Consider following two competing first ordr reactions,
`Poverset(k_(1))rarrA+B,Qoverset(k_(2))rarrC+D`
if `50%` of the reaction oof P wascompleted when `96%` of Q was complete ,then the ratio `(k_(2)//k_(1))` will be :

Options

A. `4.6`
B. `4.06`
C. `1.123`
D. `2.303`

Explanation

To solve the problem, we need to analyze the two competing first-order reactions and use the information provided to find the ratio \( \frac{k_2}{k_1} \).

### Step 1: Write down the reactions and their rate constants
We have two reactions:
1. \( P \overset{k_1}{\rightarrow} A + B \)
2. \( Q \overset{k_2}{\rightarrow} C + D \)

### Step 2: Understand the completion percentages
According to the problem:
- 50% of reaction P is completed when 96% of reaction Q is completed.
- This means that when 50% of P has reacted, 4% of Q remains unreacted.

### Step 3: Set up the equations for the first-order reactions
For a first-order reaction, the rate constant \( k \) can be expressed using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_0 - [A]} \right) \]
Where:
- \([A]_0\) is the initial concentration.
- \([A]\) is the concentration at time \( t \).

For reaction P:
- Initial concentration, \([P]_0 = 100\)
- Concentration after 50% completion, \([P] = 100 - 50 = 50\)

Thus, we can write:
\[ k_1 = \frac{2.303}{t} \log \left( \frac{100}{50} \right) = \frac{2.303}{t} \log(2) \]

For reaction Q:
- Initial concentration, \([Q]_0 = 100\)
- Concentration after 96% completion, \([Q] = 100 - 96 = 4\)

Thus, we can write:
\[ k_2 = \frac{2.303}{t} \log \left( \frac{100}{4} \right) = \frac{2.303}{t} \log(25) \]

### Step 4: Find the ratio \( \frac{k_2}{k_1} \)
Now we can find the ratio of the rate constants:
\[ \frac{k_2}{k_1} = \frac{\frac{2.303}{t} \log(25)}{\frac{2.303}{t} \log(2)} = \frac{\log(25)}{\log(2)} \]

### Step 5: Simplify the logarithm
Using the property of logarithms:
\[ \log(25) = \log(5^2) = 2 \log(5) \]
Thus:
\[ \frac{k_2}{k_1} = \frac{2 \log(5)}{\log(2)} \]

### Step 6: Calculate the numerical value
Using the approximate values:
- \( \log(5) \approx 0.699 \)
- \( \log(2) \approx 0.301 \)

We can compute:
\[ \frac{k_2}{k_1} = \frac{2 \times 0.699}{0.301} \approx \frac{1.398}{0.301} \approx 4.64 \]

### Final Answer
Thus, the ratio \( \frac{k_2}{k_1} \) is approximately \( 4.6 \).