Problem 96 - Chemistry The Central Science In Si Units

Question

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 \(823 \mathrm{~K} .(\mathbf{b})\) 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). \((\mathbf{d})\) 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?

Step-by-Step Solution

Answer

(a) \(K_{c3} = 478.57\); (b) CO is a stronger reducing agent; (c) CO concentration is 0.102 M; (d) Solve with ICE table at \(823 K\).

Step 1: Understand Given Equilibria

Two equilibria are provided with their equilibrium constants at 823 K. They involve reactions between CoO, H2, H2O, CO, and CO2.

Step 2: Break Down Target Reaction

The target reaction is: \(\mathrm{CoO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s) + \mathrm{CO_{2}}(g) \). We need to find its equilibrium constant \(K_{c3}\).

Step 3: Use Given Reactions for Target Reaction

To obtain the target reaction, reverse the second reaction so that CO and CO2 are on the desired side. This will reciprocate its equilibrium constant (giving \( \frac{1}{0.14} \)). Combine this with the first reaction. Sum the adjusted reactions to form the target reaction.

Step 4: Calculate Combined Equilibrium Constant

When reactions are combined, their equilibrium constants multiply. Therefore, \[ K_{c3} = K_{c1} \times \left(\frac{1}{K_{c2}}\right) = 67 \times \frac{1}{0.14} = 478.57 \].

Step 5: Determine Strength of Reducing Agent

Since \(K_{c3} > 1\), it indicates the forward reaction favored, suggesting CO can more effectively reduce CoO to Co compared to H2, making CO a stronger reducing agent than H2.

Step 6: Ideal Gas Law for Concentration at 298 K

Use PV = nRT to find concentration of CO: \( P = 101.3 \text{ kPa}, V = 0.25 \text{ L}, R = 0.0831 \text{ L kPa mol}^{-1}\text{ K}^{-1}, T = 298 \text{ K} \). Calculate moles: \( n = \frac{PV}{RT} \).

Step 7: Calculate moles of CO at 298 K

Plug in values: \( n = \frac{101.3 \times 0.25}{0.0831 \times 298} \). This results in approximately 0.102 mol of CO.

Step 8: Reaction at Equilibrium at 823 K

At equilibrium with heating to 823 K, use \(K_{c3}\) to determine moles of reacted CO and thereby how much CoO remains. Set up the equilibrium ice table to solve for remaining CoO.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a critical concept in understanding chemical equilibria. It provides a measure of the extent to which a reaction proceeds to form products at a given temperature. If we have a reaction: \[ aA + bB \rightleftharpoons cC + dD \] its equilibrium constant \( K_c \), expressed in terms of concentrations, is given by: \[ K_c = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \] For reactions involving gases, concentrations can be expressed as partial pressures, and this is often denoted as \( K_p \). However, in our exercise, we are dealing with \( K_c \), indicating concentrations in terms of moles per liter.

• If \( K_c > 1 \), the reaction favors the formation of products at equilibrium.
• If \( K_c < 1 \), the reaction favors the reactants.

Understanding the value of \( K_c \) helps predict the direction of the reaction under set conditions. In the exercise provided, calculating \( K_c3 \) involved manipulating two given equilibria, showing how individuals equilibria can be combined to analyze a new reaction.

Reducing Agents

Reducing agents are substances that can donate electrons to other materials, thereby reducing them. In a redox reaction, the reducing agent loses electrons and gets oxidized itself. Within the given textbook exercise, we compared the reducing abilities of carbon monoxide (CO) and hydrogen (H2) based on their equilibrium constants in different reactions.
In the context provided:

• Carbon Monoxide (CO): Acts as a reducing agent by donating electrons to the metal oxide (CoO), turning it into pure cobalt (Co) and forming carbon dioxide (CO2).
• Hydrogen (H2): Another common reducing agent which can similarly reduce metal oxides.

A larger equilibrium constant for a particular reduction reaction means that the forward reaction is more favored, indicating a stronger reducing ability. In the exercise, CO demonstrated a superior reducing capability compared to H2, as implied by the equilibrium constant \( K_{c3} \). This can be an important piece of information when choosing practical reducing agents in industrial and laboratory processes.

Ideal Gas Law

The Ideal Gas Law is an essential equation used in Chemistry to relate the conditions of gases to each other. Expressed as:\[ PV = nRT \]where:

• \(P\) stands for pressure of the gas,
• \(V\) represents volume,
• \(n\) is the amount of substance in moles,
• \(R\) is the gas constant, often given as 0.0831 L kPa mol^-1 K^-1,
• \(T\) is the temperature in Kelvin.

The Ideal Gas Law allows us to calculate any one of these variables if the others are provided. In particular cases, this law helps assess how gases behave under various conditions, assuming no intermolecular forces, which is an idealized scenario.
In part (c) of the exercise, you used this law to determine the concentration of carbon monoxide (CO) gas under specific conditions, emphasizing its utility in scenarios assuming ideal behaviour. Using ideal conditions often simplifies calculations within manageable error ranges for gases in most circumstances.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. Central to this are concepts like reactants, products, and reaction mechanisms, which describe the stepwise transformation from reactants to products.
In the exercise given, we examine the reactions:
1. \( \text{CoO} + \text{H}_2 \rightleftharpoons \text{Co} + \text{H}_2\text{O} \)2. \( \text{H}_2 + \text{CO}_2 \rightleftharpoons \text{CO} + \text{H}_2\text{O} \)
Chemical transformations are influenced by several factors:

• Temperature: Changing temperature can shift equilibrium by favoring either the forward or reverse reaction, as given by Le Chatelier’s principle.
• Pressure and Concentrations: These can also affect reaction directions and rates, especially for gaseous reactions.
• Catalysts: Speed up reactions by lowering the activation energy without being consumed in the process.

Understanding these reactions includes knowing how to manipulate conditions to favor the production of desired products or to study reaction kinetics under varying conditions. This process involves breaking and forming chemical bonds, essential in chemical engineering, synthetic chemistry, and bioengineering.