Chapter 17: Problem 22
The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?
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Chapter 17: Problem 22
The deciding factor on why HF is a weak acid and not a strong acid like the other hydrogen halides is entropy. What occurs when HF dissociates in water as compared to the other hydrogen halides?
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Acrylonitrile is the starting material used in the manufacture of acrylic fibers (U.S. annual production capacity is more than two million pounds). Three industrial processes for the production of acrylonitrile are given below. Using data from Appendix 4, calculate \(\Delta S^{\circ}, \Delta H^{p},\) and \(\Delta G^{\circ}\) for each process. For part a, assume that \(T=25^{\circ} \mathrm{C} ;\) for part \(\mathrm{b}, T=70 .^{\circ} \mathrm{C} ;\) and for part \(\mathrm{c}, T=700 .^{\circ} \mathrm{C}\) Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.
Consider the reactions $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ $$\mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text { reaction } 2}>K_{\text { reaction }}\) . Explain.
Consider the system $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\) a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol},\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C} .\) c. Show by calculations that \(\Delta G=0\) at equilibrium.
The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and \(\Delta S_{\mathrm{univ}}\) for the following process: \(\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)\)
If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is 79.14 \(\mathrm{kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\) . Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C},\) calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .^{\circ} \mathrm{C}\) (Hint: Manipulate the equation in Exercise 85.)
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