The enthalpy change $(\Delta H)$ for the neutralisation of $1 \ M \ HCl$ by caustic potash in dilute solution at $298 \ K$ is ..... $kJ$.

According to Hess's Law,the enthalpy change of a reaction depends on which of the following?

Calculate the enthalpy change in $kJ$ for the reaction: $2C_{(graphite)} + 2H_{2(g)} \to C_2H_{4(g)}$
$C_{(graphite)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = -393.5 \ kJ$
$C_2H_{4(g)} + 3O_{2(g)} \to 2CO_{2(g)} + 2H_2O_{(l)} \quad \Delta H = -1410.9 \ kJ$
$H_{2(g)} + 1/2O_{2(g)} \to H_2O_{(l)} \quad \Delta H = -285.8 \ kJ$

For the reaction $2 H_2 + O_2 \rightarrow 2 H_2 O$,$\Delta H = -571 \ kJ$. Bond energy of $H-H = 435 \ kJ$ and $O=O = 498 \ kJ$. Then the average bond energy of $O-H$ bond will be:

If $H^{+} + OH^{-} \to H_2O + 13.7 \ kcal$,then the heat of neutralization for complete neutralization of one mole of $H_2SO_4$ by base will be.......$kcal$

If heat of neutralization is $-13.7 \, KCal$ at $25 \, ^oC$ and $\Delta H_f^o (H_2O) = -68 \, KCal$,then the standard enthalpy of formation of $OH^{-}$ will be.....$KCal$. (in $.3$)