The standard heat of formation,in kcal / mol | vedclass

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Question

(A) The reaction for the crystallisation of $BaSO_4$ is:
$Ba^{2+}{_{	ext{(aq)}}} + SO_4^{2-}{_{	ext{(aq)}}} ightarrow BaSO_{4	ext{(s)}}$
The en

Vedclass · Accepted Answer

$-128.5$

Vedclass · Answer

(A) The reaction for the crystallisation of $BaSO_4$ is:
$Ba^{2+}{_{	ext{(aq)}}} + SO_4^{2-}{_{	ext{(aq)}}} ightarrow BaSO_{4	ext{(s)}}$
The enthalpy of reaction (heat of crystallisation) is given by:
$\Delta H_{	ext{crys}} = \Delta H_f(BaSO_{4	ext{(s)}}) - [\Delta H_f(Ba^{2+}{_{	ext{(aq)}}}) + \Delta H_f(SO_4^{2-}{_{	ext{(aq)}}})]$
Substituting the given values:
$-4.5 = -349 - [\Delta H_f(Ba^{2+}_{(aq)}) + (-216)]$
$-4.5 = -349 - \Delta H_f(Ba^{2+}_{(aq)}) + 216$
$-4.5 = -133 - \Delta H_f(Ba^{2+}_{(aq)})$
$\Delta H_f(Ba^{2+}_{(aq)}) = -133 + 4.5$
$\Delta H_f(Ba^{2+}_{(aq)}) = -128.5 \ kcal / mol$

The standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]

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The standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is [Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$, Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]