Chapter 7: Problem 60
For the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) the partial pressurc of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are 4 and 8 atm, respectively, then \(\mathrm{K}_{\mathrm{p}}\) for the reaction is (1) \(16 \mathrm{~atm}\) (2) \(2 \mathrm{~atm}\) (3) \(5 \mathrm{~atm}\) (4) \(4 \mathrm{~atm}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
partial pressure
For example, in our reaction \(\text{C(s) + CO}_2\text{(g)} \rightleftharpoons 2 \text{CO(g)}\), the partial pressure of \(\text{CO}_2\) is 4 atm and the partial pressure of \(\text{CO}\) is 8 atm.
The total pressure is the sum of these partial pressures. However, for the equilibrium constant calculation, we only need the individual partial pressures of the gases involved.
chemical equilibrium
In our reaction, chemical equilibrium is achieved when the rates at which \(\text{CO}_2\) reacts with carbon (C) to form \(\text{CO}\), and \(\text{CO}\) decomposes back to \(\text{CO}_2\) and C, are the same. Hence, the concentrations or partial pressures of \(\text{CO}_2\) and \(\text{CO}\) remain constant.
equilibrium expression
The equilibrium constant \(K_p\) is given by: \[ K_{p} = \frac{{(P_{\text{CO}})^2}}{{P_{\text{CO}_2}}} \] Here, we raise the partial pressure of \(\text{CO}\) to the power of 2 because 2 moles of \(\text{CO}\) are formed for every mole of \(\text{CO}_2\) consumed.
Kp calculation
Given the partial pressures: \(\text{CO}_2 = 4 \text{ atm}\) and \(\text{CO} = 8 \text{ atm}\), the equilibrium expression is: \[ K_{p} = \frac{{(8)^2}}{{4}} \] Simplifying this: \[ K_{p} = \frac{{64}}{{4}} = 16 \] Therefore, for our reaction at equilibrium, \(K_p\) is 16 atm, indicating the balance between products and reactants.