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Isolation of Group \(8 \mathrm{~B}(10)\) elements, used as industrial catalysts, involves a series of steps. For nickel, the sulfide ore is roasted in air: \(\mathrm{Ni}_{3} \mathrm{~S}_{2}(s)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{NiO}(s)+\mathrm{SO}_{2}(g) .\) The metal oxide is reduced by the \(\mathrm{H}_{2}\) in water gas \(\left(\mathrm{CO}+\mathrm{H}_{2}\right)\) to impure \(\mathrm{Ni}: \mathrm{NiO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{H}_{2} \mathrm{O}(g) .\) The \(\mathrm{CO}\) in water gas then reacts with the metal in the Mond process to form gaseous nickel carbonyl, \(\mathrm{Ni}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g),\) which is sub- sequently decomposed to the metal. (a) Balance each of the three steps, and obtain an overall balanced equation for the conversion of \(\mathrm{Ni}_{3} \mathrm{~S}_{2}\) to \(\mathrm{Ni}(\mathrm{CO})_{4}\). (b) Show that the overall \(Q_{\mathrm{c}}\) is the product of the \(Q_{c}\) 's for the individual reactions.

Short Answer

Expert verified
The overall balanced equation is: \[ \text{Ni}_{3} \text{~S}_{2}(s) + 4 \text{O}_{2}(g) + 3 \text{H}_{2}(g) + 12 \text{CO}(g) \rightleftharpoons 3 \text{Ni}(\text{CO})_{4}(g) + 2 \text{SO}_{2}(g) + 3 \text{H}_{2} \text{O}(g) \]. The overall reaction quotient is the product of the individual quotients.

Step by step solution

01

Balance the First Reaction

The given reaction is: \[ \text{Ni}_{3} \text{~S}_{2}(s) + \text{O}_{2}(g) \rightleftharpoons \text{NiO}(s) + \text{SO}_{2}(g) \] To balance this reaction, determine the stoichiometric coefficients that balance the number of atoms of each element on both sides of the equation. The balanced reaction is: \[ \text{Ni}_{3} \text{~S}_{2}(s) + 4 \text{O}_{2}(g) \rightleftharpoons 3 \text{NiO}(s) + 2 \text{SO}_{2}(g) \]
02

Balance the Second Reaction

The given reaction is: \[ \text{NiO}(s) + \text{H}_{2}(g) \rightleftharpoons \text{Ni}(s) + \text{H}_{2} \text{O}(g) \] Balance the number of atoms of each element on both sides of the equation. The balanced reaction is: \[ \text{NiO}(s) + \text{H}_{2}(g) \rightleftharpoons \text{Ni}(s) + \text{H}_{2} \text{O}(g) \]
03

Balance the Third Reaction

The given reaction is: \[ \text{Ni}(s) + \text{CO}(g) \rightleftharpoons \text{Ni}(\text{CO})_{4}(g) \] Balance the number of atoms of each element on both sides of the equation. The balanced reaction is: \[ \text{Ni}(s) + 4 \text{CO}(g) \rightleftharpoons \text{Ni}(\text{CO})_{4}(g) \]
04

Combine the Balanced Reactions

To find the overall balanced equation, add the three balanced reactions together and cancel out the common terms on both sides.\[ \begin{align*} &\left(\text{Ni}_{3} \text{~S}_{2}(s) + 4 \text{O}_{2}(g) \rightleftharpoons 3 \text{NiO}(s) + 2 \text{SO}_{2}(g) \right) + \ &3 \left(\text{NiO}(s) + \text{H}_{2}(g) \rightleftharpoons \text{Ni}(s) + \text{H}_{2} \text{O}(g)\right) + \ &3 \left( \text{Ni}(s) + 4 \text{CO}(g) \rightleftharpoons \text{Ni}(\text{CO})_{4}(g) \right) \ &\Rightarrow \text{Ni}_{3} \text{~S}_{2}(s) + 4 \text{O}_{2}(g) + 3 \text{H}_{2}(g) + 12 \text{CO}(g) \rightleftharpoons 3 \text{Ni}(\text{CO})_{4}(g) + 2 \text{SO}_{2}(g) + 3 \text{H}_{2} \text{O}(g) \end{align*} \]
05

Determine the Overall Reaction Quotient (Qc)

The reaction coefficients for each reaction are as follows: \[ Q_{c_{1}} = \frac{[\text{NiO}]^3 [\text{SO}_{2}]^2}{[\text{O}_{2}]^4} \]\[ Q_{c_{2}} = \frac{[\text{Ni}]^1 [\text{H}_{2} \text{O}]^1}{[\text{NiO}]^1 [\text{H}_{2}]^1} \]\[ Q_{c_{3}} = \frac{[\text{Ni}(\text{CO})_{4}]}{[\text{Ni}] [\text{CO}]^4} \] The overall reaction quotient is the product of the individual reaction quotients:\[ Q_{c} = Q_{c_{1}} \times Q_{c_{2}} \times Q_{c_{3}} \]Substitute the values to get:\[ Q_{c} = \left(\frac{[\text{NiO}]^3 [\text{SO}_{2}]^2}{[\text{O}_{2}]^4} \right) \times \left( \frac{[\text{Ni}]^1 [\text{H}_{2} \text{O}]^1}{[\text{NiO}]^1 [\text{H}_{2}]^1} \right) \times \left(\frac{[\text{Ni}(\text{CO})_{4}]}{[\text{Ni}] [\text{CO}]^4} \ \right) \]

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes where reactants are transformed into products. In the context of nickel extraction, we can see three distinct chemical reactions. During these reactions, bonds are broken and formed, which involves energy changes. The first reaction involves nickel sulfide reacting with oxygen to form nickel oxide and sulfur dioxide. In the second reaction, nickel oxide reacts with hydrogen to form nickel and water. In the third reaction, nickel reacts with carbon monoxide to form nickel carbonyl. Each of these reactions has specific reactants and products, and understanding how they interact is crucial. They help us isolate and purify metals from ores effectively.
Balancing Equations
Balancing chemical equations is essential to ensure the law of conservation of mass is followed. This law states that matter cannot be created or destroyed in a chemical reaction. For nickel extraction, balancing equations involves ensuring the same number of each type of atom on both sides of the reaction. For example, in the first reaction, ensuring the number of nickel, sulfur, and oxygen atoms are the same on both sides. Here's how the first reaction is balanced:
  • Original reaction: \(\text{Ni}_{3} \text{~S}_{2}(s) + \text{O}_{2}(g) \rightleftharpoons \text{NiO}(s) + \text{SO}_{2}(g)\)
  • Balanced reaction: \(\text{Ni}_{3} \text{~S}_{2}(s) + 4 \text{O}_{2}(g) \rightleftharpoons 3 \text{NiO}(s) + 2 \text{SO}_{2}(g)\)
Balancing each reaction this way ensures the overall reaction is balanced. This is important for calculating quantities of reactants and products, and for understanding the efficiency of the extraction process.
Reaction Quotients
The reaction quotient, denoted as \(Q_{c}\), is a measure of the relative amounts of products and reactants present during a reaction at a particular point in time. It helps in predicting the direction of the reaction. For nickel extraction, we use the reaction quotients for each of the individual steps to find the overall \(Q_{c}\).
  • For the first reaction: \(Q_{c_{1}} = \frac{[\text{NiO}]^3 [\text{SO}_{2}]^2}{[\text{O}_{2}]^4}\)
  • For the second reaction: \(Q_{c_{2}} = \frac{[\text{Ni}]^1 [\text{H}_{2} \text{O}]^1}{[\text{NiO}]^1 [\text{H}_{2}]^1}\)
  • For the third reaction: \(Q_{c_{3}} = \frac{[\text{Ni}(\text{CO})_{4}]}{[\text{Ni}] [\text{CO}]^4}\)
The overall reaction quotient is then the product of these individual quotients: \(Q_{c} = Q_{c_{1}} \times Q_{c_{2}} \times Q_{c_{3}}\). This allows us to combine the effects of each step into a single measure, which is useful for understanding the overall system's behavior.

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Most popular questions from this chapter

The methane used to obtain \(\mathrm{H}_{2}\) for \(\mathrm{NH}_{3}\) manufacture is impure and usually contains other hydrocarbons, such as propane, \(\mathrm{C}_{3} \mathrm{H}_{8}\). Imagine the reaction of propane occurring in two steps: \(\mathrm{C}_{3} \mathrm{H}_{8}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 3 \mathrm{CO}(g)+7 \mathrm{H}_{2}(g)\) $$ \begin{array}{r} K_{\mathrm{p}}=8.175 \times 10^{15} \text { at } 1200 . \mathrm{K} \\ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \\ K_{\mathrm{p}}=0.6944 \text { at } 1200 . \mathrm{K} \end{array} $$ (a) Write the overall equation for the reaction of propane and steam to produce carbon dioxide and hydrogen. (b) Calculate \(K_{p}\) for the overall process at \(1200 .\) K. (c) When 1.00 volume of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and 4.00 volumes of \(\mathrm{H}_{2} \mathrm{O},\) each at \(1200 . \mathrm{K}\) and \(5.0 \mathrm{~atm},\) are mixed in a container, what is the final pressure? Assume the total volume remains constant, that the reaction is essentially complete, and that the gases behave ideally. (d) What percentage of the \(\mathrm{C}_{3} \mathrm{H}_{8}\) remains unreacted?

\(\mathrm{At} 425^{\circ} \mathrm{C}, K_{\mathrm{p}}=4.18 \times 10^{-9} \mathrm{for}\) the reaction $$ 2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g) \pm \mathrm{Br}_{2}(g) $$ In one experiment, 0.20 atm of \(\mathrm{HBr}(g), 0.010\) atm of \(\mathrm{H}_{2}(g),\) and 0.010 atm of \(\mathrm{Br}_{2}(g)\) are introduced into a container. Is the reaction at equilibrium? If not, in which direction will it proceed?

A toxicologist studying mustard gas, \(\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2},\) a blistering agent, prepares a mixture of \(0.675 M \mathrm{SCl}_{2}\) and \(0.973 \mathrm{M}\) \(\mathrm{C}_{2} \mathrm{H}_{4}\) and allows it to react at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) $$ \mathrm{SCl}_{2}(g)+2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(g) $$ At equilibrium, \(\left[\mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}\right]=0.350 \mathrm{M}\). Calculate \(K_{\mathrm{p}}\)

When a chemical company employs a new reaction to manufacture a product, the chemists consider its rate (kinetics) and yield (equilibrium). How does each of these affect the usefulness of a manufacturing process?

Ammonium carbamate \(\left(\mathrm{NH}_{2} \mathrm{COONH}_{4}\right)\) is a salt of carbamic acid that is found in the blood and urine of mammals. At \(250 .{ }^{\circ} \mathrm{C}, K_{\mathrm{c}}=1.58 \times 10^{-8}\) for the following equilibrium: $$ \mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) $$ If \(7.80 \mathrm{~g}\) of \(\mathrm{NH}_{2} \mathrm{COONH}_{4}\) is put into a \(0.500-\mathrm{L}\) evacuated container, what is the total pressure at equilibrium?

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