Question

The equilibrium constant \(K_{c}\) for \(C(s)+C O_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) is \(1.9\) at \(1000 \mathrm{~K}\) and \(0.133\) at \(298 \mathrm{~K}\). (a) If excess \(\mathrm{C}\) is allowed to react with \(25.0 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) in a 3.00-L vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}_{2}\) are produced? (b) How many grams of \(C\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of \(\mathrm{CO}\) be greater or smaller? (d) If the reaction is endothermic, how does increasing the temperature affect the equilibrium constant?

Short answer

At 1000 K and in a 3.00-L vessel, when 25.0 g of CO₂ reacts with excess carbon, approximately 9.19 g of CO₂ is produced and 2.51 g of carbon is consumed. Using a smaller vessel would reduce the yield of CO due to increased concentration. If the reaction is endothermic, an increased temperature increases the value of \(K_c\), shifting the equilibrium towards the product side.

Step-by-step solution

(a) Calculate the initial moles of reactants)

First, we need to calculate the initial moles of the reactant CO₂. We are given that there are 25.0 grams of CO₂, and we will convert that to moles using the molar mass of CO₂: \[\text{moles of CO₂} = \frac{25.0 \,\text{g}}{44.01 \,\text{g/mol}} = 0.568 \,\text{moles}\] The volume of the vessel is 3.00 L and the equilibrium constant, \(K_c\), at 1000 K is 1.9.

(b) Set up ICE table)

Next, we need to set up an ICE table to find the equilibrium concentrations of CO₂ and CO. Since there is an excess of carbon, we don't have to worry about its concentration. ``` Initial Change Equilibrium CO₂ 0.568 moles -x 0.568 - x moles CO 0 +2x 2x moles ```

(c) Calculate the equilibrium concentrations)

Now, we use the equilibrium constant \(K_c\) to determine the equilibrium concentrations of reactants and products. Remember that the concentration of solid carbon, C(s), is not included in the calculation. \[K_c = \frac{[\text{CO}]^2}{[\text{CO₂}]} = \frac{(2x)^2}{0.568-x}\] Plug in the given value of equilibrium constant, 1.9, at 1000 K: \[1.9 = \frac{(2x)^2}{0.568-x}\] Next, we will solve for x.

(d) Solve for x)

After a bit of algebraic manipulation, we have a quadratic equation: \[3.802x^2 + 1.9x - 1.09 = 0\] Solving for x using the quadratic formula, we get x ≈ 0.209 moles.

(e) Calculate the grams of CO₂ produced)

Now that we have the value for x, we can calculate the grams of CO₂ produced at equilibrium; it is the same as the consumed amount: \[\text{grams of CO₂} = \text{moles} \times \text{molar mass} = (0.209 \,\text{moles})\times(44.01 \,\text{g/mol}) ≈ 9.19 \,\text{g}\]

(f) Calculate the grams of C consumed)

To find the grams of C consumed, we need first to calculate the moles of C involved in the reaction, knowing that the ratio of C consumed to CO produced is 1:1: \[\text{moles of C} = 0.209 \,\text{moles}\] Then we can convert moles to grams of C: \[\text{grams of C} = \text{moles} \times \text{molar mass} = (0.209 \,\text{moles})\times(12.01 \,\text{g/mol}) ≈ 2.51 \,\text{g}\]

(g) Predict the effect of a smaller vessel on yield)

If a smaller vessel is used, the reaction will be more concentrated. According to Le Châtelier's principle, the system will shift towards the side with the lower number of moles to minimize the change in concentration. In this case, since the number of moles of CO₂ (g) is less than the number of moles of CO (g), the equilibrium will shift to the reactant side, reducing the yield of CO.

(h) Predict the effect of increasing temperature)

As the reaction is endothermic, increasing the temperature will cause the system to attempt to absorb the additional heat. The equilibrium will shift toward the side that absorbs heat, which is the product side of the reaction. This will result in a higher value of \(K_c\) at a higher temperature.

Key concepts

Understanding the Equilibrium Constant

The equilibrium constant, represented as K_c, is crucial to the understanding of chemical equilibrium. It quantifies the ratio of concentrations of products to reactants at equilibrium, raised to the power of their respective coefficients from the balanced equation. In our exercise, the equilibrium constant K_c for C(s) + CO₂(g) ⇌ 2 CO(g) varies with temperature, illustrating that K_c is temperature-dependent.

Conceptually, this constant can be viewed as a measure of how far a reaction will proceed to reach equilibrium at a given temperature. A larger K_c indicates a greater proportion of products relative to reactants, or a reaction skewed towards the right side. Conversely, a smaller K_c signals that reactants predominate at equilibrium, with the reaction leaning to the left. For our exercise, K_c values at different temperatures will affect the amount of CO₂ that reacts, showing how temperature changes can be used to control the extent of the reaction.

Applying the ICE Table Method

The ICE table stands for Initial, Change, and Equilibrium—a systematic way to organize information about concentrations or pressures of reactants and products during a chemical reaction. It helps in figuring out the conditions at equilibrium. We start by filling in initial conditions, then express changes that occur as the reaction moves towards equilibrium, and finally state the conditions at equilibrium by balancing them with the reaction stoichiometry.

In our problem, with excess carbon, the ICE table simplifies the process of solving for the unknown change, represented by 'x', to find the equilibrium concentration. Once we have the change in concentration of CO₂, we apply stoichiometry to deduce the amounts of CO₂ produced and C consumed. This table is an invaluable tool for problems involving equilibrium and is essential for visualizing how the reaction progresses.

Le Châtelier's Principle in Action

Le Châtelier's principle is a guideline for predicting how a system at equilibrium responds to external changes. It states that if an equilibrium system is subjected to a change, the system will adjust or shift to counteract the effect of the change and re-establish equilibrium.

Impact of Volume Changes

In our problem, when the volume of the vessel decreases, the total pressure increases. Since the reaction produces more moles of gas on the product side (2 moles of CO for every 1 mole of CO₂), the system will shift towards the reactant side (fewer moles of gas) to reduce the pressure, resulting in less CO being produced.

Temperature Effects on Equilibrium

Similarly, for temperature changes in endotheric reactions, an increase in temperature is akin to adding heat, making the system shift towards the product side to 'use up' the added heat, increasing the equilibrium constant. This principle is key to understanding and controlling chemical reactions in various real-world applications.

Endothermic Reactions and Equilibrium Shifts

Endothermic reactions absorb energy (in the form of heat) from their surroundings to proceed. As a result, when the temperature is increased, the equilibrium position shifts to favor the endothermic reaction's products—this is a direct application of Le Châtelier's principle.

Our exercise demonstrates that increasing temperature will cause more CO to be produced because the reaction C(s) + CO₂(g) ⇌ 2 CO(g) is endothermic. The equilibrium constant K_c increases as temperature rises, confirming that at higher temperatures, the products of the endothermic reaction are more favored. Such temperature dependencies are invaluable in industrial chemical processes, where precise control over reaction conditions is crucial for yield optimization and energy efficient operations.