To solve the problem, we will follow these steps:
### Step 1: Understand the relationship between effusion rates and molecular weights.
According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This can be mathematically expressed as:
\[
\frac{R_1}{R_2} = \sqrt{\frac{M_2}{M_1}}
\]
where:
- \( R_1 \) is the rate of effusion of the mixture,
- \( R_2 \) is the rate of effusion of \( SO_2 \),
- \( M_1 \) is the molar mass of the mixture,
- \( M_2 \) is the molar mass of \( SO_2 \).
### Step 2: Set up the equation using the given effusion rates.
We know from the problem that the mixture effuses 1.6 times faster than \( SO_2 \). Thus, we can write:
\[
\frac{R_{mixture}}{R_{SO_2}} = 1.6
\]
Substituting this into Graham's law gives us:
\[
1.6 = \sqrt{\frac{M_{SO_2}}{M_{mixture}}}
\]
### Step 3: Calculate the molar mass of \( SO_2 \).
The molar mass of \( SO_2 \) is:
\[
M_{SO_2} = 32 \, (\text{S}) + 2 \times 16 \, (\text{O}) = 64 \, g/mol
\]
### Step 4: Solve for the molar mass of the mixture.
Now we can square both sides of the equation to eliminate the square root:
\[
(1.6)^2 = \frac{64}{M_{mixture}}
\]
Calculating \( (1.6)^2 \):
\[
2.56 = \frac{64}{M_{mixture}}
\]
Rearranging gives:
\[
M_{mixture} = \frac{64}{2.56} \approx 25 \, g/mol
\]
### Step 5: Set up the equilibrium expression.
The equilibrium reaction is:
\[
F_2(g) \rightleftharpoons 2F(g)
\]
Let \( x \) be the mole fraction of \( F_2 \) and \( 1-x \) be the mole fraction of \( F \). The molar mass of the mixture can be expressed as:
\[
M_{mixture} = x \cdot M_{F_2} + (1-x) \cdot M_F
\]
Where:
- \( M_{F_2} = 38 \, g/mol \) (since \( F \) has an atomic mass of 19),
- \( M_F = 19 \, g/mol \).
Substituting the known values:
\[
25 = x \cdot 38 + (1-x) \cdot 19
\]
### Step 6: Solve for \( x \).
Expanding the equation:
\[
25 = 38x + 19 - 19x
\]
Combining like terms:
\[
25 = 19 + 19x
\]
Subtracting 19 from both sides:
\[
6 = 19x
\]
Thus,
\[
x = \frac{6}{19} \approx 0.316
\]
### Step 7: Calculate the mole fraction of \( F \).
The mole fraction of \( F \) is:
\[
1 - x = 1 - 0.316 = 0.684
\]
### Step 8: Calculate \( K_p \).
The equilibrium constant \( K_p \) is given by:
\[
K_p = \frac{(P_F)^2}{P_{F_2}}
\]
Since the total pressure is 1 atm, the partial pressures are equal to the mole fractions:
\[
P_F = 0.684 \, atm, \quad P_{F_2} = 0.316 \, atm
\]
Substituting these values into the equation for \( K_p \):
\[
K_p = \frac{(0.684)^2}{0.316}
\]
Calculating this gives:
\[
K_p = \frac{0.467056}{0.316} \approx 1.48 \, atm
\]
### Final Answer:
Thus, the value of \( K_p \) is approximately **1.49 atm**.
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