Given the reactions:
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)

Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$

Given the following thermochemical equations:
$1) \ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}, \Delta H = -787 \ kJ$
$2) \ H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(l)}, \Delta H = -286 \ kJ$
$3) \ C_2H_{2(g)} + \frac{5}{2} O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}, \Delta H = -1310 \ kJ$
Calculate the enthalpy of formation of acetylene $(C_2H_{2(g)})$ in $kJ \ mol^{-1}$.

The enthalpy of solution of $NaOH_{(s)}$ in water is $-41.6 \ kJ \ mol^{-1}$. When $NaOH_{(s)}$ is dissolved in water,the temperature of the water will ..........

Given the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)

Which of the following substances has the highest value of standard molar enthalpy of formation at $298 \ K$?