Among the following given substances, the one with zero $\Delta_{f}H^{\circ}$ is

If $(i)$ $C + O_2 \to CO_2$,$(ii)$ $C + 1/2 O_2 \to CO$,$(iii)$ $CO + 1/2 O_2 \to CO_2$,the heats of reaction are $Q$,$-12$,and $-10$ respectively. Then $Q =$ ?

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.

The values of heat of formation of $SO_2$ and $SO_3$ are $-298.2 \ kJ$ and $-395.2 \ kJ$ respectively. The heat of reaction for the following reaction will be $... \ kJ$
$SO_2 + \frac{1}{2}O_2 \to SO_3$

The enthalpy of neutralization of $HCN$ by $NaOH$ is $-12.13 \, kJ \, mol^{-1}$. The enthalpy of ionisation of $HCN$ will be $... \, kJ \, mol^{-1}$.

Calculate the enthalpy change of the reaction for $H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$, given the bond energies (in $\text{kJ mol}^{-1}$): $H-H = 436$, $Cl-Cl = 242$, $H-Cl = 431$.