On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$

Given that the heat of reaction for $A + \frac{1}{2} O_2 \to AO$ is $-50 \ kcal$ and for $AO + \frac{1}{2} O_2 \to AO_2$ is $100 \ kcal$. Find the heat of reaction for $A + O_2 \to AO_2$ in $kcal$.

The value of enthalpy of neutralization is highest for the reaction between:

Given:
$(A) \ 2 \ CO_{(g)} + O_{2(g)} \rightarrow 2 \ CO_{2(g)} \quad \Delta H_1^\theta = -x \ kJ \ mol^{-1}$
$(B) \ C \ (\text{graphite}) + O_{2(g)} \rightarrow CO_{2(g)} \quad \Delta H_2^\theta = -y \ kJ \ mol^{-1}$
The $\Delta H^\theta$ for the reaction $C \ (\text{graphite}) + \frac{1}{2} O_{2(g)} \rightarrow CO_{(g)}$ is:

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.

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)
$H_2O_{(g)} \rightarrow H_{(g)} + OH_{(g)} \quad \Delta H^{\circ} = X \ kJ \ mol^{-1}$