Question

$\Delta H_f^{\circ }$ of water is $-285.8{\text{kJ mol}}^{-1}$. If the enthalpy of neutralisation of monoacidic strong base is $-57.3{\text{kJ mol}}^{-1}$, $\Delta H_f^{\circ }$ of $O{H}^{-}$ ion will be:

• A. $-114.25{\text{kJmol}}^{-1}$
• B. $114.25{\text{kJmol}}^{-1}$
• C. $228.5{\text{kJmol}}^{-1}$
• D. $-228.5{\text{kJmol}}^{-1}$

Answer 1

The correct option is D $-228.5{\text{kJmol}}^{-1}$

$H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)\therefore \Delta H_f^{\circ }=-285.8{\text{kJ mol}}^{-1}\dots \dots \left(1\right)$
$H^{+}(aq.)+O{H}^{-}(aq.)\to H_{2}O\left(l\right);\Delta H_{\text{neut}}^{\circ }=-57.3{\text{kJ mol}}^{-1}\dots \dots \left(2\right)$
$\frac{1}{2}{H}_2\left(g\right)\to H^{+}(aq.)+e^{-};\Delta H_f^{\circ }=0\text{(by convention)}\dots \dots \left(3\right)$

On reversing equation (2) and (3) and then adding (1), (2) , (3) gives

$\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)+e^{-}\to O{H}^{-}(aq.)$
$\Delta H_f^{\circ }$ of $O{H}^{-}=-285.8+57.3$
$=-228.5{\text{kJ mol}}^{-1}$