At 500 K, the half-life period of a gaseous reaction at the initial pressure of 80 kPa is 350 sec. When the pressure is 40 kPa, the half life period is 175 sec. The order of reaction is

Options

A. zero
B. one
C. two
D. three

Answer

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

### Step 1: Understand the relationship between half-life and pressure
The half-life of a reaction can be influenced by the order of the reaction. For different orders of reactions, the relationship between half-life (t₁/₂) and initial concentration (or pressure in this case) varies.

### Step 2: Analyze the given data
We have two sets of data:
1. At an initial pressure of 80 kPa, the half-life (t₁/₂) is 350 seconds.
2. At an initial pressure of 40 kPa, the half-life (t₁/₂) is 175 seconds.

### Step 3: Compare the half-lives
Notice that when the pressure is halved from 80 kPa to 40 kPa, the half-life also halves from 350 seconds to 175 seconds. This indicates a direct relationship between the half-life and the initial pressure.

### Step 4: Relate this observation to reaction order
For a zero-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{[A]_0}{2k} \]
where \([A]_0\) is the initial concentration (or pressure) and \(k\) is the rate constant.

From the observation that halving the pressure results in halving the half-life, we can conclude that:
\[ t_{1/2} \propto [A]_0 \]
This relationship holds true for a zero-order reaction.

### Step 5: Conclusion
Since the half-life is directly proportional to the initial pressure, we conclude that the reaction is a zero-order reaction.

### Final Answer:
The order of the reaction is **zero**.

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