The general expression for half life period of an nth order reaction
`t_(1//2)=(2^(n-1)-1)/(k(n-1)a^(n-1))` is

Options

A. valid for all reactions
B. not valid for fractional or negative order reactions
C. not valid for first order reactions
D. not valid for zero order reaction

Answer

To derive the general expression for the half-life period of an nth order reaction, we start with the definition of half-life and the integrated rate laws for different orders of reactions.

### Step 1: Understanding Half-Life
The half-life (\( t_{1/2} \)) of a reaction is the time required for the concentration of a reactant to decrease to half of its initial concentration.

### Step 2: Rate Law for nth Order Reaction
For an nth order reaction, the rate of reaction can be expressed as:
\[ -\frac{d[A]}{dt} = k[A]^n \]
where \( [A] \) is the concentration of the reactant, \( k \) is the rate constant, and \( n \) is the order of the reaction.

### Step 3: Integrated Rate Law for nth Order Reaction
The integrated rate law for an nth order reaction is given by:
\[ \frac{1}{[A]^{n-1}} = \frac{1}{[A_0]^{n-1}} + (n-1)kt \]
where \( [A_0] \) is the initial concentration of the reactant.

### Step 4: Finding Half-Life Expression
At half-life, the concentration of the reactant is half of its initial concentration:
\[ [A] = \frac{[A_0]}{2} \]
Substituting this into the integrated rate law:
\[ \frac{1}{(\frac{[A_0]}{2})^{n-1}} = \frac{1}{[A_0]^{n-1}} + (n-1)kt_{1/2} \]

### Step 5: Simplifying the Equation
This can be simplified as:
\[ \frac{2^{n-1}}{[A_0]^{n-1}} = \frac{1}{[A_0]^{n-1}} + (n-1)kt_{1/2} \]
Multiplying through by \( [A_0]^{n-1} \):
\[ 2^{n-1} = 1 + (n-1)k[A_0]^{n-1}t_{1/2} \]

### Step 6: Isolating \( t_{1/2} \)
Rearranging gives:
\[ (n-1)k[A_0]^{n-1}t_{1/2} = 2^{n-1} - 1 \]
Thus,
\[ t_{1/2} = \frac{2^{n-1} - 1}{k(n-1)[A_0]^{n-1}} \]

### Conclusion
The general expression for the half-life period of an nth order reaction is:
\[ t_{1/2} = \frac{2^{n-1} - 1}{k(n-1)[A_0]^{n-1}} \]