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 use the relationship between half-life and pressure for reactions of different orders.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Half-life at 1 atm (T_half1) = 100 seconds
- Half-life at 0.5 atm (T_half2) = 50 seconds
- Initial pressure (P1) = 1 atm
- New pressure (P2) = 0.5 atm

2. **Use the Relationship Between Half-life and Pressure:**
For a reaction of order \( n \), the relationship between half-life and pressure can be expressed as:
\[ \frac{T_{half1}}{T_{half2}} = \left(\frac{P_2}{P_1}\right)^{(n-1)} \]

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ \frac{100 \, \text{sec}}{50 \, \text{sec}} = \left(\frac{0.5 \, \text{atm}}{1 \, \text{atm}}\right)^{(n-1)} \]

4. **Simplify the Left Side:**
Simplifying the left side gives:
\[ 2 = \left(0.5\right)^{(n-1)} \]

5. **Express 0.5 as a Fraction:**
Recognize that \( 0.5 = \frac{1}{2} \):
\[ 2 = \left(\frac{1}{2}\right)^{(n-1)} \]

6. **Convert to Exponential Form:**
Rewrite the equation in exponential form:
\[ 2^1 = 2^{-(n-1)} \]

7. **Set the Exponents Equal:**
Since the bases are the same, we can set the exponents equal to each other:
\[ 1 = -(n - 1) \]

8. **Solve for n:**
Rearranging gives:
\[ n - 1 = -1 \implies n = 0 \]

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