At $1 \ bar$ and $298 \ K$,the standard molar enthalpy of formation of which substance is zero?

One mole of $C_2H_5OH_{(l)}$ was completely burnt in oxygen to form $CO_{2(g)}$ and $H_2O_{(l)}$. The standard enthalpy of formation $\Delta_fH^{\ominus}$ of $C_2H_5OH_{(l)}, CO_{2(g)}$ and $H_2O_{(l)}$ is $x, y, z \ kJ \ mol^{-1}$ respectively. What is $\Delta_rH^{\ominus}$ (in $kJ \ mol^{-1}$) for this reaction?

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

If $C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -X$,and $CO_{(g)} + \frac{1}{2} O_{2(g)} \rightarrow CO_{2(g)}$,$\Delta H = -Y$,calculate $\Delta_f H$ for $CO_{(g)}$ formation.

In an exothermic reaction,$\Delta H$ is

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.