The enthalpy of formation of H_2O is -68 k.c | vedclass

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Question

(A) The neutralization reaction is: $H^+_{(aq)} + OH^-_{(aq)} \rightleftharpoons H_2O_{(l)}$The enthalpy change for this reaction is $\Delta H = -13.7

Vedclass · Accepted Answer

$-54.3 \ k.cal/mol$

Vedclass · Answer

(A) The neutralization reaction is: $H^+_{(aq)} + OH^-_{(aq)} \rightleftharpoons H_2O_{(l)}$The enthalpy change for this reaction is $\Delta H = -13.7 \ k.cal/mol$.Using the formula: $\Delta H = \Delta_f H^{\ominus}(H_2O) - [\Delta_f H^{\ominus}(H^+) + \Delta_f H^{\ominus}(OH^-)]$Given $\Delta_f H^{\ominus}(H^+) = 0 \ k.cal/mol$ (by convention) and $\Delta_f H^{\ominus}(H_2O) = -68 \ k.cal/mol$.Substituting the values: $-13.7 = -68 - [0 + \Delta_f H^{\ominus}(OH^-)]$$\Delta_f H^{\ominus}(OH^-) = -68 + 13.7 = -54.3 \ k.cal/mol$.

The enthalpy of formation of $H_2O$ is $-68 \ k.cal/mol$. Calculate the enthalpy of formation of $OH^-$. Given that the enthalpy of neutralization of $H^+$ and $OH^-$ is $-13.7 \ k.cal/mol$.

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