For a zero order reaction `'[R]_0` is the initial concentration

Options

A. `t_(1//2) propR_0`
B. `t_(1//2) prop1//R_0`
C. `t_(1//2) propR_0^2`
D. `t_(1//2) prop1//R_0^2`

Answer

To solve the question regarding the relationship between the half-life of a zero-order reaction and its initial concentration, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Zero-Order Reactions:**
- In a zero-order reaction, the rate of reaction is constant and does not depend on the concentration of the reactants. The general form of a zero-order reaction can be written as:
\[ A \rightarrow B \]

2. **Integrated Rate Law for Zero-Order Reactions:**
- The integrated rate law for a zero-order reaction is given by:
\[ [A]_t = [A]_0 - kt \]
- Where:
- \([A]_t\) is the concentration of the reactant at time \(t\),
- \([A]_0\) is the initial concentration,
- \(k\) is the rate constant,
- \(t\) is the time.

3. **Finding the Half-Life:**
- The half-life (\(t_{1/2}\)) of a reaction is the time required for the concentration of the reactant to decrease to half of its initial concentration. For a zero-order reaction, we set \([A]_t = \frac{[A]_0}{2}\):
\[ \frac{[A]_0}{2} = [A]_0 - kt_{1/2} \]

4. **Rearranging the Equation:**
- Rearranging the equation gives:
\[ kt_{1/2} = [A]_0 - \frac{[A]_0}{2} \]
- Simplifying this, we find:
\[ kt_{1/2} = \frac{[A]_0}{2} \]

5. **Solving for Half-Life:**
- Now, we can solve for \(t_{1/2}\):
\[ t_{1/2} = \frac{[A]_0}{2k} \]

6. **Substituting Initial Concentration:**
- If we denote the initial concentration as \([R]_0\), we can write:
\[ t_{1/2} = \frac{[R]_0}{2k} \]

### Final Result:
Thus, the half-life of a zero-order reaction in terms of the initial concentration \([R]_0\) is given by:
\[ t_{1/2} = \frac{[R]_0}{2k} \]