Enthalpy of sublimation of a substance is equal to

$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)}$
$B.E. (H-H) = x_1$; $B.E. (O=O) = x_2$;
$B.E. (O-H) = x_3$
Heat of vaporisation of water $= x_4$,then $\Delta H_f$ [heat of formation of liquid water] is:

Under similar conditions,the enthalpy of freezing is exactly opposite to:

Using the data provided,calculate the bond energy $(kJ \ mol^{-1})$ of a $C \equiv C$ bond in $C_{2}H_{2}$. (Take the bond energy of a $C-H$ bond as $350 \ kJ \ mol^{-1}$)
$2C_{(s)} + H_{2(g)} \longrightarrow C_{2}H_{2(g)} \quad \Delta H = 225 \ kJ \ mol^{-1}$
$2C_{(s)} \longrightarrow 2C_{(g)} \quad \Delta H = 1410 \ kJ \ mol^{-1}$
$H_{2(g)} \longrightarrow 2H_{(g)} \quad \Delta H = 330 \ kJ \ mol^{-1}$

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......$kJ$.

$H^{+}_{(aq)} + OH^{-}_{(aq)} \rightarrow H_2O_{(l)} + 56 \text{ kJ/mol}$. The heat of neutralization for the complete neutralization of $1 \text{ mole}$ of $H_2SO_4$ will be ...... $\text{kJ}$.