Problem 87 - Chemistry: The Central Science in SI Units, Global Edition

Question

\(\mathrm{NiO}\) 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?

Step-by-Step Solution

Answer

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 1: 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.

Step 2: 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: $$P_{CO} = 20 \mathrm{kPa}$$$$P_{CO_2} = 0 \mathrm{kPa}$$

Step 3: 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:$$Q = \frac{0 \mathrm{kPa}}{20 \mathrm{kPa}} = 0$$

Step 4: 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.

Step 5: 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.

Key Concepts

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.