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Chapter 15: Problem 90

In 1899 the German chemist Ludwig Mond developed a process for purifying nickel by converting it to the volatile nickel tetracarbonyl \(\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]\) (b.p. \(=\) \(\left.42.2^{\circ} \mathrm{C}\right)\) $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ (a) Describe how you can separate nickel and its solid impurities. (b) How would you recover nickel? \(\left[\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(\left.-602.9 \mathrm{kj} / \mathrm{mol} .\right]\)

Short Answer

Expert verified
In the Mond process, Nickel reacts with Carbon monoxide to form nickel tetracarbonyl. The nickel tetracarbonyl is volatile, so it separates from the solid impurities when heated. To recover the nickel, the nickel tetracarbonyl is heated, causing it to decompose back into nickel and carbon monoxide.

Step by step solution

01

Understand Mond Process

The Mond process is a method for purifying nickel. In this process, Nickel reacts with Carbon monoxide to form nickel tetracarbonyl. The reaction is a reversible reaction, which means it can also be reversed under certain conditions. The nickel tetracarbonyl is a volatile compound, which means it easily evaporates and can be separated from other solid impurities.
02

Perform Separation

In the process, the impure nickel reacts with the carbon monoxide gas to form nickel tetracarbonyl. The mixture is heated and nickel tetracarbonyl gas forms and separates from the solid impurities because it is volatile. This forms the basis of the separation technique.
03

Recovery of Nickel

To recover nickel, the temperature is increased, nickel tetracarbonyl decomposes into nickel and carbon monoxide. The nickel obtained in this step is the pure nickel. The reaction for recovery is: \[ \mathrm{Ni}(\mathrm{CO})_{4}(g) \rightarrow \mathrm{Ni}(s) + 4 \mathrm{CO}(g) \] This phenomenon is facilitated by the fact that formation of nickel tetracarbonyl is exothermic, thus the reaction is reversed by heating.

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Most popular questions from this chapter

Consider the dissociation of iodine: $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ A \(1.00-\mathrm{g}\) sample of \(\mathrm{I}_{2}\) is heated at \(1200^{\circ} \mathrm{C}\) in a 500 -mL flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in problem \(15.65(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ is found to be \(2 \times 10^{-42}\) at \(25^{\circ} \mathrm{C}\). (a) What is \(K_{\mathrm{c}}\) for the reaction at the same temperature? (b) The very small value of \(K_{P}\left(\right.\) and \(\left.K_{\mathrm{c}}\right)\) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

At \(1024^{\circ} \mathrm{C}\), the pressure of oxygen gas from the decomposition of copper(II) oxide \((\mathrm{CuO})\) is \(0.49 \mathrm{~atm}\) $$ 4 \mathrm{CuO}(s) \rightleftharpoons 2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{O}_{2}(g) $$ (a) What is the \(K_{P}\) for the reaction? (b) Calculate the fraction of \(\mathrm{CuO}\) decomposed if 0.16 mole of it is placed in a 2.0 - \(L\) flask at \(1024^{\circ} \mathrm{C}\). (c) What would be the fraction if a 1.0 -mole sample of \(\mathrm{CuO}\) were used? (d) What is the smallest amount of \(\mathrm{CuO}\) (in moles) that would establish the equilibrium?

Consider this reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ If the equilibrium partial pressures of \(\mathrm{N}_{2}, \mathrm{O}_{2}\), and NO are 0.15 atm, 0.33 atm, and 0.050 atm, respectively, at \(2200^{\circ} \mathrm{C}\), what is \(K_{P} ?\)

Heating solid sodium bicarbonate in a closed vessel established this equilibrium: \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}_{2}(g)\) What would happen to the equilibrium position if (a) some of the \(\mathrm{CO}_{2}\) were removed from the system, (b) some solid \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) were added to the system, (c) some of the solid \(\mathrm{NaHCO}_{3}\) were removed from the system? The temperature remains constant.
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