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\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{~K}\), the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2} .\) If a CO pressure of \(20 \mathrm{kPa}\) is to be employed in the furnace and total pressure never exceeds \(101.3 \mathrm{kPa}\), will reduction occur?

Short Answer

Expert verified
Given that the equilibrium constant \(K_p = 6.0 \times 10^2\), the initial CO pressure is 20 kPa, and the initial CO2 pressure is 0 kPa, we can calculate the initial reaction quotient as \(Q = \frac{P_{CO_2}}{P_{CO}} = \frac{0}{20} = 0\). Since \(Q = 0 < K_p = 6.0 \times 10^2\), the reaction will proceed in the forward direction, leading to the reduction of NiO to Ni. Thus, the reduction will occur under the given conditions.

Step by step solution

01

Write the expression for the reaction quotient

The reaction quotient, Q, can be calculated by dividing the partial pressure of products by the partial pressure of reactants:$$Q = \frac{P_{CO_2}}{P_{CO}}$$ where \(P_{CO_2}\) is the partial pressure of CO2 and \(P_{CO}\) is the partial pressure of CO.
02

Determine the initial partial pressures of CO and CO2

We are given that the CO pressure to be employed in the furnace is 20 kPa, and the total pressure never exceeds 101.3 kPa. Since no partial pressure is given for CO2, we can assume that the initial partial pressure of CO2 is zero. Therefore, we have: $$P_{CO} = 20 \mathrm{kPa}$$$$P_{CO_2} = 0 \mathrm{kPa}$$
03

Calculate the initial reaction quotient, Q

Now, plug in the initial partial pressures of CO and CO2 that we found in step 2 into Q equation:$$Q = \frac{0 \mathrm{kPa}}{20 \mathrm{kPa}} = 0$$
04

Compare Q to Kp

We have calculated the Q value as 0, and the equilibrium constant Kp as \(6.0 \times 10^2\). Since $$Q = 0 < K_p = 6.0 \times 10^2$$ this means that the reaction will proceed forward, towards the products.
05

Determine if reduction occurs

Since Q < Kp, the reaction will proceed in the forward direction, which is the reduction of NiO to Ni. Therefore, the reduction of NiO will occur under the given conditions.

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Key Concepts

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

Reaction Quotient
The concept of a reaction quotient, denoted as \( Q \), is vital in understanding how a chemical reaction proceeds. It's a mathematical expression that helps us determine the direction in which a reaction is likely to go at a given set of conditions. Specifically, it compares the concentrations or pressures of the products and reactants at any point in time.

In our example with \( \mathrm{NiO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s) + \mathrm{CO_2}(g) \), we use the partial pressures of the gases involved to find \( Q \).
  • The reaction quotient formula is \( Q = \frac{P_{\text{CO}_2}}{P_{\text{CO}}} \), where \( P_{\text{CO}_2} \) is the partial pressure of carbon dioxide and \( P_{\text{CO}} \) is the partial pressure of carbon monoxide.
  • If \( Q < K_p \), the reaction will proceed towards the products to reach equilibrium.
  • If \( Q = K_p \), the system is at equilibrium, meaning no net movement occurs.
  • If \( Q > K_p \), the reaction will shift towards the reactants.
In this case, since initial \( Q = 0 \) and \( K_p = 600 \), the reaction proceeds towards producing more \( \mathrm{Ni} \) and \( \mathrm{CO}_2 \).
Equilibrium Constant
The equilibrium constant, \( K_p \), is a critical concept when studying chemical equilibrium. It quantifies the position of equilibrium in terms of partial pressures (hence the "p" in \( K_p \)) for gaseous reactions.

For the reaction of \( \mathrm{NiO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s) + \mathrm{CO_2}(g) \), the equilibrium constant \( K_p \) at 1600 K is 600. The immense value of \( K_p \) suggests that, under equilibrium conditions, the concentration of the products is much greater than that of the reactants.
  • This large \( K_p \) value indicates a reaction highly product-favored at equilibrium.
  • In industrial processes, like the reduction of nickel, this means efficient conversion of reactants to products, which is often desired.
  • In our scenario, starting with a very low \( Q \), the large \( K_p \) further emphasizes that the reaction will move toward producing \( \mathrm{Ni} \) and \( \mathrm{CO}_2 \) for that given temperature and pressure.
Understanding \( K_p \) is invaluable for predicting how a reaction behaves or in designing optimal conditions for industrial reactions like reducing \( \mathrm{NiO} \) to \( \mathrm{Ni} \).
Reduction Reaction
A reduction reaction involves the gain of electrons by a substance, often accompanied by a decrease in oxidation state. In the context of industrial chemistry, such as reducing nickel oxide (\( \mathrm{NiO} \)) to nickel (\( \mathrm{Ni} \)), reduction reactions play a vital role.

Here, nickel oxide is reduced by carbon monoxide:
\( \mathrm{NiO}(s) + \mathrm{CO}(g) \rightarrow \mathrm{Ni}(s) + \mathrm{CO}_2(g) \)
  • \( \mathrm{NiO} \) gains electrons, becoming \( \mathrm{Ni} \), a process indicative of reduction.
  • Simultaneously, \( \mathrm{CO} \) is oxidized to \( \mathrm{CO}_2 \), making this a redox (oxidation-reduction) reaction.
  • In reducing metal oxides, carbon monoxide is favored as it effectively acts as a reducing agent, donating electrons to the metal oxide.
Reduction reactions like this are essential for obtaining pure metals from their ores and are widely utilized in metallurgical industries. Understanding these processes helps in creating more efficient and environmentally favorable industrial procedures.

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Most popular questions from this chapter

The following equilibria were measured at \(823 \mathrm{~K}\) : $$\begin{array}{l} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{array}$$ (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at \(823 \mathrm{~K}\). (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{~K} ?(\mathbf{c})\) If you were to place \(5.00 \mathrm{~g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of \(250 \mathrm{~mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of \(101.3 \mathrm{kPa}\) and a temperature of \(298 \mathrm{~K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to \(823 \mathrm{~K}\) and allowed to come to equilibrium, how much \(\operatorname{CoO}(s)\) remains?

As shown in Table \(15.2,\) the equilibrium constant for the reaction \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is \(K_{p}=4.23 \times 10^{-7}\) at \(300^{\circ} \mathrm{C}\). Pure \(\mathrm{NH}_{3}\) is placed in a 1.00-L flask and allowed to reach equilibrium at this temperature. There are \(1.05 \mathrm{~g} \mathrm{NH}_{3}\) in the equilibrium mixture. (a) What are the masses of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) in the equilibrium mixture? (b) What was the initial mass of ammonia placed in the vessel? (c) What is the total pressure in the vessel?

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) at \(1300 \mathrm{~K} K_{c}=0.57\) (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K} K_{p}=0.0572\)

Bromine and hydrogen react in the gas phase to form hydrogen bromide: \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) .\) The reaction enthalpy is \(\Delta H^{\circ}=-6 \mathrm{~kJ} .\) (a) To increase the equilibrium yield of hydrogen bromide would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen bromide by controlling the pressure of this reaction? If so, would high or low pressure favor formation of \(\mathrm{HBr}(g) ?\)

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