When initial concentration of the reactant is doubled, the half-life period of a zero order reaction

Options

A. is tripled
B. is doubled
C. is halved
D. remains unchanged

Explanation

To solve the problem regarding the effect of doubling the initial concentration on the half-life period of a zero-order reaction, we can follow these steps:

### Step 1: Understand the Half-Life Formula for Zero-Order Reactions
The half-life (T_half) of a zero-order reaction is given by the formula:
\[ T_{1/2} = \frac{C_0}{2k} \]
where:
- \( C_0 \) = initial concentration of the reactant
- \( k \) = rate constant of the reaction

### Step 2: Determine the New Half-Life When Initial Concentration is Doubled
If the initial concentration \( C_0 \) is doubled, the new concentration becomes \( 2C_0 \). We can calculate the new half-life \( T_{1/2}' \) using the same formula:
\[ T_{1/2}' = \frac{2C_0}{2k} \]

### Step 3: Simplify the New Half-Life Expression
Now, simplify the expression for the new half-life:
\[ T_{1/2}' = \frac{2C_0}{2k} = \frac{C_0}{k} \]

### Step 4: Relate the New Half-Life to the Original Half-Life
From the original half-life formula, we know:
\[ T_{1/2} = \frac{C_0}{2k} \]
Now, we can express the new half-life in terms of the original half-life:
\[ T_{1/2}' = 2 \times \frac{C_0}{2k} = 2 \times T_{1/2} \]

### Conclusion
Thus, when the initial concentration of the reactant is doubled, the half-life period of a zero-order reaction also doubles.

### Final Answer
The half-life period of a zero-order reaction when the initial concentration is doubled is:
\[ T_{1/2}' = 2 \times T_{1/2} \]