At 300 K the half-life of 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 :

Options

A. 0
B. 1
C. 2
D. 3

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 concentration
The half-life (\(T_{1/2}\)) of a reaction can vary depending on the order of the reaction. The general relationships are:
- **Zero-order reaction**: \(T_{1/2} = \frac{[A_0]}{2k}\)
- **First-order reaction**: \(T_{1/2} = \frac{0.693}{k}\) (independent of concentration)
- **Second-order reaction**: \(T_{1/2} = \frac{1}{k[A_0]}\)

Here, \([A_0]\) is the initial concentration, and \(k\) is the rate constant.

### Step 2: Relate pressure to concentration
For gases, pressure is directly proportional to concentration. Thus, we can express the initial concentration in terms of pressure:
- \([A_0] \propto P\)

### Step 3: Set up the equations
From the problem:
- At \(P = 1 \, \text{atm}\), \(T_{1/2} = 100 \, \text{s}\)
- At \(P = 0.5 \, \text{atm}\), \(T_{1/2} = 50 \, \text{s}\)

Using the relationship derived from the order of reaction:
- For some order \(n\), we can express the half-life as:
\[ T_{1/2} \propto P^{-(n-1)} \]

### Step 4: Write the equations for the two scenarios
1. For \(P = 1 \, \text{atm}\):
\[ T_{1/2} = k \cdot 1^{-(n-1)} = k \]
So, \(100 = k\) (Equation 1)

2. For \(P = 0.5 \, \text{atm}\):
\[ T_{1/2} = k \cdot (0.5)^{-(n-1)} \]
So, \(50 = k \cdot (0.5)^{-(n-1)}\) (Equation 2)

### Step 5: Divide Equation 1 by Equation 2
Dividing the two equations gives:
\[ \frac{100}{50} = \frac{k}{k \cdot (0.5)^{-(n-1)}} \]
This simplifies to:
\[ 2 = (0.5)^{(n-1)} \]

### Step 6: Solve for \(n\)
We can express \(2\) as \(2^1\) and \(0.5\) as \(2^{-1}\):
\[ 2^1 = 2^{-(n-1)} \]
This implies:
\[ 1 = -(n-1) \]
Solving for \(n\):
\[ n - 1 = -1 \\ n = 0 \]

### Conclusion
The order of the reaction is **0** (zero-order reaction).