At 300 K, half life of a gaseous reactant initially at 58 KPa is 320 min. When the pressure is 29 KPa, the half life is 160 mm. The order of the reaction is

Options

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

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 a reaction of order \( n \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The half-life \( t_{1/2} \) of a reaction can be expressed in terms of the initial pressure \( P_0 \) and the order of the reaction \( n \). The formula for half-life in terms of pressure is:
\[ t_{1/2} \propto \frac{1}{P_0^{n-1}} \]
This means that the half-life is inversely proportional to the pressure raised to the power of \( n-1 \).

2. **Setting Up the Equation**:
We have two scenarios:
- For the first half-life:
\[ t_{1/2,1} = 320 \text{ min}, \quad P_{0,1} = 58 \text{ kPa} \]
- For the second half-life:
\[ t_{1/2,2} = 160 \text{ min}, \quad P_{0,2} = 29 \text{ kPa} \]
Using the relationship, we can set up the equation:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left(\frac{P_{0,2}}{P_{0,1}}\right)^{1-n} \]

3. **Substituting Values**:
Substitute the known values into the equation:
\[ \frac{320}{160} = \left(\frac{29}{58}\right)^{1-n} \]
Simplifying the left side gives:
\[ 2 = \left(\frac{1}{2}\right)^{1-n} \]

4. **Solving the Equation**:
Now we can solve for \( n \):
\[ 2 = 2^{-(1-n)} \]
Taking logarithm (base 2) on both sides:
\[ 1 = -(1-n) \]
Rearranging gives:
\[ 1 = -1 + n \implies n = 2 \]

5. **Conclusion**:
The order of the reaction is \( n = 2 \).

### Final Answer:
The order of the reaction is 2.