/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 In 1899 the German chemist Ludwi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
Nickel can be separated from its solid impurities by heating it in a stream of carbon monoxide which forms Nickel tetracarbonyl, a gas that can be physically separated from the impurities. The nickel is then recovered from Nickel tetracarbonyl by decomposing it at a temperature of around 180°C producing pure nickel. The process is favored due to the exothermic nature of the reaction.

Step by step solution

01

Separating Nickel from its solid impurities

Nickel is first heated in a stream of carbon monoxide (CO), producing Nickel tetracarbonyl gas \(\mathrm{Ni}(\mathrm{CO})_{4}\), which is volatile and thus separates from the solid impurities.
02

Understand the reversible nature

The reaction \(\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)\) is reversible, meaning it can proceed in both directions. When cooled to room temperature Nickel tetracarbonyl decomposes back to nickel and carbon monoxide.
03

Recovering Nickel from Nickel tetracarbonyl

Once the Nickel tetracarbonyl gas \(\mathrm{Ni}(\mathrm{CO})_{4}\) is formed and it is separated from the solid impurities, it is then carefully decomposed at a temperature of around 180°C (higher than the boiling point of Nickel tetracarbonyl) to produce pure Nickel.
04

Understanding Enthalpy Change

The reaction is exothermic \(\left[\Delta H_{\mathrm{f}}^{\circ}\right.\) for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is $\left.-602.9 \mathrm{~kJ} / \mathrm{mol}.\right]$ meaning energy is released during the formation of Nickel tetracarbonyl. This indicates that the reaction will shift towards the right (since exothermic reactions are favored) producing more of Nickel tetracarbonyl.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nickel Tetracarbonyl
Nickel tetracarbonyl is a noteworthy compound in the field of inorganic chemistry, especially when discussing the Mond process for nickel purification. A complex of nickel with carbon monoxide, \( \mathrm{Ni}(\mathrm{CO})_{4} \), this compound is a volatile liquid at room temperature, with a boiling point of just 42.2°C. Its low boiling point is the key property utilized in the Mond process to separate nickel from its impurities. When nickel reacts with carbon monoxide under controlled conditions, a gaseous form of nickel tetracarbonyl is produced. By exploiting the different physical states, this process effectively isolates pure nickel from solid contaminants.

During the purification process, the formation of nickel tetracarbonyl is crucial, since it's a reversible step. Also interesting to note is that the formation of the compound is exothermic, releasing energy in form of heat, which is indicative of the stability of the product formed in this chemical reaction. Understanding the physical and chemical properties of nickel tetracarbonyl, such as its volatility and reaction kinetics, is central to mastering the concept of nickel purification through the Mond process.
Separation Techniques in Chemistry
The purification of chemical substances often relies on a plethora of separation techniques that leverage differences in physical properties—such as boiling point, volatility, or solubility—to isolate a target compound. In the context of the Mond process, the ability of nickel to form a volatile compound with carbon monoxide is exploited. This technique of separation based on volatility is also known as 'distillation.' When nickel tetracarbonyl is formed, it can be removed from the non-volatile impurities simply by allowing it to vaporize.

The beauty of this separation lies in the reversible nature of nickel tetracarbonyl's formation, allowing for the recovery of nickel after its separation. This example of the Mond process showcases just one of the many separation techniques in chemistry, underlying a fundamental concept: the careful selection of appropriate separation methods is essential for achieving highly pure substances in both laboratory and industrial settings.
Reversible Chemical Reactions
Reversible chemical reactions are characterized by their ability to proceed in both forward and reverse directions. The reaction involved in the Mond process where nickel reacts with carbon monoxide to form nickel tetracarbonyl is an exemplary reversible reaction, denoted by the symbol \( \rightleftharpoons \). This equilibrium allows for the initial separation of nickel from its impurities and subsequently facilitates its recovery.

At a high temperature, the reaction favors the formation of nickel tetracarbonyl, but as the temperature decreases, the equilibrium shifts in the reverse direction, causing the compound to decompose back to nickel and carbon monoxide. This reversible nature is a cornerstone of chemical reaction concepts, as it implies that the products of the reaction can be converted back into the reactants under certain conditions, making the process efficient and less wasteful.
Exothermic Reactions
Exothermic reactions are those that release energy, typically in the form of heat, upon proceeding. This is a central concept in thermodynamics and reaction kinetics. In the context of the Mond process, the formation of nickel tetracarbonyl is an exothermic process with an enthalpy change (\(\Delta H_{f}^{\circ}\)) of -602.9 kJ/mol. This indicates that when nickel and carbon monoxide react to form nickel tetracarbonyl, a substantial amount of energy is liberated.

The exothermic nature of this reaction can be linked to the discussion of reaction spontaneity and equilibrium. In general, exothermic reactions tend to be spontaneous because they result in a lower energy state, which is naturally favored. In the Mond process, the exothermic nature contributes to the forward reaction being favored at certain temperatures, thus enhancing the efficiency of the nickel purification process. Recognizing exothermic reactions is vital in understanding energy changes within chemical reactions, as well as in designing and controlling chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium constant \(\left(K_{P}\right)\) for the formation of the air pollutant nitric oxide (NO) in an automobile engine at \(530^{\circ} \mathrm{C}\) is \(2.9 \times 10^{-11}\) : $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ (a) Calculate the partial pressure of NO under these conditions if the partial pressures of nitrogen and oxygen are 3.0 atm and 0.012 atm, respectively. (b) Repeat the calculation for atmospheric conditions where the partial pressures of nitrogen and oxygen are 0.78 atm and 0.21 atm and the temperature is \(25^{\circ} \mathrm{C}\). (The \(K_{P}\) for the reaction is \(4.0 \times 10^{-31}\) at this temperature.) (c) Is the formation of NO endothermic or exothermic? (d) What natural phenomenon promotes the formation of NO? Why?

About 75 percent of hydrogen for industrial use is produced by the steam- reforming process. This process is carried out in two stages called primary and secondary reforming. In the primary stage, a mixture of steam and methane at about 30 atm is heated over a nickel catalyst at \(800^{\circ} \mathrm{C}\) to give hydrogen and carbon monoxide: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=260 \mathrm{~kJ} /\mathrm{mol}\end{array}$$ The secondary stage is carried out at about \(1000^{\circ} \mathrm{C}\), in the presence of air, to convert the remaining methane to hydrogen: $$\begin{array}{r}\mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \\\\\Delta H^{\circ}=35.7 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ (a) What conditions of temperature and pressure would favor the formation of products in both the primary and secondary stage? (b) The equilibrium constant \(K_{\mathrm{c}}\) for the primary stage is 18 at \(800^{\circ} \mathrm{C}\). (i) Calculate \(K_{P}\) for the reaction. (ii) If the partial pressures of methane and steam were both 15 atm at the start, what are the pressures of all the gases at equilibrium?

In this chapter we learned that a catalyst has no effect on the position of an equilibrium because it speeds up both the forward and reverse rates to the same extent. To test this statement, consider a situation in which an equilibrium of the type $$2 \mathrm{~A}(g) \rightleftharpoons \mathrm{B}(g)$$ is established inside a cylinder fitted with a weightless piston. The piston is attached by a string to the cover of a box containing a catalyst. When the piston moves upward (expanding against atmospheric pressure), the cover is lifted and the catalyst is exposed to the gases. When the piston moves downward, the box is closed. Assume that the catalyst speeds up the forward reaction \((2 \mathrm{~A} \longrightarrow \mathrm{B})\) but does not affect the reverse process \((\mathrm{B} \longrightarrow 2 \mathrm{~A}) .\) Suppose the catalyst is suddenly exposed to the equilibrium system as shown here. Describe what would happen subsequently. How does this "thought" experiment convince you that no such catalyst can exist?

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right) .\) The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$\mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right)$$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(\mathrm{I}_{2}\) with another \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4} .\) Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

Write the expressions for the equilibrium constants \(K_{P}\) of the following thermal decomposition reactions: (a) \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons\) $$\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (b) \(2 \mathrm{CaSO}_{4}(s) \rightleftharpoons\) $$2 \mathrm{CaO}(s)+2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.