If the ratio of bond dissociation energies of $XY$,$X_2$,and $Y_2$ is $1 : 1 : 0.5$ and the enthalpy of formation of $XY$ is $\Delta_f H = -200 \ kJ \ mol^{-1}$,then the bond dissociation energy of $X_2$ in $kJ \ mol^{-1}$ is:

The heat of solution of $Na_2SO_{4(s)}$ and $Na_2SO_4 \cdot 10H_2O_{(s)}$ are $-2.34 \ kJ \ mol^{-1}$ and $78.87 \ kJ \ mol^{-1}$ respectively. The heat of hydration of $Na_2SO_{4(s)}$ is:

The standard enthalpies of combustion of $C_6H_{6(l)}$,$C(graphite)$ and $H_{2(g)}$ are respectively $-3270 \ kJ \ mol^{-1}$,$-394 \ kJ \ mol^{-1}$ and $-286 \ kJ \ mol^{-1}$. What is the standard enthalpy of formation of $C_6H_{6(l)}$ in $kJ \ mol^{-1}$?

The standard enthalpy of formation of $H_2O_{(g)}$ at $298 \ K$ is $-241.82 \ kJ \ mol^{-1}$. Calculate the enthalpy of formation of $H_2O_{(g)}$ at $373 \ K$,assuming $C_p$ is independent of temperature.
$C_p$ of $H_2O_{(g)} = 33.58 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $H_{2(g)} = 28.84 \ J \ K^{-1} \ mol^{-1}$
$C_p$ of $O_{2(g)} = 29.37 \ J \ K^{-1} \ mol^{-1}$

Explain the change in enthalpy related to a chemical reaction.

The heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be: