If a reaction with `t_(1//2)=69.3` seconds , has a rate constant value of `10^(-2)` per second , the order is

Options

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

Answer

To determine the order of the reaction given the half-life \( t_{1/2} = 69.3 \) seconds and the rate constant \( k = 10^{-2} \) s\(^{-1}\), we can follow these steps:

### Step 1: Understand the relationship between half-life and order of reaction
The half-life of a reaction depends on its order. For different orders, the half-life expressions are as follows:
- **Zero-order**: \( t_{1/2} = \frac{[A]_0}{2k} \)
- **First-order**: \( t_{1/2} = \frac{0.693}{k} \)
- **Second-order**: \( t_{1/2} = \frac{1}{k[A]_0} \)

### Step 2: Identify the given values
From the question, we have:
- \( t_{1/2} = 69.3 \) seconds
- \( k = 10^{-2} \) s\(^{-1}\)

### Step 3: Check if the half-life corresponds to first-order kinetics
For a first-order reaction, we can use the formula for half-life:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the value of \( k \):
\[ t_{1/2} = \frac{0.693}{10^{-2}} = \frac{0.693}{0.01} = 69.3 \text{ seconds} \]

### Step 4: Conclusion
Since the calculated half-life matches the given half-life, we can conclude that the reaction is first-order.

### Final Answer
The order of the reaction is **first-order**.

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