The reaction SO2Cl2(g) → SO2(g) + Cl2(g) is a first order reaction carried out at constant volume
at 600 K whose t1/2 is 4.1 hr. The initial pressure of SO2Cl2 is 1.25 atm. What will be the total
pressure of the system after 1 hour?
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To determine the total pressure of the system after 1 hour, we need to calculate the concentration of SO2Cl2 remaining at that time.
Given that the half-life of the reaction is 4.1 hours, we can use the first-order integrated rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
Since the reaction is carried out at constant volume, the concentration is directly proportional to the pressure. Therefore, we can also use pressure in this equation.
ln(Pt/P0) = -kt
Rearranging the equation to solve for Pt:
Pt = P0 * e^(-kt)
Where Pt is the total pressure at time t and P0 is the initial pressure.
Now, let's substitute the known values into the equation:
P0 = 1.25 atm (initial pressure)
t = 1 hour
k = ln(2)/t1/2 = ln(2)/4.1 hr (rate constant)
Pt = 1.25 * e^(-ln(2)/4.1 * 1)
Calculating this expression:
Pt = 1.25 * e^(-ln(2)/4.1)
Pt ≈ 1.135 atm
Therefore, the total pressure of the system after 1 hour will be approximately 1.135 atm.
To determine the total pressure of the system after 1 hour in a first-order reaction, we need to use the concept of half-life.
Since the half-life of the reaction is given as 4.1 hours, we can use the following equation to calculate the remaining concentration of SO2Cl2 after 1 hour:
ln(N₀/N) = kt
Where:
N₀ is the initial concentration of SO2Cl2
N is the concentration of SO2Cl2 after 1 hour
k is the rate constant of the first-order reaction
t is the time (in this case, 1 hour)
First, we need to calculate the rate constant (k). The half-life (t1/2) can be related to the rate constant using the following equation:
t1/2 = ln(2)/k
Rearranging the equation, we can solve for k:
k = ln(2)/t1/2
Substituting the given t1/2 (4.1 hours) into the equation, we can calculate k:
k = ln(2)/4.1
Now, let's calculate the rate constant k:
k = 0.169 hr^-1 (approximately)
Next, we can use the rate constant to calculate the remaining concentration of SO2Cl2 after 1 hour:
ln(N₀/N) = kt
ln(1.25/N) = (0.169 hr^-1)(1 hr)
Simplifying the equation, we get:
ln(1.25/N) = 0.169
Using the properties of logarithms, we can rearrange the equation to solve for N:
1.25/N = e^0.169
N = 1.25 / e^0.169
N = 1.09
Therefore, the concentration of SO2Cl2 after 1 hour is 1.09 atm.
Since the reaction occurs at constant volume, we can assume that the moles of the reaction remain constant. Therefore, the moles of SO2(g) and Cl2(g) produced at 1 hour will also be 1.09 atm.
Knowing that the initial pressure of SO2Cl2 is 1.25 atm, the total pressure of the system after 1 hour will be the sum of the partial pressures of SO2Cl2, SO2, and Cl2:
Total pressure = 1.25 atm + 1.09 atm + 1.09 atm
Total pressure = 3.43 atm
Therefore, the total pressure of the system after 1 hour is approximately 3.43 atm.