Which of the following is correct representation of the variation of half - life with initial concentration of a zero order reaction ?

Options

A.
B.
C.
D.

Answer

To solve the problem regarding the variation of half-life with initial concentration in a zero-order reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Zero-Order Reaction Equation**:
The rate law for a zero-order reaction can be expressed as:
\[ [A]_t = [A]_0 - kt \]
where:
- \([A]_t\) is the concentration at time \(t\),
- \([A]_0\) is the initial concentration,
- \(k\) is the rate constant,
- \(t\) is time.

2. **Determine the Half-Life Expression**:
The half-life (\(t_{1/2}\)) for a zero-order reaction is the time required for the concentration of the reactant to decrease to half of its initial concentration. At half-life:
\[ [A]_{t_{1/2}} = \frac{[A]_0}{2} \]

3. **Set Up the Equation for Half-Life**:
Substitute \([A]_{t_{1/2}}\) into the zero-order reaction equation:
\[ \frac{[A]_0}{2} = [A]_0 - kt_{1/2} \]

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

5. **Solve for Half-Life**:
Now, we can express the half-life in terms of the initial concentration:
\[ t_{1/2} = \frac{[A]_0}{2k} \]

6. **Analyze the Relationship**:
From the equation \(t_{1/2} = \frac{[A]_0}{2k}\), we can see that:
- \(t_{1/2}\) is directly proportional to the initial concentration \([A]_0\) since \(2k\) is a constant.

7. **Graphical Representation**:
Since \(t_{1/2}\) increases linearly with \([A]_0\), the graph of half-life (\(t_{1/2}\)) versus initial concentration (\([A]_0\)) will be a straight line.

### Conclusion:
The correct representation of the variation of half-life with initial concentration of a zero-order reaction is a straight line graph, indicating that half-life is directly proportional to the initial concentration.