The enthalpy of formation of $H_2O_{(g)}$ at $25 \, ^\circ C$ is $-241.8 \, kJ \, mol^{-1}$. For the reaction ${H_2}_{(g)} + \frac{1}{2} {O_2}_{(g)} \to {H_2}O_{(g)}$,the enthalpy change will be ....... $kJ \, mol^{-1}$?

Heat of combustion of two isomers $x$ and $y$ are $17 \ kJ/mol$ and $12 \ kJ/mol$ respectively. From this information,it may be concluded that:

The enthalpy of combustion of ${C_6H_6}_{(l)}$ is $-3250 \, kJ/mol$. When $0.39 \, g$ of benzene is burnt in excess of oxygen in an open vessel,the amount of heat evolved is:

Given: $S + O_2 \rightarrow SO_2 : \Delta H_1 = -298.2 \ kJ$,$SO_2 + 1/2 \ O_2 \rightarrow SO_3 : \Delta H_2 = -98.7 \ kJ$,$SO_3 + H_2O \rightarrow H_2SO_4 : \Delta H_3 = -130.2 \ kJ$,and $H_2 + 1/2 \ O_2 \rightarrow H_2O : \Delta H_4 = -287.3 \ kJ$. Calculate the heat of formation of $H_2SO_4$ 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?

Molar enthalpy of combustion of $C_2H_{2(g)}$,$C_{(s)}$,and $H_{2(g)}$ are $-1300$,$-394$,and $-286 \ kJ/mol$ respectively. The standard enthalpy of formation of $C_2H_{2(g)}$ is $:$