Chapter 22
Chemistry and Chemical Reactivity · 54 exercises
Problem 61
Titanium is the seventh most abundant metal in the earth's crust. It is strong, lightweight, and resistant to corrosion; these properties lead to its use in aircraft engines. To obtain metallic titanium, ilmenite (FeTiOs), an ore of titanium, is first treated with sulfuric acid to form FesO, and \(\mathrm{Ti}\left(\mathrm{SO}_{4}\right)_{2}\). After separating these compounds, the latter substance is converted to \(\mathrm{TiO}_{2}\) in basic solution: \(\mathrm{FeTiO}_{3}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow\) $$ \mathrm{Ti}^{4+}(\mathrm{aq})+4 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{TiO}_{2}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq})+\mathrm{Ti}\left(\mathrm{SO}_{4}\right)_{2}(\mathrm{aq})+3 \mathrm{H}_{2} \mathrm{O}(\ell) $$ What volume of \(18.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is required to react completely with \(1.00 \mathrm{kg}\) of ilmenite? What mass of \(\mathrm{TiO}_{2} \mathrm{can}\) theoretically be produced by this sequence of reactions?
4 step solution
Problem 63
A \(A\) 0.213-g sample of uranyl(VI) nitrate, UO,(NO,), is dissolved in \(20.0 \mathrm{mL}\) of \(1.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) and shaken with Zn. The zinc reduces the uranyl ion, \(\mathrm{UO}_{2}^{2+}\), to a uranium ion, U". To determine the value of \(n,\) this solution is titrated with KMnO_. Permanganate is reduced to Mn \(^{2+}\) and \(\mathrm{U}^{n+}\) is oxidized back to \(\mathrm{UO}_{2}^{2+}\) (a) In the titration, \(12.47 \mathrm{mL}\) of \(0.0173 \mathrm{M} \mathrm{KMnO}_{4}\) was required to reach the equivalence point. Use this information to determine the charge on the ion \(\mathrm{U}^{n+}\) (b) With the identity of \(U^{n+}\) now established, write a balanced net ionic equation for the reduction of \(\mathrm{UO}_{2}^{2+}\) by zinc (assume acidic conditions). (c) Write a balanced net ionic equation for the oxidation of \(\mathrm{U}^{n+}\) to \(\mathrm{UO}_{2}^{2+}\) by \(\mathrm{MnO}_{4}^{-}\) in acid.
6 step solution
Problem 64
A The transition metals form a class of compounds called metal carbonyls, an example of which is the teurahedral complex \(\mathrm{Ni}(\mathrm{CO})_{4}\). Given the following thermodynamic data: $$\begin{array}{lcc} \hline & \Delta H_{f}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol}) \\ \hline \mathrm{Ni} & 0 & 29.87 \\ \mathrm{CO}(\mathrm{g}) & -110.525 & +197.674 \\ \mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{g}) & -602.9 & +410.6 \\ \hline \end{array}$$ (a) Calculate the equilibrium constant for the formation of \(\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{g})\) from nickel metal and CO gas. (b) Is the reaction of \(\mathrm{Ni}(\mathrm{s})\) and \(\mathrm{CO}(\mathrm{g})\) product-or reactant-favored? (c) Is the reaction more or less product-favored at higher temperatures? How could this reaction be used in the purification of nickel metal?
5 step solution
Problem 66
An this question, we explore the differences between metal coordination by monodentate and bidentate ligands. Formation constants, \(K_{f},\) for \(\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq})\) and \(\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq})\) are as follows: $$\begin{aligned} \mathrm{Ni}^{2+}(\mathrm{aq})+6 \mathrm{NH}_{3}(\mathrm{aq}) & \longrightarrow\left[\mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}(\mathrm{aq}) & & K_{f}=10^{8} \\ \mathrm{Ni}^{2+}(\mathrm{aq})+3 \mathrm{en}(\mathrm{aq}) & \longrightarrow\left[\mathrm{Ni}(\mathrm{en})_{3}\right]^{2+}(\mathrm{aq}) & & K_{f}=10^{18} \end{aligned}$$ The difference in \(K_{f}\) between these complexes indicates a higher thermodynamic stability for the chelated complex, caused by the chelate effect. Recall that \(K\) is related to the standard free energy of the reaction by \(\Delta G^{\circ}=-R T \ln K\) and \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) We know from experiment that \(\Delta H^{\circ}\) for the \(\mathrm{NH}_{3}\) reaction is \(-109 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta H^{\circ}\) for the ethylenediamine reaction is \(-117 \mathrm{kJ} / \mathrm{mol}\). Is the difference in \(\Delta H^{\circ}\) sufficient to account for the \(10^{10}\) difference in \(K_{f}\) ? Comment on the role of entropy in the second reaction.
5 step solution