Problem 97 - 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 that follows shows the relative mole percentages of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) at a total pressure of 1 atm for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$\begin{array}{rll}\hline{\text { Temperature }\left({ }^{\circ} \mathrm{C}\right)} & \mathrm{CO}_{2}(\mathrm{~mol} \%) & \mathrm{CO}(\mathrm{mol} \%) \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\\1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

The equilibrium constant expression \(K_{p}\) for the given reaction is \(K_{p} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}\). Upon calculating the values of \(K_{p}\) at different temperatures, we find that it increases as the temperature rises, indicating that the reaction is endothermic.

Step 1: Write the equilibrium constant expression

For the given reaction: $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)$$ We can write the equilibrium constant expression (\(K_{p}\)) using partial pressures as follows: $$K_{p} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}$$

Step 2: Convert mole percentages to partial pressures

We are given the mole percentages of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\), and the total pressure is 1 atm. We can convert these mole percentages to partial pressures using the fractions of the total pressure: $$P_{\mathrm{CO}_2} = (\text{mol % of CO}_2) \times 1~\text{atm}$$ $$P_{\mathrm{CO}} = (\text{mol % of CO}) \times 1~\text{atm}$$

Step 3: Calculate the value of \(K_{p}\) at each temperature

Using the partial pressures calculated in Step 2, we can substitute them into the \(K_{p}\) equation for each temperature: $$850^\circ C: \, K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}} = \frac{(0.9377)^2}{0.0623} \approx 14.17$$ $$950^\circ C: \, K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}} = \frac{(0.9868)^2}{0.0132} \approx 73.31$$ $$1050^\circ C: \, K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}} = \frac{(0.9963)^2}{0.0037} \approx 267.80$$ $$1200^\circ C: \, K_p = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}} = \frac{(0.9994)^2}{0.0006} \approx 1666.02$$

Step 4: Determine if the reaction is exothermic or endothermic

To determine if the reaction is exothermic or endothermic, we need to consider the effect of temperature on the equilibrium constant \(K_p\). If an increase in temperature causes the value of \(K_p\) to increase, the reaction is endothermic. If an increase in temperature causes the value of \(K_p\) to decrease, the reaction is exothermic. From our calculations in Step 3, we can see that the value of \(K_p\) is increasing with rising temperature. Therefore, the reaction is endothermic.

Key Concepts

Partial Pressure

In a gaseous chemical equilibrium, the Partial Pressure of a gas refers to the pressure that particular gas would exert if it were alone in a container. It's crucial for determining the equilibrium constant when dealing with gases. In our equation,

• Partial pressure for \(\mathrm{CO}_2\) is calculated by multiplying its mole fraction by the total pressure, in this case, 1 atm.
• Likewise, calculate the partial pressure of \(\mathrm{CO}\), using its mole fraction.

By understanding each gas's partial pressure, we can calculate the equilibrium constant, \(K_p\). This happens because \(K_p\) relies on the ratios of these pressures rather than the actual numbers, simplifying calculations in practical applications. Remember, partial pressures are fundamental for reactions involving gases.

Endothermic Reaction

Endothermic reactions absorb energy from their surroundings. Whether a chemical reaction is endothermic or exothermic depends greatly on the reaction's energy change. Here, a clue comes from the behavior of the equilibrium constant, \(K_p\).

• In endothermic reactions, increasing the temperature typically increases \(K_p\).
• In contrast, a decreasing \(K_p\) with rising temperature suggests an exothermic reaction.

In the given chemical reaction, \(K_p\) grows larger as the temperature rises. This trend signifies that we are observing an endothermic reaction in which heat is absorbed. Thus, knowing the behavior of \(K_p\) over different temperatures provides insight into the reaction nature.

Kp Calculation

Calculating \(K_p\) involves substituting the partial pressures into the equilibrium expression. For our reaction, \(\mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g)\), \(K_p\) is given by the formula:

• \[K_{p} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_2}}\]

Using this formula, we can plug in the partial pressures derived from the mole percentages at each temperature: 850°C, 950°C, 1050°C, and 1200°C. For instance, at 850°C:

• \(P_{\mathrm{CO}} = 0.9377\)
• \(P_{\mathrm{CO}_2} = 0.0623\)
• Thus, \(K_p = \frac{(0.9377)^2}{0.0623} \approx 14.17\)

Repeat for other temperatures to illustrate how \(K_p\) changes. These calculations highlight how the approach leverages partial pressures, essential for assessing gas-phase equilibria.

Chemical Equilibrium

Chemical equilibrium refers to a state where the rates of the forward and reverse reactions are equal, leading to no net change in the system. Although molecules continue to react, their concentrations remain constant over time. In the context of gases:

• Equilibrium constants like \(K_p\) measure the equilibrium state's favorability towards products or reactants.
• In dynamic equilibrium, reactants and products are formed at the same rate.

In our exercise, the equilibrium involves the reaction \(\mathrm{C}(s) + \mathrm{CO}_2(g) \rightleftharpoons 2 \mathrm{CO}(g)\). This equation implies that the system has reached equilibrium when the ratio of product pressure squared to the reactant pressure has stabilized at a given \(K_p\). Understanding these principles helps in grasping how reactions behave in closed systems.