Question

The standard heat of formation, in kcal/mol, of $Ba^{2+}$ is:
Given: Standard heat of formation of SO₄²⁻(aq) = -216 kcal/mol, standard heat of crystallization of BaSO₄(s) = -4.5 kcal/mol, standard heat of formation of BaSO₄(s) = -349 kcal/mol.

• A. 133.0
• B. 133.0 + 133.0
• C. 220.5
• D. -128.5

Hint:

Always apply Hess's Law when you have multiple reactions and want to find the heat of formation of an intermediate species.

Answer 1

The Correct Option is D

We are given the following information:

\(\Delta H_f^{\circ} (\text{SO}_4^{2-}(aq)) = -216 \text{ kcal/mol}\)

\(\Delta H_{\text{crystallization}} (\text{BaSO}_4(s)) = -4.5 \text{ kcal/mol}\)

\(\Delta H_f^{\circ} (\text{BaSO}_4(s)) = -349 \text{ kcal/mol}\)

We need to find \(\Delta H_f^{\circ} (\text{Ba}^{2+}(aq))\).

The formation reaction for BaSO₄(s) is:

Ba²⁺(aq) + SO₄^2-(aq) \(\rightarrow\) BaSO₄(s)

Using Hess's Law, we can write:

\(\Delta H_f^{\circ} (\text{BaSO}_4(s)) = \Delta H_f^{\circ} (\text{Ba}^{2+}(aq)) + \Delta H_f^{\circ} (\text{SO}_4^{2-}(aq)) + \Delta H_{\text{crystallization}}(\text{BaSO}_4(s))\)

Rearranging to solve for \(\Delta H_f^{\circ} (\text{Ba}^{2+}(aq))\):

\(\Delta H_f^{\circ} (\text{Ba}^{2+}(aq)) = \Delta H_f^{\circ} (\text{BaSO}_4(s)) - \Delta H_f^{\circ} (\text{SO}_4^{2-}(aq)) - \Delta H_{\text{crystallization}}(\text{BaSO}_4(s))\)

Substituting the given values:

\(\Delta H_f^{\circ} (\text{Ba}^{2+}(aq)) = -349 - (-216) - (-4.5) = -349 + 216 + 4.5 = -128.5 \text{ kcal/mol}\)

Therefore, the standard heat of formation of Ba²⁺ is -128.5 kcal/mol.

The correct answer is (4).