To calculate the half-life of the first-order reaction \( C_2H_4O(g) \rightarrow CH_4(g) + CO(g) \), we will follow these steps:
### Step 1: Understand the Reaction and Given Data
We have the initial pressure of \( C_2H_4O(g) \) as \( 80 \, mm \) and the total pressure at the end of \( 20 \, min \) as \( 120 \, mm \).
### Step 2: Calculate the Change in Pressure
The change in pressure (\( x \)) of \( C_2H_4O \) can be calculated as follows:
- Initial pressure of \( C_2H_4O \) = \( 80 \, mm \)
- Total pressure after \( 20 \, min \) = \( 120 \, mm \)
The increase in pressure is due to the formation of \( CH_4 \) and \( CO \). Since each mole of \( C_2H_4O \) produces one mole of \( CH_4 \) and one mole of \( CO \), the total increase in pressure is \( 2x \), where \( x \) is the change in pressure of \( C_2H_4O \).
Thus, we can write:
\[
80 \, mm + 2x = 120 \, mm
\]
Solving for \( x \):
\[
2x = 120 \, mm - 80 \, mm = 40 \, mm \implies x = 20 \, mm
\]
### Step 3: Calculate the Remaining Pressure of \( C_2H_4O \)
The remaining pressure of \( C_2H_4O \) after \( 20 \, min \) is:
\[
P_{C_2H_4O} = 80 \, mm - 20 \, mm = 60 \, mm
\]
### Step 4: Use the First-Order Reaction Formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[
k = \frac{2.303}{t} \log \left( \frac{P_0}{P_t} \right)
\]
Where:
- \( P_0 \) = initial pressure = \( 80 \, mm \)
- \( P_t \) = pressure at time \( t \) = \( 60 \, mm \)
- \( t = 20 \, min \)
Substituting the values:
\[
k = \frac{2.303}{20} \log \left( \frac{80}{60} \right)
\]
### Step 5: Calculate the Logarithm
Calculating the logarithm:
\[
\log \left( \frac{80}{60} \right) = \log \left( \frac{4}{3} \right) \approx 0.1249
\]
### Step 6: Calculate \( k \)
Now substituting back into the equation for \( k \):
\[
k = \frac{2.303}{20} \times 0.1249 \approx 0.0144 \, min^{-1}
\]
### Step 7: Calculate the Half-Life
The half-life (\( t_{1/2} \)) for a first-order reaction is given by:
\[
t_{1/2} = \frac{0.693}{k}
\]
Substituting the value of \( k \):
\[
t_{1/2} = \frac{0.693}{0.0144} \approx 48.1 \, min
\]
### Conclusion
Thus, the half-life of the reaction is approximately \( 48.1 \, min \).
