Calculate the enthalpy of solution of potassium chloride $(KCl)$ if its lattice enthalpy $\Delta_{L} H = 700 \ kJ \ mol^{-1}$ and hydration enthalpy $\Delta_{hyd} H = -680 \ kJ \ mol^{-1}$.

The heat evolved in the combustion of benzene is given by the equation $C_6H_{6(l)} + 7.5 O_{2(g)} \to 3H_2O_{(l)} + 6CO_{2(g)}$,$\Delta H = -781.0 \ kcal \ mol^{-1}$. Which of the following quantities of heat energy will be evolved when $39 \ g$ of benzene is burnt in an open container?

The enthalpy change for the reaction,$H_{2(g)} + C_2H_{4(g)} \to C_2H_{6(g)}$ is $......$ $kcal \ mol^{-1}$. The bond energies are,$[e_{H-H} = 103, e_{C-H} = 99, e_{C-C} = 80]$ and $[e_{C=C} = 145] \ kcal \ mol^{-1}$.

The heat of combustion of $C_xH_y$,carbon,and hydrogen are $a, b$,and $c \ cal/mole$ respectively. The heat of formation of $C_xH_y$ will be:

The bond enthalpies of $H_2$,$X_2$,and $HX$ are in the ratio of $2 : 1 : 2$. If the standard enthalpy of formation of $HX$ is $-50 \ kJ \ mol^{-1}$,the bond enthalpy of $H_2$ is .... $kJ \ mol^{-1}$.

The heat of neutralization of a strong acid with a strong base is constant and is equal to: