Find the enthalpy of formation of the OH^- io | vedclass

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(B) The enthalpy of formation of $H_2O_{(l)}$ is given as $\Delta H_f(H_2O, l) = -286.20 \ KJ/mol$.For the reaction: $H_2O_{(l)} 	o H^+_{(aq)} + OH^-

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$-228.88$

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(B) The enthalpy of formation of $H_2O_{(l)}$ is given as $\Delta H_f(H_2O, l) = -286.20 \ KJ/mol$.For the reaction: $H_2O_{(l)} 	o H^+_{(aq)} + OH^-_{(aq)}$,the enthalpy change is $\Delta H_r = 57.32 \ KJ/mol$.The enthalpy of reaction is calculated as:$\Delta H_r = [\Delta H_f(H^+, aq) + \Delta H_f(OH^-, aq)] - \Delta H_f(H_2O, l)$By convention,the enthalpy of formation of $H^+_{(aq)}$ is $0 \ KJ/mol$.Substituting the values:$57.32 = [0 + \Delta H_f(OH^-, aq)] - (-286.20)$$57.32 = \Delta H_f(OH^-, aq) + 286.20$$\Delta H_f(OH^-, aq) = 57.32 - 286.20 = -228.88 \ KJ/mol$.

Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:
$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$

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Find the enthalpy of formation of the $OH^-$ ion in $KJ$ at $25^\circ C$ from the following data:$H_2O_{(l)} \to H^+_{(aq)} + OH^-_{(aq)} ; \Delta H = 57.32 \ KJ$$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)} ; \Delta H = -286.20 \ KJ$