In a reaction, the time required to complete half of the reaction was found to increase 16 times when the initial concentration of the reactant was reduced to 1/4th. What is the order of the reaction?

Answer

To determine the order of the reaction based on the information provided, we can follow these steps:

### Step 1: Understand the relationship between half-life and concentration
The half-life (\( T_{1/2} \)) of a reaction is related to the initial concentration (\( [A_0] \)) and the order of the reaction (\( n \)). The general relationship can be expressed as:
\[ T_{1/2} \propto [A_0]^{(1-n)} \]

### Step 2: Set up the equations based on the problem statement
According to the problem, when the initial concentration is reduced to \( \frac{1}{4} \) of its original value, the half-life increases 16 times. Let's denote:
- \( T_{1/2} \) as the original half-life at concentration \( [A_0] \)
- \( T'_{1/2} = 16 T_{1/2} \) as the new half-life at concentration \( [A_0]/4 \)

Using the relationship from Step 1, we can write two equations:
1. For the original concentration:
\[ T_{1/2} = k [A_0]^{(1-n)} \]
2. For the new concentration:
\[ T'_{1/2} = k \left(\frac{[A_0]}{4}\right)^{(1-n)} = k [A_0]^{(1-n)} \cdot \left(\frac{1}{4}\right)^{(1-n)} \]

### Step 3: Relate the two half-lives
From the equations, we have:
\[ 16 T_{1/2} = k [A_0]^{(1-n)} \cdot \left(\frac{1}{4}\right)^{(1-n)} \]
Substituting \( T_{1/2} \) from the first equation:
\[ 16 (k [A_0]^{(1-n)}) = k [A_0]^{(1-n)} \cdot \left(\frac{1}{4}\right)^{(1-n)} \]

### Step 4: Simplify the equation
We can cancel \( k [A_0]^{(1-n)} \) from both sides (assuming \( k \) and \( [A_0]^{(1-n)} \) are not zero):
\[ 16 = \left(\frac{1}{4}\right)^{(1-n)} \]

### Step 5: Rewrite \( \frac{1}{4} \) as a power of 4
We know that \( \frac{1}{4} = 4^{-1} \), so we can rewrite the equation:
\[ 16 = 4^{-1(1-n)} \]
Since \( 16 = 4^2 \), we can equate the exponents:
\[ 2 = -1(1-n) \]

### Step 6: Solve for \( n \)
Expanding the equation gives:
\[ 2 = -1 + n \]
\[ n = 3 \]

### Conclusion
The order of the reaction is **3**.

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