Understanding the System
Given an equilibrium mixture with 20% by volume of A in a total pressure of 1 atm, we need to find the equilibrium constant \( K_p \) for the reaction involving the molecule A, represented as \( A_2 \rightleftharpoons 2A \).
Establishing the Composition
- The volume percentage indicates that in a total volume of 100 L, there are 20 L of A and 80 L of \( A_2 \).
- In terms of partial pressures:
- \( P_A = \frac{20}{100} \times 1 \text{ atm} = 0.2 \text{ atm} \)
- \( P_{A_2} = \frac{80}{100} \times 1 \text{ atm} = 0.8 \text{ atm} \)
Writing the Equilibrium Expression
For the reaction \( A_2 \rightleftharpoons 2A \), the expression for \( K_p \) is given by:
\[ K_p = \frac{(P_A)^2}{P_{A_2}} \]
Calculating \( K_p \)
- Substitute the partial pressures into the equilibrium expression:
\[ K_p = \frac{(0.2 \text{ atm})^2}{0.8 \text{ atm}} \]
- Compute the values:
\[ K_p = \frac{0.04 \text{ atm}^2}{0.8 \text{ atm}} = 0.05 \text{ atm} \]
Conclusion
The equilibrium constant \( K_p \) for the reaction at the given conditions is:
\[ K_p = 0.05 \text{ atm} \]
This value indicates the relationship between the concentrations of reactants and products at equilibrium, providing insight into the extent of the reaction under the specified conditions.