Problem 24 - Chemistry and Chemical Reactivity

Question

Calculate \(K\) for the reaction $$ \mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftarrows \mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) $$ given the following information: $$ \begin{array}{ll} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) & K=1.6 \\ \mathrm{FeO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) & K=0.67 \end{array} $$

Step-by-Step Solution

Answer

The equilibrium constant \( K \) for the target reaction is approximately 2.39.

Step 1: Analyze Given Reactions

We have two known reactions, each with its own equilibrium constant:1. Reaction 1: \( \text{H}_2\text{O}(\text{g}) + \text{CO}(\text{g}) \rightleftarrows \text{H}_2(\text{g}) + \text{CO}_2(\text{g}), K_1 = 1.6 \).2. Reaction 2: \( \text{FeO}(\text{s}) + \text{CO}(\text{g}) \rightleftarrows \text{Fe}(\text{s}) + \text{CO}_2(\text{g}), K_2 = 0.67 \).

Step 2: Manipulate Reactions to Match Target Reaction

We need to manipulate the given reactions to derive the target reaction, which is:\( \text{Fe}(\text{s}) + \text{H}_2\text{O}(\text{g}) \rightleftarrows \text{FeO}(\text{s}) + \text{H}_2(\text{g}) \).

Step 3: Reverse Reaction 2

Reverse reaction 2 to get:\( \text{Fe}(\text{s}) + \text{CO}_2(\text{g}) \rightleftarrows \text{FeO}(\text{s}) + \text{CO}(\text{g}) \).When a reaction is reversed, the equilibrium constant becomes the reciprocal:\[ K_2' = \frac{1}{0.67} \approx 1.493 \].

Step 4: Add Reversed Reaction 2 to Reaction 1

Add the reversed version of reaction 2 to reaction 1:\[\text{H}_2\text{O}(\text{g}) + \text{CO}(\text{g}) + \text{Fe}(\text{s}) + \text{CO}_2(\text{g}) \rightleftarrows \text{H}_2(\text{g}) + \text{CO}_2(\text{g}) + \text{FeO}(\text{s}) + \text{CO}(\text{g})\]Cancel out \( \text{CO}(\text{g}) \) and \( \text{CO}_2(\text{g}) \) on both sides:\( \text{Fe}(\text{s}) + \text{H}_2\text{O}(\text{g}) \rightleftarrows \text{FeO}(\text{s}) + \text{H}_2(\text{g}) \).

Step 5: Calculate Equilibrium Constant for Target Reaction

The equilibrium constant for the target reaction is the product of the equilibrium constants from the modified reactions:\[ K = K_1 \times K_2' = 1.6 \times 1.493 \approx 2.3888 \].

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state where the concentrations of reactants and products remain constant over time. This doesn't mean the reactants and products are equal, but rather that their rates of formation and conversion are balanced. At equilibrium, the forward and reverse reaction rates are equal. The equilibrium constant, denoted as \( K \), is a ratio of the concentrations of products to reactants. Each concentration is raised to the power of its coefficient in the balanced equation.
In this context, substances in solid form, like iron (Fe) or iron oxide (FeO), are not included in the equilibrium expression. This is because their concentrations remain constant. The value of \( K \) gives us insight into the relative amounts of reactants and products at equilibrium. A large \( K \) means a high product concentration, suggesting the reaction favors the creation of products, whereas a small \( K \) indicates a reaction that favors the reactants.

Reaction Manipulation

Reaction manipulation involves changing a given reaction to derive another reaction with the desired outcome. This technique is crucial in chemical equilibrium problems where we use the values of known equilibrium constants to find the constants of other reactions. It can involve reversing reactions, changing stoichiometric coefficients, or even adding reactions together.

Reversing Reactions

When we reverse a reaction, the equilibrium constant for the new reaction becomes the reciprocal of the original constant. For example, if the equilibrium constant for a reaction is \( K_2 = 0.67 \), reversing the reaction would result in \( K_2' = \frac{1}{0.67} \).

Adding Reactions

When combining reactions, their equilibrium constants are multiplied. This makes sense because we're effectively considering how the overall system behaves, given we are summing up their respective equilibria. By manipulating the original reactions accordingly, we can deduce the equilibrium constant of our new target reaction with ease.

Reversible Reactions

Reversible reactions are reactions where the reactants convert to products and vice versa, under the same conditions. These reactions can proceed in both directions, forward and backward, reaching a state of equilibrium in which the rates of the forward and reverse reactions are equal.
An example is the reaction between water vapor and carbon monoxide used in our exercise. This reaction can proceed to form hydrogen gas and carbon dioxide or reverse to reform the reactants. The equilibrium constant \( K \) helps determine the extent to which each side of the reaction is favored at equilibrium. Reversible reactions are common in both nature and industry, where balance is key to efficient and controlled results.
Being aware of these characteristics and constants allows chemists to predict reaction behaviors and manipulate conditions to favor desired pathways, optimizing the yield of products.