For first order reaction: `t_(99.9)/t_50`=x , Here x is

Options

A. 3
B. 5
C. 7
D. 10

Answer

To solve the problem, we need to find the ratio \( \frac{t_{99.9}}{t_{50}} \) for a first-order reaction. Let's break it down step by step.

### Step 1: Understand the First-Order Reaction Kinetics
For a first-order reaction, the time taken to reach a certain percentage of completion can be calculated using the formula:
\[ t = \frac{1}{k} \cdot 2.303 \log \left( \frac{A_0}{A_t} \right) \]
where:
- \( t \) is the time taken,
- \( k \) is the rate constant,
- \( A_0 \) is the initial concentration,
- \( A_t \) is the concentration at time \( t \).

### Step 2: Calculate \( t_{99.9} \)
For \( t_{99.9} \), the concentration remaining \( A_t \) is \( 0.001 A_0 \) (since 99.9% has reacted).
Substituting into the formula:
\[ t_{99.9} = \frac{1}{k} \cdot 2.303 \log \left( \frac{A_0}{0.001 A_0} \right) \]
This simplifies to:
\[ t_{99.9} = \frac{1}{k} \cdot 2.303 \log(1000) = \frac{1}{k} \cdot 2.303 \cdot 3 \]
Thus,
\[ t_{99.9} = \frac{6.909}{k} \]

### Step 3: Calculate \( t_{50} \)
For \( t_{50} \), the concentration remaining \( A_t \) is \( 0.5 A_0 \) (since 50% has reacted).
Substituting into the formula:
\[ t_{50} = \frac{1}{k} \cdot 2.303 \log \left( \frac{A_0}{0.5 A_0} \right) \]
This simplifies to:
\[ t_{50} = \frac{1}{k} \cdot 2.303 \log(2) \]
Thus,
\[ t_{50} = \frac{2.303 \cdot 0.3010}{k} \quad (\text{using } \log(2) \approx 0.3010) \]

### Step 4: Find the Ratio \( \frac{t_{99.9}}{t_{50}} \)
Now we can find the ratio:
\[ \frac{t_{99.9}}{t_{50}} = \frac{\frac{6.909}{k}}{\frac{2.303 \cdot 0.3010}{k}} = \frac{6.909}{2.303 \cdot 0.3010} \]
Calculating the denominator:
\[ 2.303 \cdot 0.3010 \approx 0.693 \]
Thus,
\[ \frac{t_{99.9}}{t_{50}} = \frac{6.909}{0.693} \approx 10 \]

### Final Answer
Therefore, the value of \( x \) is:
\[ x = 10 \]

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