Problem 86 - Chemistry The Central Science In Si Units

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 \(298 \mathrm{~K} .(\mathbf{a})\) If excess 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 CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of \(\mathrm{CO}\) be greater or smaller? (d) Is the reaction endothermic or exothermic?

Step-by-Step Solution

Answer

(a) 16.98 g \( \mathrm{CO} \); (b) 1.21 g \( \mathrm{C} \); (c) Smaller yield; (d) Exothermic.

Step 1: Understanding the Reaction

The given reaction is \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \). The equilibrium constant \( K_c \) is provided for this reaction at two different temperatures: 1.9 at 1000 K and 0.133 at 298 K.

Step 2: Determine Initial Moles of \( \mathrm{CO}_2 \)

To find out how many moles of \( \mathrm{CO}_2 \) we start with, convert the mass to moles using the formula: \( n = \frac{m}{M} \) where \( m \) is the mass and \( M \) is the molar mass. Here, \( M_{\mathrm{CO}_2} = 44.01 \text{ g/mol} \), so \( n = \frac{25.0 \text{ g}}{44.01 \text{ g/mol}} \approx 0.568 \text{ mol} \).

Step 3: Calculate Initial Concentration of \( \mathrm{CO}_2 \)

The initial concentration \( [\mathrm{CO}_2]_0 \) is calculated using the formula \( [\mathrm{CO}_2]_0 = \frac{n}{V} \) where \( n \) is moles and \( V \) is volume. \( V = 3.00 \text{ L} \), so \( [\mathrm{CO}_2]_0 = \frac{0.568 \text{ mol}}{3.00 \text{ L}} \approx 0.189 \text{ M} \).

Step 4: Write the Expression for \( K_c \)

The expression for \( K_c \) is \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \). At equilibrium, let the change in \([\mathrm{CO}_2]\) be \(-x\); hence, \([\mathrm{CO}] = 2x\) and \([\mathrm{CO}_2] = 0.189 - x\). Substitute these into the \( K_c \) expression.

Step 5: Solve for \( x \)

Using \( K_c = 1.9 \), we get \( 1.9 = \frac{(2x)^2}{0.189 - x} = \frac{4x^2}{0.189 - x} \). Solve for \( x \). This results in the quadratic equation \( 4x^2 = 1.9(0.189 - x) \). Solving gives \( x \approx 0.101 \) M.

Step 6: Calculate Moles of \( \mathrm{CO} \) Produced

Since \( [\mathrm{CO}] = 2x = 2(0.101 \text{ M}) = 0.202 \text{ M} \), convert the concentration to moles: \( n_{\mathrm{CO}} = \text{concentration} \times \text{volume} = 0.202 \text{ M} \times 3.00 \text{ L} = 0.606 \text{ mol} \).

Step 7: Convert Moles of \( \mathrm{CO} \) to Grams

The molar mass of \( \mathrm{CO} \) is 28.01 g/mol, so \( \text{grams of } \mathrm{CO} = 0.606 \text{ mol} \times 28.01 \text{ g/mol} \approx 16.98 \text{ g} \).

Step 8: Determine Grams of \( \mathrm{C} \) Consumed

Using stoichiometry, for every mole of \( \mathrm{CO}_2 \) consumed, \( 1 \) mole of \( \mathrm{C} \) is consumed and \( 2 \) moles of \( \mathrm{CO} \) are produced. Since \( 0.101 \) moles of \( \mathrm{CO}_2 \) are consumed (\( 0.189 - x \)), the same moles of \( \mathrm{C} \) are consumed. With \( M_{\mathrm{C}} = 12.01 \text{ g/mol} \), \( \text{grams of } \mathrm{C} = 0.101 \text{ mol} \times 12.01 \text{ g/mol} \approx 1.21 \text{ g} \).

Step 9: Analyze Effect of Vessel Size Change

Decreasing the volume of the vessel increases the pressure, shifting the equilibrium towards the side with fewer gas molecules. Thus, the yield of \( \mathrm{CO} \) decreases in a smaller vessel (2 moles on product side compared to 1 mole on reactant side).

Step 10: Determine Reaction Type (Endothermic or Exothermic)

Since \( K_c \) decreases with a decrease in temperature from 1000 K to 298 K, the reaction is exothermic. According to Le Châtelier's principle, an exothermic reaction tends to shift left (producing fewer products) when cooled.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a numerical value that indicates the ratio of products to reactants in a chemical reaction at equilibrium. It's crucial to realize that \( K_c \) provides insight into the position of equilibrium—a high \( K_c \) means the equilibrium lies towards the products, while a low \( K_c \) favors the reactants.
For the reaction \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the equilibrium constant varies with temperature, being 1.9 at 1000 K and 0.133 at 298 K. This indicates that the reaction favors the formation of CO more at higher temperatures.

• The expression for \( K_c \) for this reaction is: \( K_c = \frac{[\mathrm{CO}]^2}{[\mathrm{CO}_2]} \).
• The changes in concentration as the reaction approaches equilibrium are used to solve for unknown quantities, helping us determine the extent to which the reaction has proceeded.

This concept is fundamental in predicting how a system behaves when subjected to changes in conditions, such as temperature and pressure.

Le Châtelier's Principle

Le Châtelier's Principle is a fundamental concept in chemical equilibrium. It states that if a dynamic equilibrium system experiences a change in concentration, temperature, or pressure, the system will adjust itself to counteract that change and reach a new equilibrium state.
For instance, when the volume of the reaction vessel is decreased, thus increasing pressure, the equilibrium for the reaction \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \) shifts. Since there are more moles of gas on the right side (2 moles of CO compared to 1 mole of CO₂), the equilibrium shifts toward the reactants to reduce the pressure.

• This means in a smaller vessel, the yield of \( \mathrm{CO} \) would decrease.
• Adjustments in temperature lead to shifts depending on whether the reaction is exothermic or endothermic.

Understanding Le Châtelier's Principle allows us to predict how a reactant or product concentration may change in response to external alterations.

Stoichiometry

Stoichiometry is the section of chemistry that involves calculating the amounts of reactants and products in a chemical reaction. It relies on the principle of the conservation of mass and the mole ratio derived from a balanced chemical equation.
In the given reaction, stoichiometry tells us that one mole of \( \mathrm{C}(s) \) reacts with one mole of \( \mathrm{CO}_2(g) \) to produce two moles of \( \mathrm{CO}(g) \).

• Knowing the initial moles of \( \mathrm{CO}_2 \), we can deduce the moles of \( \mathrm{CO} \) produced and the \( \mathrm{C} \) consumed using the mole ratio.
• For example, from 0.101 moles of \( \mathrm{CO}_2 \) consumed, we produce 0.202 moles of \( \mathrm{CO} \) and consume the same 0.101 moles of \( \mathrm{C} \).

Stoichiometry serves as a quantitative backbone in chemical equations, helping us gain a deeper understanding of how chemical quantities interact in a reaction.

Endothermic vs Exothermic Reactions

Chemical reactions can either absorb or release energy, classifying them as endothermic or exothermic, respectively. For the equilibrium reaction \( \mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the temperature dependence of \( K_c \) provides clues about the reaction's energy nature.
Here, as the temperature decreases from 1000 K to 298 K, \( K_c \) also decreases. This behavior is typical of an exothermic reaction, which releases heat. According to Le Châtelier's Principle, lowering the temperature would shift equilibrium to the left (towards the reactants), as less heat would push the reaction backward.

• In exothermic reactions, products are favored at lower temperatures, but higher temperatures provide energy driving reactions toward more products.
• Understanding whether a reaction is endothermic or exothermic helps predict how temperature changes will affect equilibrium positions.

Recognizing these types of reactions enables us to foresee how changes in conditions can influence the products we might obtain.