Problem 72 - 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 \(3.00\) -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?

Step-by-Step Solution

Answer

(a) At the equilibrium, there are 3.92 g of CO₂ produced. (b) 5.75 g of C is consumed in the reaction. (c) If a smaller vessel is used, the yield of CO will be smaller. (d) For an endothermic reaction, increasing temperature will increase the equilibrium constant.

Step 1: (a) Calculate the amount of CO₂ produced at equilibrium

First, we will write the reaction in its proper form: \(C(s) + CO_2(g) \rightleftharpoons 2 CO(g)\) We are given the initial amount of CO2, 25.0 g, and the volume of the reaction vessel, 3.00 L. Assuming that the reaction goes to completion, there would be 25.0 g / 44.01 g/mol CO₂ = 0.568 mol of CO₂ initially present. The equilibrium constant (K₄) for this reaction at 1000 K is 1.9. We can use this information to set up an ICE (Initial, Change, Equilibrium) table: | | C | CO2 | 2 CO | |---------|---|-----|------| | Initial | - |0.568| 0 | | Change | - |-x | 2x | | Equil. | - |0.568-x| 2x | The equilibrium expression for this reaction is: \(K_{c} = \frac{[CO]^2}{[CO₂]}\) We will plug the data in the formula: \(1.9 = \frac{(2x)^2}{(0.568-x)}\) Solve for x: \(x = 0.479\) This value of x represents the mole of CO₂ that reacts. Now, we can find the amount of CO₂ at equilibrium: \(0.568 - 0.479 = 0.089\:mol\) Now, we can calculate the mass of the CO₂_remaining: mass = mol × molar_mass mass = 0.089 × 44.01 g/mol ≈ 3.92 g (a) Therefore, 3.92 g of CO₂ is present at equilibrium.

Step 2: (b) Calculate the amount of C consumed

Since excess carbon is present, we can assume that the equilibrium amount of C does not affect the equilibrium constant. Therefore, we can use the value of x to determine the amount of C consumed in the reaction. The stoichiometry of the reaction is 1:1 between CO₂ and C, hence the amount of C consumed is equal to the amount of CO₂ reacted: Amount of C consumed = 0.479 mol To find the mass of C consumed, we multiply the molar amount by the molar mass of C: mass = mol × molar_mass mass = 0.479 ×12.01 g/mol ≈ 5.75 g (b) Therefore, 5.75 g of C is consumed in the reaction.

Step 3: (c) Effect of a smaller vessel on the yield of CO

According to Le Châtelier's principle, a change in the system's environment, such as volume or pressure, will shift the equilibrium position to counteract this change. In this case, decreasing the volume of the reaction vessel (i.e., increasing pressure) will shift the reaction towards the side with fewer gas moles. In our reaction, we have: \(C(s) + CO₂(g) \rightleftharpoons 2 CO(g)\) On the left-hand side, there is 1 mole of gas, and on the right-hand side, there are 2 moles of gas. Therefore, decreasing the volume (increasing pressure) will shift the reaction to the left and produce less CO. (c) If a smaller vessel is used, the yield of CO will be smaller.

Step 4: (d) Effect of increasing the temperature on the equilibrium constant

To determine the effect of increasing temperature on the equilibrium constant, we will first need to find if the reaction is endothermic or exothermic. We are given that the reaction is endothermic, meaning that it absorbs heat. According to Le Châtelier's principle, an endothermic reaction will shift towards the products when the temperature is increased. It means the equilibrium constant will increase when the temperature is increased. (d) For an endothermic reaction, increasing temperature will increase the equilibrium_constant.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a mathematical expression that provides insight into the ratio of concentrations of products and reactants at equilibrium in a reversible reaction. For the reaction \( C(s) + CO_2(g) \rightleftharpoons 2 CO(g) \), the equilibrium constant is calculated using the formula:
\[ K_c = \frac{[CO]^2}{[CO_2]} \]This represents the concentration of carbon monoxide (\( CO \)) squared over the concentration of carbon dioxide (\( CO_2 \)). The equilibrium constant is dimensionless and is temperature-specific. Therefore, a change in temperature will alter its value.

• At 1000 K, \( K_c \) is given as 1.9, indicating the favorability of products (\( CO \)).
• At 298 K, \( K_c \) is 0.133, indicating a shift towards reactants (\( CO_2 \)).

The magnitude of \( K_c \) gives an indication of how far the reaction proceeds to the right (products) or to the left (reactants). A large \( K_c \) suggests a high product concentration at equilibrium.

Le Châtelier's Principle

Le Châtelier's Principle is a fundamental concept in chemistry that describes how a system at equilibrium responds to disturbances. If an external change is imposed on a system at equilibrium, the system adjusts itself to counteract the effect of this change and restore a new equilibrium.
For example, in the reaction \( C(s) + CO_2(g) \rightleftharpoons 2 CO(g) \), if the volume of the container is decreased, the pressure increases. According to Le Châtelier's Principle, the system will shift towards the side with fewer gas moles to decrease pressure.

• In this reaction, shifting to the left (more \( CO_2 \)) makes more "less gas" moles, following a decrease in volume.
• Conversely, increasing temperature for this endothermic reaction will shift the equilibrium toward products to absorb the added heat.

Endothermic Reaction

An endothermic reaction involves the absorption of heat, meaning that energy is taken in from the surroundings. In the context of our reaction \( C(s) + CO_2(g) \rightleftharpoons 2 CO(g) \), it is given that it is endothermic.
In endothermic reactions, the temperature plays a crucial role:

• Increasing the temperature provides more energy, favoring the formation of products as the system absorbs heat to maintain equilibrium.
• This results in an increased equilibrium constant \( K_c \) with rising temperature, as more product formation is favored.

It's important to note that endothermic reactions are often associated with drops in surrounding temperature, as energy is drawn from the environment and incorporated into the system.

Reaction Stoichiometry

Stoichiometry is a branch of chemistry that studies the quantitative relationships and ratios in chemical reactions. It helps us determine the amount of reactants and products involved in a given chemical equation.
For the reaction \( C(s) + CO_2(g) \rightleftharpoons 2 CO(g) \), stoichiometry provides the mole ratio between reactants and products:

• From the balanced equation, 1 mole of \( C \) reacts with 1 mole of \( CO_2 \) to produce 2 moles of \( CO \).
• This informs us that the amount of \( C \) consumed is equivalent to the amount of \( CO_2 \) reacted based on their 1:1 mole ratio, and each mole of \( CO_2 \) forms twice as many moles of \( CO \).
• Using stoichiometry, one can calculate mass changes and convert between moles and grams using molar masses.

This quantitative analysis is crucial for predicting how much of each substance is needed or produced in chemical reactions.