A system at equilibrium contains I2(g) at a pressure of 0.23 atm and I(g) at a pressure of 0.24 atm. The system is then compressed to half its volume.
Find the pressure of I2 when the system returns to equilibrium.
Find the pressure of I when the system returns to equilibrium.
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To find the pressure of I2 when the system returns to equilibrium, we need to use the concept of the ideal gas law and the equilibrium constant expression.
1. Start by writing the balanced chemical equation for the reaction:
I2(g) ⇌ 2I(g)
2. Set up an expression for the equilibrium constant, Kp, using the partial pressures of the gases:
Kp = (P(I))^2 / P(I2)
3. Use the given initial pressures to calculate the initial equilibrium constant:
Kp_initial = (0.24 atm)^2 / 0.23 atm = 0.2507
4. Since the system is compressed to half its volume, the total number of gas molecules remains the same. However, the partial pressures of the gases change. Let's assume the pressure of I2 after compression is P(I2)', and the pressure of I after compression is P(I)'.
5. Using the ideal gas law, we know that the pressure and volume are inversely proportional at constant temperature. Therefore, the ratio of the initial pressure to the final pressure is equal to the ratio of the final volume to the initial volume:
(P(I2) / P(I2)') = (V(I2)' / V(I2))
(0.23 atm / P(I2)') = (0.5 / 1)
P(I2)' = 0.23 atm / 0.5
P(I2)' = 0.46 atm
6. Now, we can use the equilibrium constant expression to find the equilibrium pressure of I2:
Kp = (P(I))^2 / P(I2)'
0.2507 = (P(I))^2 / 0.46 atm
P(I)^2 = 0.2507 * 0.46 atm
P(I) = √(0.2507 * 0.46 atm)
P(I) = 0.369 atm (approximately)
Therefore, the pressure of I2 when the system returns to equilibrium is 0.46 atm, and the pressure of I is 0.369 atm.