Problem 84 - Chemistry The Central Science

Question

The equilibrium constant \(K_{c}\) for \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) \(2 \mathrm{CO}(g)\) 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 \(3.00-\mathrm{L}\) vessel at \(1000 \mathrm{~K}\), how many grams of \(\mathrm{CO}\) are produced? (b) How many grams of \(\mathrm{C}\) are consumed? \((\mathrm{c})\) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

Step-by-Step Solution

Answer

(a) 11.87 g of CO is produced, (b) 5.08 g of C is consumed, (c) The yield of CO will be smaller in a smaller vessel, and (d) The reaction is endothermic.

Step 1: Write the balanced chemical equation

The balanced chemical equation for the given reaction is: \[C(s)+CO_{2}(g) \rightleftharpoons 2CO(g)\]

Step 2: Calculate moles of CO\(_2\) and equilibrium concentration

First, calculate the moles of CO\(_2\) initially present. 25.0 g of CO\(_2\) (molar mass = 44.01 g/mol) will have: \[\frac{25.0 \, \text{g}}{44.01 \, \text{g/mol}}=0.568\, \text{mol}\] Next, calculate the initial concentration of CO\(_2\). It is given that the reaction takes place in a 3.00 L vessel. \[\frac{0.568\, \text{mol}}{3.00\, \text{L}}=0.189 \, \text{M}\]

Step 3: Set up the ICE table for the reaction

An ICE table will be used to represent the equilibrium concentrations of the species involved in the reaction: Initial, Change and Equilibrium concentrations. Since C is a solid, it is not included in the table. Reaction species | Initial (M)| Change (M)| Equilibrium (M) -------------------|------------|------------|--------------- CO\(_2\) | 0.189 | -x | 0.189 - x CO | 0 | +2x | 2x Set up the equation for \(K_c\) at 1000 K given as 1.9. \[1.9 = \frac{(2x)^2}{0.189 - x}\]

Step 4: Solve for x (moles of CO formed)

Solve the above equation for x which gives the moles of CO formed. \[1.9(0.189 - x) = 4x^2\] \[4x^2+1.9x-0.3591=0\] Solve the quadratic equation for x (taking the positive value as concentration cannot be negative): \[x \approx 0.141\, \text{M}\]

Step 5: Calculate grams of CO produced and grams of C consumed

Convert moles of CO to grams (molar mass of CO = 28.01 g/mol): \[0.141\, \text{M} \times 3.00 \, \text{L} \times 28.01\, \text{g/mol} \approx 11.87\, \text{g}\, \text{CO}\] For each mole of CO produced, a mole of C is consumed. \[0.141\, \text{M} \times 3.00 \, \text{L} \times 12.01\, \text{g/mol} \approx 5.08\, \text{g}\, \text{C}\]

Step 6: Determine the effect of a smaller vessel and the nature of the reaction

(c) If a smaller vessel is used, the concentrations of the reactants and the products will increase. However, according to Le Chatelier's principle, when the concentration is increased, the reaction shifts to the side with fewer moles of gas, which is the left (reactants) in this case. Thus, the yield of CO will be smaller. (d) We can determine the nature of the reaction by comparing the \(K_c\) values at different temperatures. An increase in temperature results in an increase in \(K_c\) value (1.9 at 1000 K and 0.133 at 298 K). This indicates that the reaction shifts towards the products with an increase in temperature, suggesting that the reaction is endothermic. Answers: (a) 11.87 g of CO is produced, (b) 5.08 g of C is consumed, (c) The yield of CO will be smaller in a smaller vessel, and (d) The reaction is endothermic.

Key Concepts

Le Chatelier's principle

Le Chatelier's principle is a crucial concept in understanding how a chemical system at equilibrium responds to changes in concentration, temperature, or pressure. It offers a predictable way for us to foresee the direction in which an equilibrium will shift, making it extremely valuable in chemical engineering and industrial processes.

Essentially, Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts to counteract the change and re-establish equilibrium.

• If the concentration of a reactant is increased, the equilibrium shifts towards the products to reduce the added concentration.
• If the concentration of a product is increased, the equilibrium will shift towards the reactants.
• An increase in pressure causes the equilibrium to shift towards the side with fewer moles of gas, reducing pressure.
• An increase in temperature for an endothermic reaction (like the one in our problem) will shift the equilibrium towards the products to absorb the added heat.

In our exercise, when the vessel size is reduced, concentrations of the reactants increase, causing the equilibrium to shift towards the reactants, as per Le Chatelier’s principle. As a result, the yield of CO decreases in a smaller vessel.

Endothermic Reaction

An endothermic reaction is a type of chemical reaction that absorbs energy from its surroundings, usually in the form of heat. This means that for the reaction to proceed, it requires an input of energy. In the context of equilibrium systems, this absorption of energy has critical implications for how the equilibrium shifts with changes in temperature.

• In an endothermic reaction such as the one given in the exercise, an increase in temperature results in a higher equilibrium constant ( K_c greater) because the reaction consumes heat. Thus, the reaction favors the formation of products as the temperature rises.

To determine the nature of a reaction energetically, observe the equilibrium constant at different temperatures. In our problem, the equilibrium constant at 1000 K is higher than at 298 K, indicating that the product-forming reaction becomes more favorable at higher temperatures, which is characteristic of endothermic reactions.

Chemical Equilibrium

Chemical equilibrium refers to a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal. Thus, the concentrations of reactants and products remain constant over time. This does not indicate that the concentrations are equal, but rather that they no longer change because the forward and backward reactions occur at the same rate.

The equilibrium constant ( K_c ) is a vital indicator in these reactions. It expresses the ratio of product concentrations to reactant concentrations at equilibrium. The value of K_c gives insight into the position of equilibrium:

• If K_c is much greater than 1, it means that at equilibrium, the reaction heavily favors the formation of products.
• If K_c is much less than 1, more reactants are present at equilibrium, indicating that the forward reaction is less favored.

For the reaction in the exercise, K_c changes with temperature, reflecting a change in the balance between reactants and products and further confirming the endothermic nature of the process. The higher K_c at elevated temperature (1000 K) aligns with the increased production of CO, demonstrating how chemical equilibrium is sensitive to temperature changes.