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 given information about the half-life and concentration, 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 formula for half-life for a reaction of order \(n\) is:

\[ t_{1/2} \propto A_0^{(1-n)} \]

### Step 2: Set up the equations for two scenarios
1. Let \(t_1\) be the half-life when the initial concentration is \(A_0\).
\[ t_1 \propto A_0^{(1-n)} \]

2. Let \(t_2\) be the half-life when the concentration is reduced to \(\frac{A_0}{4}\).
\[ t_2 \propto \left(\frac{A_0}{4}\right)^{(1-n)} \]

### Step 3: Relate the two half-lives
According to the problem, \(t_2\) is 16 times \(t_1\):
\[ t_2 = 16 t_1 \]

### Step 4: Substitute the expressions for half-lives
Substituting the expressions we have:
\[ \left(\frac{A_0}{4}\right)^{(1-n)} = 16 \cdot A_0^{(1-n)} \]

### Step 5: Simplify the equation
We can simplify this equation:
\[ \frac{A_0^{(1-n)}}{4^{(1-n)}} = 16 \cdot A_0^{(1-n)} \]

Dividing both sides by \(A_0^{(1-n)}\) (assuming \(A_0 \neq 0\)):
\[ \frac{1}{4^{(1-n)}} = 16 \]

### Step 6: Express 16 in terms of base 4
We know that \(16 = 4^2\), so we can rewrite the equation as:
\[ \frac{1}{4^{(1-n)}} = 4^2 \]

### Step 7: Combine the powers
This gives us:
\[ 4^{-(1-n)} = 4^2 \]

### Step 8: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal:
\[ -(1-n) = 2 \]

### Step 9: Solve for \(n\)
Rearranging gives us:
\[ 1 - n = -2 \implies n = 3 \]

### Conclusion
The order of the reaction is \(n = 3\).