The standard enthalpy of formation of $CO_{(g)}$, $CO_{2(g)}$, $N_2O_{(g)}$ and $N_2O_{4(g)}$ are respectively $-110$, $-393$, $81$ and $-10 \ kJ \ mol^{-1}$. The enthalpy change (in $kJ$) of the following reaction is $N_2O_{4(g)} + 3CO_{(g)} \longrightarrow N_2O_{(g)} + 3CO_{2(g)}$

The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$

For the reaction $\frac{1}{2} X_2O_{(s)} \to X_{(s)} + \frac{1}{4} O_{2(g)}$ with $\Delta H = 90 \ kJ$,the enthalpy change for the reaction of $1 \ mol$ of $O_2$ with metal $X$ is ................ $kJ$.

The standard enthalpies of formation for $CH_{4(g)}$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-75$,$-393.5$,and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$ is ......... $kJ$.

Given that $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(g)}; \Delta H_1$ and $H_{2(g)} + 1/2 O_{2(g)} \to H_2O_{(l)}; \Delta H_2$,which of the following is correct?

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. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$