At 300 k the half - life a sample of a gaseous compound initially at 1 atm is 100 sec. When the pressure is 0.5 atm the half - life is 50 sec. The order of reaction is

Answer

To determine the order of the reaction based on the given half-lives at different pressures, we can follow these steps:

### Step 1: Understand the relationship between half-life and order of reaction
For a reaction of order \( n \), the half-life \( t_{1/2} \) can be expressed in terms of the initial concentration (or pressure for gases) \( P_0 \):
- For a zero-order reaction:
\[ t_{1/2} = \frac{P_0}{2k} \]
- For a first-order reaction:
\[ t_{1/2} = \frac{0.693}{k} \]
- For a second-order reaction:
\[ t_{1/2} = \frac{1}{kP_0} \]

### Step 2: Set up the ratio of half-lives
Given:
- \( t_{1/2,1} = 100 \) seconds at \( P_0 = 1 \) atm
- \( t_{1/2,2} = 50 \) seconds at \( P_0 = 0.5 \) atm

We can use the formula for half-life in terms of pressure:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{P_{0,1}}{P_{0,2}}\right)^{1-n} \]

### Step 3: Substitute the values into the equation
Substituting the known values:
\[ \frac{100}{50} = \left(\frac{1}{0.5}\right)^{1-n} \]
This simplifies to:
\[ 2 = (2)^{1-n} \]

### Step 4: Solve for \( n \)
Taking logarithm on both sides:
\[ 2^1 = 2^{1-n} \]
This implies:
\[ 1 = 1 - n \]
Thus, solving for \( n \):
\[ n = 0 \]

### Conclusion
The order of the reaction is **zero-order**.

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