphite is the thermodynamically most stable allotrope of carbon.
The correct answer is:

Consider the following reactions:
$(i)$ $H_{(aq)}^{+} + OH^{-}_{(aq)} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_1 \ kJ \ mol^{-1}$
$(ii)$ $H_{2_{(g)}} + \frac{1}{2} O_{2_{(g)}} \longrightarrow H_2O_{(l)}$,$\Delta H = -X_2 \ kJ \ mol^{-1}$
$(iii)$ $CO_{2_{(g)}} + H_{2_{(g)}} \longrightarrow CO_{(g)} + H_2O_{(l)}$,$\Delta H = -X_3 \ kJ \ mol^{-1}$
$(iv)$ $C_2H_{2_{(g)}} + \frac{5}{2} O_{2_{(g)}} \longrightarrow 2CO_{2_{(g)}} + H_2O_{(l)}$,$\Delta H = -X_4 \ kJ \ mol^{-1}$
Enthalpy of formation of $H_2O_{(l)}$ is

If $\Delta H_f (H_2O) = X$,then the heat of neutralization of $CH_3COOH$ and $NaOH$ will be:

Equal volumes of $1 \, M \, HCl$ and $1 \, M \, H_2SO_4$ are neutralized by a dilute $NaOH$ solution,and $x$ and $y \, kcal$ of heat are liberated,respectively. Which of the following is correct?