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. $$ \begin{array}{lll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \mathrm{CO}_{2}(\text { mol } \%) & \text { CO (mol \%) } \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\ 1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\ \hline \end{array} $$

Short answer

The equilibrium constant expression for the given reaction is: \[K_p = \frac{P_{CO}^2}{P_{CO_2}}\] The calculated values of Kp at different temperatures are as follows: - At 850ºC: \(K_p \approx 14.16\) - At 950ºC: \(K_p \approx 73.53\) - At 1050ºC: \(K_p \approx 268.63\) - At 1200ºC: \(K_p \approx 1665.67\) Since the value of Kp increases with increasing temperature, the reaction is endothermic.

Step-by-step solution

According to the given reaction: \[C(s) + CO_2(g) \rightleftharpoons 2CO(g)\] We can write the equilibrium constant expression (Kp) as follows: \[K_p = \frac{P_{CO}^2}{P_{CO_2}}\] #Step 2: Calculate partial pressures for each temperature#

We are given mole percentages and we know that the total pressure is 1 atm. We can calculate the partial pressure of CO2 and CO at each temperature using their mole percentages and the given total pressure. \[P_{CO_2} = \frac{\text{mole percentage of }CO_2}{100} \times P_{total}\] \[P_{CO} = \frac{\text{mole percentage of } CO}{100} \times P_{total}\] #Step 3: Calculate Kp at each temperature#

Now that we have the partial pressures for each temperature, we can plug them into the equilibrium constant expression (Kp) from Step 1 and calculate Kp at each temperature. #Step 4: Determine if the reaction is exothermic or endothermic#

To determine if the reaction is exothermic or endothermic, observe how the equilibrium constant (Kp) changes with increasing temperature. If Kp increases with increasing temperature, then the reaction is endothermic. If Kp decreases with increasing temperature, then the reaction is exothermic. Now, let's calculate the values of Kp and determine the nature of the reaction: For T = 850ºC: \(P_{CO_2} = \frac{6.23}{100} \times 1\, atm = 0.0623\, atm\), \(P_{CO} = \frac{93.77}{100} \times 1\, atm = 0.9377\, atm\), \(K_p = \frac{(0.9377)^2}{0.0623} \approx 14.16\) For T = 950ºC: \(P_{CO_2} = \frac{1.32}{100} \times 1\, atm = 0.0132\, atm\), \(P_{CO} = \frac{98.68}{100} \times 1\, atm = 0.9868\, atm\), \(K_p = \frac{(0.9868)^2}{0.0132} \approx 73.53\) For T = 1050ºC: \(P_{CO_2} = \frac{0.37}{100} \times 1\, atm = 0.0037\, atm\), \(P_{CO} = \frac{99.63}{100} \times 1\, atm = 0.9963\, atm\), \(K_p = \frac{(0.9963)^2}{0.0037} \approx 268.63\) For T = 1200ºC: \(P_{CO_2} = \frac{0.06}{100} \times 1\, atm = 0.0006\, atm\), \(P_{CO} = \frac{99.94}{100} \times 1\, atm = 0.9994\, atm\), \(K_p = \frac{(0.9994)^2}{0.0006} \approx 1665.67\) As we can see, the value of Kp increases with increasing temperature, which indicates that the reaction is endothermic.

Key concepts

Le Chatelier's Principle

Le Chatelier's Principle is a key concept in chemistry that helps us predict how a system at equilibrium will respond to changes in concentration, temperature, or pressure. Simply put, it states that if an external stress is applied to a system at equilibrium, the system will adjust in a way that counteracts the change, thereby establishing a new equilibrium.

When we consider the exercise, we see a chemical reaction where carbon, in the presence of carbon dioxide, creates carbon monoxide. Changes in temperature have impacted the equilibrium position. According to Le Chatelier's Principle, increasing the temperature of an endothermic reaction will shift the equilibrium to the right, thus producing more products, in this case, carbon monoxide. This is reflected in the increasing values of the equilibrium constant, Kp, with increasing temperature, which aligns with the observation that the reaction is endothermic.

Partial Pressure

Partial pressure is the pressure exerted by a single gas in a mixture of gases. It's a crucial concept when dealing with gas-phase reactions at equilibrium. The partial pressure of a gas is directly proportional to its mole fraction in the gas mixture. When the total pressure of the system is known, as in our exercise where it's given as 1 atm, partial pressures can be calculated using the mole percentages.

In our step-by-step solution, partial pressures for CO and CO2 at each temperature were determined this way, which then were plugged into the equilibrium-constant expression to find the value of Kp. It's essential to understand that the equilibrium constant is directly influenced by the partial pressures of the reactants and products. As Le Chatelier's Principle would suggest, changes in the partial pressure can shift the equilibrium, affecting the concentrations of reactants and products in the reaction.

Endothermic Reactions

Now, let’s focus on endothermic reactions, where heat is absorbed from the surroundings, resulting in the cooling of the environment. This process requires energy, and the reaction's enthalpy is positive. When a reaction is endothermic, increasing the temperature will favor the formation of products, as reflected in Le Chatelier's Principle.

In our exercise, we determined the reaction's endothermic nature by observing the increase in Kp values with rising temperature. This meant that as heat was added, it acted as a reactant, thus shifting the equilibrium to produce more CO. It’s important to recognize that the temperature dependence of Kp signifies the reaction's enthalpy. Students should note that for endothermic reactions, as the temperature increases, Kp will also increase, indicating a higher concentration of products at equilibrium.