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  • The enthalpy of neutralization of a strong acid by a strong base is <img alt="\mathrm{-57.32 \, \, \mathrm{kJ\, \, eqv}^{-1}. }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B-57

The enthalpy of neutralization of a strong acid by a strong base is \mathrm{-57.32 \, \, \mathrm{kJ\, \, eqv}^{-1}. } The enthalpy of formation of water is \mathrm{-285.84\, \, \mathrm{kj\, \, eqv}^{-1}} The enthalpy of formation of hydroxyl ion is -

Option: 1

\mathrm{+228.5 \mathrm{~kJ} / \mathrm{mol}}


Option: 2

\mathrm{-114.26 \mathrm{~kJ} / \mathrm{mol}}


Option: 3

\mathrm{+114.26 \mathrm{~kJ} / \mathrm{mol}}


Option: 4

\mathrm{-228.52 \mathrm{~kg} / \mathrm{mm}}


Answers (1)

best_answer

The process of neutralization is

\mathrm{ \begin{aligned} & \mathrm{H}_{(99)}^{+}+\mathrm{OH}_{(\text {(aq) }}^{-} \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \quad \therefore \Delta H^{\circ}=257.32 \\\\ & \therefore \Delta H_r^0=\frac{\Delta H_f^0}{H_2 O(l)}-\left[\frac{\Delta H_f^0}{H_{(a q)}^{+}}+\frac{\Delta H_f^0}{O H_{(a q)}{ }^0}\right] \\ & \end{aligned} }

 

\mathrm{\begin{aligned} \Rightarrow-57.32 & =-285.84-(0+x) \\\\ \Rightarrow x & =-285.84+57.32 \\\\ & =-228.52 \mathrm{k}) / \mathrm{mol} \text { (option d) } \end{aligned}}

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Pankaj Sanodiya

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