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.
A).Find the pressure of I2 when the system returns to equilibrium.
B).Find the pressure of I when the system returns to equilibrium.
0.52
To solve this problem, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since we are dealing with the same system at equilibrium, the number of moles and temperature remain constant. So, we can write:
(P1)(V1) = (P2)(V2)
where:
P1 = initial pressure of I2
V1 = initial volume
P2 = pressure of I2 at equilibrium
V2 = final volume (half the initial volume)
A). Find the pressure of I2 when the system returns to equilibrium.
We'll assume that the initial volume is V, so the final volume is V/2.
From the information given,
P1 = 0.23 atm
V1 = V
P2 = pressure of I2 at equilibrium (unknown)
V2 = V/2
Using the equation (P1)(V1) = (P2)(V2), we can solve for P2:
(0.23 atm)(V) = (P2)(V/2)
0.23 atm = P2/2
Multiplying both sides by 2, we have:
P2 = 0.23 atm × 2
P2 = 0.46 atm
Therefore, the pressure of I2 when the system returns to equilibrium is 0.46 atm.
B). Find the pressure of I when the system returns to equilibrium.
Since the system is still at equilibrium, the pressure of I remains the same. Therefore, the pressure of I at equilibrium is 0.24 atm.
To answer these questions, we can apply the concept of the equilibrium constant, Kp. The equilibrium constant is defined as the ratio of the partial pressures of the products to the partial pressures of the reactants, each raised to the power of their stoichiometric coefficients.
The balanced chemical equation for the reaction is:
I2(g) ⇌ 2I(g)
A). Find the pressure of I2 when the system returns to equilibrium:
Step 1: Calculate the initial value of Kp.
Kp = (P(I))^2 / P(I2)
= (0.24)^2 / 0.23
= 0.25
Step 2: Apply the ideal gas law to the compressed system at equilibrium.
(P(I2))^2 / P(I) = Kp
Since the volume is halved, the new pressure can be determined using Boyle's Law:
(P(I2))^2 / P(I) = Kp
(P(I2))^2 / P(I) = 0.25 (using the value of Kp from step 1)
Now, let's consider the change in the pressure of I2 and I upon compression.
When the system is compressed to half its volume, the pressure will be doubled for both I2 and I.
So, (P(I2))^2 / (2 * P(I)) = 0.25
(P(I2))^2 / P(I) = 0.5
Now, we can solve this equation for the pressure of I2 when the system returns to equilibrium:
(P(I2))^2 / P(I) = 0.5
(P(I2))^2 = 0.5 * P(I)
P(I2) = sqrt(0.5 * P(I))
Substituting the given value of P(I) = 0.24 atm into the equation, we can calculate the pressure of I2 when the system returns to equilibrium.
B). Find the pressure of I when the system returns to equilibrium:
Using the equation (P(I2))^2 / P(I) = 0.5, we can solve for P(I) since we know the pressure of I2 when the system returns to equilibrium.
(P(I2))^2 = 0.5 * P(I)
P(I) = (P(I2))^2 / 0.5
Substituting the given value of P(I2) = 0.23 atm into the equation, we can calculate the pressure of I when the system returns to equilibrium.