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(A) Given,for reaction:$(I) \quad H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \quad \Delta H_r = 57.32 \, kJ$$(II) \quad H_{2(g)} + \frac{1

Vedclass · Accepted Answer

$-228.88$

Vedclass · Answer

(A) Given,for reaction:$(I) \quad H_2O_{(\ell)} \rightarrow H^{+}_{(aq)} + OH^{-}_{(aq)} \quad \Delta H_r = 57.32 \, kJ$$(II) \quad H_{2(g)} + \frac{1}{2} O_{2(g)} \rightarrow H_2O_{(\ell)} \quad \Delta H_r = -286.20 \, kJ$For reaction $(I)$:$\Delta H_r = \Delta H_f^o(H^{+}, aq) + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell)$Since $\Delta H_f^o(H^{+}, aq) = 0$,we have:$57.32 = 0 + \Delta H_f^o(OH^{-}, aq) - \Delta H_f^o(H_2O, \ell) \quad \dots (III)$For reaction $(II)$:$\Delta H_f^o(H_2O, \ell) = -286.20 \, kJ$Substituting this value in equation $(III)$:$57.32 = \Delta H_f^o(OH^{-}, aq) - (-286.20)$$\Delta H_f^o(OH^{-}, aq) = 57.32 - 286.20$$\Delta H_f^o(OH^{-}, aq) = -228.88 \, kJ$

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