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 and pressures, we can follow these steps:

### Step 1: Understand the relationship between half-life and pressure
The half-life (t₁/₂) of a reaction is related to the pressure of the reactant. For a reaction of order n, the half-life can be expressed as:
\[ t_{1/2} \propto P^{(1-n)} \]
where P is the pressure of the reactant.

### Step 2: Set up the ratio of half-lives and pressures
Given:
- \( t_{1/2,1} = 320 \) min at \( P_1 = 58 \) KPa
- \( t_{1/2,2} = 160 \) min at \( P_2 = 29 \) KPa

Using the relationship established:
\[ \frac{t_{1/2,1}}{t_{1/2,2}} = \left( \frac{P_1}{P_2} \right)^{(1-n)} \]

### Step 3: Substitute the values into the equation
Substituting the known values:
\[ \frac{320}{160} = \left( \frac{58}{29} \right)^{(1-n)} \]

### Step 4: Simplify the equation
This simplifies to:
\[ 2 = \left( 2 \right)^{(1-n)} \]

### Step 5: Equate the exponents
Since the bases are the same, we can equate the exponents:
\[ 1 = 1 - n \]

### Step 6: Solve for n
Rearranging gives:
\[ n = 0 \]

### Conclusion
The order of the reaction is 0.

### Final Answer
The order of the reaction is 0.

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