A first order reaction has the rate constant, `k=4.6times10^(-3)s^(-1)`.The number of correct statement/s from the following is/are
Given: `log3=0.48`
A.Reaction completes in `1000s`.
B.The reaction has a half-life of `500s`.
C.The time required for `10%` completion is `25` times the time required for `90%` completion.
D.The degree of dissociation is equal to `(1-e^(-kt))`
E.The rate and the rate constant have the same unit.

Answer

To solve the problem, we will analyze each statement regarding the first-order reaction with the given rate constant \( k = 4.6 \times 10^{-3} \, s^{-1} \).

### Step 1: Analyze Statement A
**Statement A: Reaction completes in 1000s.**

For a first-order reaction, the completion of the reaction theoretically takes infinite time. However, we can calculate how much of the reaction is complete after 1000 seconds using the formula for first-order kinetics:

\[ \text{Completion} = 1 - e^{-kt} \]

Substituting \( k = 4.6 \times 10^{-3} \, s^{-1} \) and \( t = 1000 \, s \):

\[ \text{Completion} = 1 - e^{-4.6 \times 10^{-3} \times 1000} \]
\[ = 1 - e^{-4.6} \approx 1 - 0.0101 \approx 0.9899 \]

This means that after 1000 seconds, the reaction is approximately 98.99% complete, but it does not complete fully. Therefore, **Statement A is incorrect.**

### Step 2: Analyze Statement B
**Statement B: The reaction has a half-life of 500s.**

The half-life \( t_{1/2} \) for a first-order reaction is given by:

\[ t_{1/2} = \frac{0.693}{k} \]

Substituting the value of \( k \):

\[ t_{1/2} = \frac{0.693}{4.6 \times 10^{-3}} \approx 150.65 \, s \]

Since 150.65 seconds is not equal to 500 seconds, **Statement B is incorrect.**

### Step 3: Analyze Statement C
**Statement C: The time required for 10% completion is 25 times the time required for 90% completion.**

Let \( t_{10} \) be the time for 10% completion and \( t_{90} \) be the time for 90% completion.

Using the first-order kinetics formula:

\[ t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right) \]

For 10% completion:

\[ t_{10} = \frac{1}{k} \ln\left(\frac{100}{90}\right) = \frac{1}{k} \ln\left(\frac{10}{9}\right) \]

For 90% completion:

\[ t_{90} = \frac{1}{k} \ln\left(\frac{100}{10}\right) = \frac{1}{k} \ln(10) \]

Now, we check if \( t_{10} = 25 \times t_{90} \):

\[ t_{10} = \frac{1}{k} \ln\left(\frac{10}{9}\right) \quad \text{and} \quad t_{90} = \frac{1}{k} \ln(10) \]

Calculating the ratio:

\[ \frac{t_{10}}{t_{90}} = \frac{\ln\left(\frac{10}{9}\right)}{\ln(10)} \]

Using \( \ln(10) \approx 2.303 \) and \( \ln\left(\frac{10}{9}\right) \approx 0.1054 \):

\[ \frac{t_{10}}{t_{90}} \approx \frac{0.1054}{2.303} \approx 0.0458 \]

Since \( 0.0458 \) is not equal to \( \frac{1}{25} \), **Statement C is incorrect.**

### Step 4: Analyze Statement D
**Statement D: The degree of dissociation is equal to \( (1 - e^{-kt}) \)**

This statement is true for first-order reactions. The degree of dissociation \( \alpha \) is defined as:

\[ \alpha = 1 - e^{-kt} \]

Thus, **Statement D is correct.**

### Step 5: Analyze Statement E
**Statement E: The rate and the rate constant have the same unit.**

For a first-order reaction, the unit of the rate constant \( k \) is \( s^{-1} \) and the unit of rate is \( \text{mol L}^{-1} s^{-1} \). Therefore, they do not have the same unit. **Statement E is incorrect.**

### Summary of Statements
- A: Incorrect
- B: Incorrect
- C: Incorrect
- D: Correct
- E: Incorrect

### Final Answer
The number of correct statements is **1** (only Statement D).

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