Problem 86 - Chemistry: The Central Science

Question

Write the equilibrium-constant expression for the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The table included below shows the relative mole percentages of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CO}(\mathrm{g})\) at a total pressure of \(1 \mathrm{~atm}\) for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$ \begin{array}{lll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \mathrm{CO}_{2} \text { (mol \%) } & \text { CO (mol \%) } \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\ 1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The equilibrium constant expression for the given reaction is \(K_p = \frac{P_{CO}^2}{P_{CO_2}}\). By calculating the partial pressures of CO₂ and CO at each temperature using the given mole percentages, we found the values of Kₚ for each temperature. As Kₚ decreases with increasing temperature, the reaction is exothermic.

Step 1: Write the equilibrium constant expression#index_content#For the given reaction, C (s) + CO₂ (g) ⇌ 2 CO (g) we can write the equilibrium constant expression in terms of the partial pressures of the gases: \[K_p = \frac{P_{CO}^2}{P_{CO_2}}\]

Step 2: Calculate the partial pressures of CO₂ and CO at each temperature#index_content#The mole percentages of CO₂ and CO are provided for various temperatures. To calculate the partial pressures at each temperature, multiply the mole percentages by the total pressure (which is given as 1 atm). For example, at 850 °C: Partial Pressure of CO₂: 6.23 mol% × 1 atm = 0.0623 atm Partial Pressure of CO: 93.77 mol% × 1 atm = 0.9377 atm Repeat the same process for each of the other temperatures to obtain the corresponding partial pressures.

Step 3: Calculate Kₚ at each temperature#index_content#Use the equilibrium constant expression and the calculated partial pressures to find Kₚ at each temperature. For example, at 850 °C, we have: \[K_p = \frac{P_{CO}^2}{P_{CO_2}} = \frac{(0.9377\ \text{atm})^2}{0.0623\ \text{atm}} \approx 14.18\] Repeat this process for each of the other temperatures to obtain the corresponding Kₚ values.

Step 4: Determine if the reaction is exothermic or endothermic#index_content#Now, let's analyze the trend in Kₚ values as temperature increases. If Kₚ increases with increasing temperature, the reaction is endothermic. If Kₚ decreases with increasing temperature, the reaction is exothermic. Comparing the Kₚ values calculated at each temperature, we can see that Kₚ decreases as temperature increases. This means that the reaction is exothermic. In conclusion, we have calculated the equilibrium constant, Kₚ, for the given reaction at several temperatures and determined that the reaction is exothermic.

Key Concepts

Partial Pressures and Their Role in Equilibrium

In the realm of chemical equilibria, partial pressures play a pivotal role when dealing with gaseous reactions. The pressure exerted by a single gas in a mixture is known as its partial pressure. The equilibrium constant expression for reactions involving gases can be written in terms of partial pressures, symbolized as \( K_p \). The relationship is established based on the mole fraction of each gas and the total pressure of the gas mixture.

For instance, given the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the equilibrium constant expression in terms of partial pressures is \( K_p = \frac{P_{CO}^2}{P_{CO_2}} \). In practice, this means that if we know the mole percentages and the total pressure (as in our exercise), we can compute the partial pressures and subsequently the value of \( K_p \) at different temperatures.

Understanding how to calculate partial pressures is crucial because it allows students to determine not only the direction in which a reaction will proceed at equilibrium but also the extent to which the reactants are converted into products under specific conditions.

Le Chatelier's Principle in Action

Le Chatelier's principle is a cornerstone concept in chemical equilibrium that predicts how a system at equilibrium will respond to changes in concentration, temperature, or pressure. When a system at equilibrium is disturbed, it adjusts to minimize the disturbance and re-establish equilibrium.

Let's apply this principle to the chemical reaction at hand. As the temperature rises, we observe changes in the mole percentages of \( \mathrm{CO}_{2} \) and \( \mathrm{CO} \), which in turn affect the partial pressures and the equilibrium constant, \( K_p \). If the external temperature increases and the equilibrium shifts towards the reactants, it suggests that the reaction favors the formation of reactants to absorb the excess heat – a signature trait of an exothermic reaction.

For the provided exercise, the value of \( K_p \) decreases as the temperature increases, indicating that the reaction releases heat, or in other words, it is exothermic. Le Chatelier's principle helps students comprehend the dynamic nature of equilibrium and the directional shift that occurs in response to external changes.

Exothermic and Endothermic Reactions

Reactions that release heat into the surrounding environment are termed exothermic, while those that absorb heat are classified as endothermic. Identifying the nature of a reaction is not just about labeling; it's about predicting how the reaction will behave when subject to changes in temperature.

In an exothermic reaction, like the one where \( \mathrm{C}(s) + \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \), the release of heat as a product implies that raising the temperature will shift the equilibrium to the left, favoring the formation of reactants. Conversely, in endothermic reactions, an increase in temperature shifts the equilibrium to the right, favoring the formation of products because the system absorbs the added heat.

This understanding is crucial for students, as it aids in predicting the outcome of temperature changes on the position of equilibrium. By observing the trend in \( K_p \) values with temperature, as highlighted in our exercise, we can confirm the exothermic nature of the reaction being studied, providing a practical application of thermodynamics to chemical equilibrium.