The standard heats of formation in $kcal \ mol^{-1}$ of $NO_{2(g)}$ and $N_2O_{4(g)}$ are $8.0$ and $2.0$ respectively. The heat of dimerization of $NO_2$ in $kcal$ for the reaction $2NO_{2(g)} \rightarrow N_2O_{4(g)}$ is:

Given the following thermochemical equations:
$(1) \ S + O_2 \rightarrow SO_2 ; \Delta H = -298.2 \ kJ$
$(2) \ SO_2 + \frac{1}{2} O_2 \rightarrow SO_3 ; \Delta H = -98.7 \ kJ$
$(3) \ SO_3 + H_2O \rightarrow H_2SO_4 ; \Delta H = -130.2 \ kJ$
$(4) \ H_2 + \frac{1}{2} O_2 \rightarrow H_2O ; \Delta H = -287.3 \ kJ$
Calculate the enthalpy of formation of $H_2SO_4$ at $298 \ K$ in $kJ$.

$18.0 \ g$ of water completely vaporises at $100^{\circ}C$ and $1 \ bar$ pressure and the enthalpy change in the process is $40.79 \ kJ \ mol^{-1}$. What will be the enthalpy change for vaporising two moles of water under the same conditions? What is the standard enthalpy of vaporisation for water?

If the value of $\Delta H_{O-H}$ is $109 \ kcal \ mol^{-1}$,then the formation of one mole of water from $H_{(g)}$ and $O_{(g)}$ is associated with:

Calculate the enthalpy change for the reaction $H_{2(g)} + Br_{2(g)} \rightarrow 2HBr_{(g)}$ in $kJ$. The bond energies of $H-H$,$Br-Br$,and $H-Br$ are $435$,$192$,and $364 \, kJ \, mol^{-1}$ respectively.

Enthalpy of sublimation of a substance is equal to