The heat of neutralization for the reaction $NaOH + HCl \to NaCl + H_2O$ is $57.1 \ kJ \ mol^{-1}$. What will be the heat released when $0.25 \ mol$ of $NaOH$ is titrated against $0.25 \ mol$ of $HCl$ (in $kJ$)?

With the help of the following data,find out the change in heat content for the reaction in $kJ$:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

If the heats of combustion of $C$,$S$,and $CS_2$ are $x$,$y$,and $z \, kJ \, mol^{-1}$ respectively,what will be the heat of formation of $CS_2$?

Given: $C + O_2 \rightarrow CO_2$ : $\Delta H = -395 \ kJ$,$S + O_2 \rightarrow SO_2$ : $\Delta H = -295 \ kJ$,$CS_2 + 3O_2 \rightarrow CO_2 + 2SO_2$ : $\Delta H = -1110 \ kJ$. Calculate the heat of formation of $CS_2$ in $kJ/mol$.

The heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. The heat of combustion at constant pressure is $(-321.30 - x R) \ kJ$. The value of $x$ is . . . . . . .

If $S + O_2 \to SO_2; (\Delta H = -298.2 \ kJ)$,$SO_2 + \frac{1}{2} O_2 \to SO_3; (\Delta H = -98.2 \ kJ)$,$SO_3 + H_2O \to H_2SO_4; (\Delta H = -130.2 \ kJ)$,$H_2 + \frac{1}{2} O_2 \to H_2O; (\Delta H = -287.3 \ kJ)$,then the enthalpy of formation of $H_2SO_4$ at $298 \ K$ will be......$kJ$.