Chapter 15: Problem 47
Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C},\) the total pressure of the system is found to be \(4.50 \mathrm{~atm}\). If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and CO.
Short Answer
Expert verified
\( P_{\mathrm{CO}_2} \approx 1.00 \text{ atm}, P_{\mathrm{CO}} \approx 3.50 \text{ atm} \).
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Heterogeneous Equilibrium
In chemistry, a heterogeneous equilibrium involves reactants and products that exist in different phases. For example, in the given reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) \), we have solid carbon and gases carbon dioxide and carbon monoxide. Gaseous and solid phases do not uniformly mix, which is why their interaction is termed heterogeneous.
Heterogeneous equilibrium is significant because only the gaseous components' pressures are included in the equilibrium constant expression, not the solids. This exclusion simplifies calculations since the concentration of solids remains constant and does not impact the equilibrium position. Therefore, the calculation focuses on the pressures of \( \mathrm{CO}_2 \) and \( \mathrm{CO} \) for solving equilibrium problems.
Heterogeneous equilibrium is significant because only the gaseous components' pressures are included in the equilibrium constant expression, not the solids. This exclusion simplifies calculations since the concentration of solids remains constant and does not impact the equilibrium position. Therefore, the calculation focuses on the pressures of \( \mathrm{CO}_2 \) and \( \mathrm{CO} \) for solving equilibrium problems.
Partial Pressure
Partial pressure is the pressure exerted by a single component of a gas mixture. In equilibrium calculations, understanding partial pressure helps derive relationships among the reactants and products. For the given reaction, the total pressure is the sum of the partial pressures of the gaseous components, \( P_{\mathrm{CO}_2} \) and \( P_{\mathrm{CO}} \).
To understand how partial pressure works, consider each gas contributing a fraction to the total pressure equivalent to its proportion in the mixture. This is critical for calculating orifice pressure and manipulating it using laws such as Dalton's Law of Partial Pressure, which states that the total pressure of a mixture is the sum of its components' partial pressures. By assigning the partial pressures and solving the equations, we can determine how each gas influences the system.
To understand how partial pressure works, consider each gas contributing a fraction to the total pressure equivalent to its proportion in the mixture. This is critical for calculating orifice pressure and manipulating it using laws such as Dalton's Law of Partial Pressure, which states that the total pressure of a mixture is the sum of its components' partial pressures. By assigning the partial pressures and solving the equations, we can determine how each gas influences the system.
Quadratic Equation
A quadratic equation features prominently when determining unknown variables in equilibrium pressure calculations. The formation of a quadratic arises naturally from mathematical manipulations, especially when balancing products and reactants under given pressures, as seen in equilibrium constant problems.
The generic form of a quadratic equation is \( ax^2 + bx + c = 0 \). Solving it typically involves using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For the equilibrium problem, transforming the pressures into this equation allows for solving the problem efficiently by finding appropriate \( x \) values. The correct \( x \) value then translates into meaningful physical values like the partial pressures in our context.
The generic form of a quadratic equation is \( ax^2 + bx + c = 0 \). Solving it typically involves using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For the equilibrium problem, transforming the pressures into this equation allows for solving the problem efficiently by finding appropriate \( x \) values. The correct \( x \) value then translates into meaningful physical values like the partial pressures in our context.
Reaction Stoichiometry
Understanding reaction stoichiometry is vital for solving equilibrium problems as it provides the mole relationship between reactants and products. It offers insight into how much of one substance reacts with another helping to simplify equations.
For this reaction, the stoichiometry indicates that one mole of \( \mathrm{CO}_2 \) yields two moles of \( \mathrm{CO} \). When placed in equilibrium expressions, these mole ratios translate into coefficients in the partial pressure-related equilibrium constant expression \( K_P = \frac{{P^2_{\mathrm{CO}}}}{{P_{\mathrm{CO}_2}}} \).
By understanding these stoichiometric relationships, we can convert between amounts of substances and relate changes in concentrations or pressures to discern how the reaction progresses to equilibrium.
For this reaction, the stoichiometry indicates that one mole of \( \mathrm{CO}_2 \) yields two moles of \( \mathrm{CO} \). When placed in equilibrium expressions, these mole ratios translate into coefficients in the partial pressure-related equilibrium constant expression \( K_P = \frac{{P^2_{\mathrm{CO}}}}{{P_{\mathrm{CO}_2}}} \).
By understanding these stoichiometric relationships, we can convert between amounts of substances and relate changes in concentrations or pressures to discern how the reaction progresses to equilibrium.
