A substance decomposes according to zero order kinetics. If the rate constant is k and initial concentration is ‘a’ then half life period of the reaction will be:

Options

A. `1//ak`
B. `0.693//k`
C. `k//2a`
D. `a//2k`

Answer

To find the half-life period of a substance that decomposes according to zero-order kinetics, we can follow these steps:

### Step 1: Understand Zero-Order Kinetics
In zero-order kinetics, the rate of reaction is constant and does not depend on the concentration of the reactant. The rate law can be expressed as:
\[ -\frac{d[A]}{dt} = k \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 2: Rearrange the Rate Law
We can rearrange the rate law to express it in terms of concentration:
\[ d[A] = -k \, dt \]

### Step 3: Integrate Both Sides
To find the relationship between concentration and time, we integrate both sides. The integration limits for concentration will be from the initial concentration \( a \) to the concentration at time \( t \) (which we denote as \([A]_t\)), and for time from 0 to \( t \):
\[ \int_{a}^{[A]_t} d[A] = -k \int_{0}^{t} dt \]
This gives us:
\[ [A]_t - a = -kt \]
or rearranging it,
\[ a - [A]_t = kt \]

### Step 4: Define Half-Life
The half-life (\( t_{1/2} \)) is the time required for the concentration to decrease to half of its initial value. Therefore, we set \([A]_t = \frac{a}{2}\):
\[ a - \frac{a}{2} = kt_{1/2} \]

### Step 5: Solve for Half-Life
Substituting \(\frac{a}{2}\) into the equation gives:
\[ \frac{a}{2} = kt_{1/2} \]
Now, we can solve for \( t_{1/2} \):
\[ t_{1/2} = \frac{a}{2k} \]

### Final Answer
Thus, the half-life period of the reaction is:
\[ t_{1/2} = \frac{a}{2k} \]

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