Problem 23 - Chemistry: The Central Science

Question

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\).

Step-by-Step Solution

Answer

To find the equilibrium constant for the reaction H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) at 823 K, first reverse the second given reaction and invert its equilibrium constant: \( K_{c(-2)} = \frac{1}{490} \). Then, multiply the equilibrium constants of the first given reaction and the reversed second reaction: \( K_{c3} = K_{c1} * K_{c(-2)} = 67 * \frac{1}{490} \approx 0.1367 \). Thus, the equilibrium constant for the desired reaction is approximately 0.1367.

Step 1: Write down the given reactions and their equilibrium constants

We have the following reactions along with their equilibrium constants at 823 K: 1) CoO(s) + H₂(g) ⇌ Co(s) + H₂O(g); \( K_{c1} = 67\) 2) CoO(s) + CO(g) ⇌ Co(s) + CO₂(g); \( K_{c2} = 490\) We want to find the equilibrium constant for this reaction: 3) H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

Step 2: Combine the reactions

We can obtain the desired reaction by reversing the second reaction and adding that new reaction to the first reaction. -2) Co(s) + CO₂(g) ⇌ CoO(s) + CO(g) Add the 1) and -2) reactions: 1) + -2): H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g)

Step 3: Combine the equilibrium constants

Since we have reversed the second reaction, we need to invert the equilibrium constant for the second reaction: \( K_{c(-2)} = \frac{1}{K_{c2}} = \frac{1}{490} \) Now, we multiply the new equilibrium constant for the reversed second reaction with the equilibrium constant for the first reaction: \( K_{c3} = K_{c1} * K_{c(-2)} = 67 * \frac{1}{490} \)

Step 4: Calculate the equilibrium constant for the desired reaction

Finally, we calculate the equilibrium constant for the desired reaction: \( K_{c3} = 67 * \frac{1}{490} \approx 0.1367 \) Therefore, the equilibrium constant for the reaction H₂(g) + CO₂(g) ⇌ CO(g) + H₂O(g) at 823 K is approximately 0.1367.

Key Concepts

Chemical Equilibria

Chemical equilibria occur in reversible reactions where the rates of the forward and backward reactions are equal. This results in no net change in the concentrations of reactants and products. It's important to note that at equilibrium, a reaction doesn't "stop"; both the forward and backward reactions continue to occur, but they do so at the same rate. This is why it's called a "dynamic" equilibrium. The ratio of the concentrations of products to reactants is defined by the equilibrium constant, denoted as \(K_c\) for reactions in terms of concentrations.

• In the given problem, we have two separate equilibria involving cobalt compounds: one with hydrogen and one with carbon monoxide.
• Each equilibrium has a specific constant (\(K_{c1} = 67\) and \(K_{c2} = 490\)) at a given temperature (823 K).
• The equilibrium constant provides insight into the extent of the reaction; a large \(K_c\) indicates a reaction that favors the formation of products.

Understanding how to combine these constants to arrive at new equilibrium conditions is crucial for solving related problems.

Reaction Mechanisms

Reaction mechanisms offer a detailed step-by-step view of the individual stages a chemical reaction undergoes. Knowing a mechanism helps chemists understand how and why reactions proceed the way they do. In equilibrium problems, especially those involving multiple steps or reversals, understanding the sequence can guide you through combining reactions correctly.

• In our problem, we started with two reactions and needed to derive a third. By reversing one reaction and combining it with another, we formed the desired equilibrium equation.
• Reversing a reaction requires inverting its equilibrium constant. This is because the process is simply going backwards, so the products and reactants swap roles.

This approach underpins more complex reaction analyses and can be used to predict the presence of intermediate compounds, aiding in constructing correct and complete reaction mechanisms.

Temperature and Equilibrium

Temperature is a pivotal factor affecting equilibria. Changes in temperature can shift the position of equilibrium, often altering the equilibrium constant \(K_c\). In endothermic reactions, increasing temperature will typically shift the equilibrium towards the products, while in exothermic reactions, it shifts towards the reactants.

• The equilibrium constants given in the problem are specific to \(823 \mathrm{~K}\), suggesting that these reactions have reached a balance at that particular temperature.
• Without temperature details, calculating or predicting \(K_c\) becomes complex as each system may react differently to temperature changes.

Thus, temperature must be kept constant during experiments unless assessing the effect of temperature on \(K_c\). Understanding this will help in investigating both potential energy changes and the thermodynamic feasibility of chemical reactions.