A reaction with respect to X is zero order till the concentration is reduced to half of initial concentration.Then this reaction become first order with respect to it.If the value of rate constants for the zero order and first order are equal to k (in magnitude), then find the time taken to reduce the concentration of X to `1/16th` of its original concentration `X_0`.

Options

A. `1/(2k)+(3ln2)/k`
B. `x_0/(2k)+(3ln2)/k`
C. `x_0/(2k)+(4ln2)/k`
D. `1/(2k)+(4ln2)/k`

Explanation

To solve the problem, we need to find the total time taken to reduce the concentration of X from its initial concentration \(X_0\) to \( \frac{X_0}{16} \). The process involves two stages: a zero-order reaction followed by a first-order reaction.

### Step-by-Step Solution:

1. **Zero-Order Reaction Phase:**
- The reaction is zero-order until the concentration is reduced to half of the initial concentration.
- The half-life (\(t_{1/2}\)) for a zero-order reaction is given by:
\[ t_{1/2} = \frac{[X_0]}{2k} \]
- Here, the initial concentration is \(X_0\) and the rate constant is \(k\).
- After this phase, the concentration of X becomes:
\[ [X] = \frac{X_0}{2} \]

2. **First-Order Reaction Phase:**
- Now, the concentration \( \frac{X_0}{2} \) will undergo a first-order reaction until it reaches \( \frac{X_0}{16} \).
- We need to find out how many half-lives are required to go from \( \frac{X_0}{2} \) to \( \frac{X_0}{16} \).
- The first half-life will reduce the concentration from \( \frac{X_0}{2} \) to \( \frac{X_0}{4} \).
- The second half-life will reduce the concentration from \( \frac{X_0}{4} \) to \( \frac{X_0}{8} \).
- The third half-life will reduce the concentration from \( \frac{X_0}{8} \) to \( \frac{X_0}{16} \).
- Therefore, it takes 3 half-lives to go from \( \frac{X_0}{2} \) to \( \frac{X_0}{16} \).

3. **Calculating the Time for First-Order Reaction:**
- The half-life for a first-order reaction is given by:
\[ t_{1/2} = \frac{\ln 2}{k} \]
- Thus, the total time taken for the first-order reaction to reduce the concentration from \( \frac{X_0}{2} \) to \( \frac{X_0}{16} \) is:
\[ t = 3 \times t_{1/2} = 3 \times \frac{\ln 2}{k} = \frac{3 \ln 2}{k} \]

4. **Total Time Calculation:**
- The total time taken to reduce the concentration from \(X_0\) to \( \frac{X_0}{16} \) is the sum of the time taken in both phases:
\[ \text{Total Time} = t_{\text{zero-order}} + t_{\text{first-order}} = \frac{X_0}{2k} + \frac{3 \ln 2}{k} \]

5. **Final Expression:**
- Therefore, the total time taken to reduce the concentration of X to \( \frac{X_0}{16} \) is:
\[ \text{Total Time} = \frac{X_0}{2k} + \frac{3 \ln 2}{k} \]