Standard enthalpy of formation of water is $-286 \ kJ \ mol^{-1}$. When $1800 \ mg$ of water is formed from its constituent elements in their standard states,the amount of energy liberated is: (in $kJ$)

Enthalpy of formation of two compounds $X$ and $Y$ are $-84 \ kJ$ and $-156 \ kJ$ respectively. Which of the following statements is correct?

Determine the enthalpy of formation for $H_2O_2(\ell)$,using the listed enthalpies of reaction:
$N_2H_{4(\ell)} + 2H_2O_{2(\ell)} \to N_{2(g)} + 4H_2O_{(\ell)}; \Delta _r H_1^o = -818 \, kJ/mol$
$N_2H_{4(\ell)} + O_{2(g)} \to N_{2(g)} + 2H_2O_{(\ell)}; \Delta _r H_2^o = -622 \, kJ/mol$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(\ell)}; \Delta _r H_3^o = -285 \, kJ/mol$
Calculate the value in $kJ/mol$.

Calculate the heat of reaction for the process: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$ given the following data:
$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H = -8.4 \, Kcal$
$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H = -17.3 \, Kcal$
$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H = -12.5 \, Kcal$
$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H = +3.9 \, Kcal$ (in $, Kcal$)

$C(\text{diamond}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -395 \text{ kJ}$
$C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta H = -393.5 \text{ kJ}$
If graphite is converted into diamond,then the $\Delta H$ for the process is . . . . . . $\text{kJ}$.

If at $298 \, K$ the bond energies of $C-H, C-C, C=C$ and $H-H$ bonds are respectively $414, 347, 615$ and $435 \, kJ \, mol^{-1}$,the value of enthalpy change for the reaction $H_2C=CH_{2(g)} + H_{2(g)} \to H_3C-CH_{3(g)}$ at $298 \, K$ will be $.... \, kJ$.