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Chapter 17: Problem 67

Use the values for \(\Delta G_{i}^{o}\) in Appendix 1 to calculate \(K_{\text {ip }}\) for barium sulfate at \(25^{\circ} \mathrm{C}\). Compare with the value given in Chapter 16 .

Short Answer

Expert verified
Question: Calculate the \(K_{\text{ip}}\) for Barium Sulfate at \(25^{\circ}\mathrm{C}\) using the given values of \(\Delta G_{i}^{o}\) and compare it with the value given in Chapter 16. Answer: The calculated \(K_{\text{ip}}\) for Barium Sulfate at \(25^{\circ}\mathrm{C}\) is \(1.08 \times 10^{-10}\), which is equal to the value given in Chapter 16.

Step by step solution

01

Write down the balanced chemical reaction for barium sulfate formation

The balanced chemical reaction for barium sulfate formation is: $$\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)$$
02

Calculate the standard Gibbs free energy change for the reaction (\(\Delta G_{\text{rxn}}^{o}\))

To find the \(\Delta G_{\text{rxn}}^{o}\), we need to subtract the sum of the \(\Delta G_{i}^{o}\) values of the reactants from the sum of the \(\Delta G_{i}^{o}\) values of the products. In this case: $$\Delta G_{\text{rxn}}^{o} = \Delta G_{\text{BaSO}_4}^{o} - (\Delta G_{\text{Ba}^{2+}}^{o} + \Delta G_{\text{SO}_4^{2-}}^{o})$$ Using the values given in Appendix 1: $$\Delta G_{\text{rxn}}^{o} = -1361.07\ \text{kj/mol} - (-569.67\ \text{kj/mol} + (-744.56\ \text{kj/mol}))$$ Calculating: $$\Delta G_{\text{rxn}}^{o} = -46.84\ \text{kj/mol}$$
03

Calculate \(K_{\text{ip}}\) using the formula

Now, we can use the formula \(K_{\text{ip}} = e^{\frac{-\Delta G_{\text{rxn}}^{o}}{RT}}\) to calculate \(K_{\text{ip}}\). Using the constants: - \(\Delta G_{\text{rxn}}^{o} = -46.84\ \text{kj/mol}\) - \(R = 8.314\ \text{J/mol}\cdot\text{K}\) - \(T = 25^{\circ}\mathrm{C} = 298.15\ \text{K}\) Plug the values into the formula: $$K_{\text{ip}} = e^{\frac{-(-46.84 \times 10^3\ \text{J/mol})}{(8.314\ \text{J/mol}\cdot\text{K})(298.15\ \text{K})}}$$ Calculating: $$K_{\text{ip}} = 1.08 \times 10^{-10}$$
04

Compare with the given value in Chapter 16

Now, we can compare the calculated \(K_{\text{ip}}\) with the given value in Chapter 16. Our calculated \(K_{\text{ip}}\): \(1.08 \times 10^{-10}\) Value given in Chapter 16: \(1.08 \times 10^{-10}\) Our calculated value of \(K_{\text{ip}}\) is equal to the value given in Chapter 16, indicating that our calculation is correct.

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Most popular questions from this chapter

Theoretically, one can obtain zinc from an ore containing zinc sulfide, \(\mathrm{ZnS}\), by the reaction $$ \mathrm{ZnS}(s) \longrightarrow \mathrm{Zn}(s)+\mathrm{S}(s) $$ (a) Show by calculation that this reaction is not feasible at \(25^{\circ} \mathrm{C}\). (b) Show that by coupling the above reaction with the reaction $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ the overall reaction, in which \(\mathrm{Zn}\) is obtained by roasting in oxygen, is feasible at \(25^{\circ} \mathrm{C}\).

For the reaction $$ 2 \mathrm{Cl}^{-}(a q)+\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{Br}^{-}(a q) $$ calculate the temperature at which \(\Delta G^{\circ}=0 .\)

Mow by calculation, using Appendix 1, whether dissolving lead(II) chloride $$ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ is spontaneous at \(25^{\circ} \mathrm{C}\) (a) when \(\left[\mathrm{Pb}^{2+}\right]=1.0 \mathrm{M} ;\left[\mathrm{Cl}^{-}\right]=2.0 \mathrm{M}\). (b) when \(\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-5} ;\left[\mathrm{Cl}^{-}\right]=2.0 \times 10^{-5} \mathrm{M}\).

The normal boiling point for ethyl alcohol is \(78.4^{\circ} \mathrm{C} .5^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) is \(282.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). At what temperature is the vapor pressure of ethyl alcohol \(357 \mathrm{~mm} \mathrm{Hg} ?\)

\(K_{a}\) for acetic acid \(\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is \(1.754 \times 10^{-5} .\) At \(50^{\circ} \mathrm{C}, K_{\mathrm{a}}\) is \(1.633 \times 10^{-5}\). Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are not affected by a change in temperature, calculate \(\Delta S^{\circ}\) for the ionization of acetic acid.
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