Find the ratio of `t_(75%)` and `t_(50%)` of first order reaction?

Answer

To find the ratio of \( t_{75\%} \) and \( t_{50\%} \) for a first-order reaction, we can follow these steps:

### Step 1: Understand the Definitions
- \( t_{50\%} \) is the time taken for 50% of the reactant to be consumed.
- \( t_{75\%} \) is the time taken for 75% of the reactant to be consumed.

### Step 2: Use the First-Order Reaction Formula
For a first-order reaction, the relationship between the rate constant \( k \), initial concentration \( A_0 \), and concentration at time \( t \) (denoted as \( A \)) is given by the equation:
\[ k = \frac{2.303}{t} \log \frac{A_0}{A} \]

### Step 3: Calculate \( t_{50\%} \)
- When 50% of the reactant is consumed, the concentration remaining \( A \) is \( 0.5 A_0 \).
- Thus, we can write:
\[ t_{50\%} = \frac{2.303}{k} \log \frac{A_0}{0.5 A_0} = \frac{2.303}{k} \log 2 \]

### Step 4: Calculate \( t_{75\%} \)
- When 75% of the reactant is consumed, the concentration remaining \( A \) is \( 0.25 A_0 \).
- Thus, we can write:
\[ t_{75\%} = \frac{2.303}{k} \log \frac{A_0}{0.25 A_0} = \frac{2.303}{k} \log 4 \]

### Step 5: Find the Ratio \( \frac{t_{75\%}}{t_{50\%}} \)
Now, we can find the ratio of \( t_{75\%} \) to \( t_{50\%} \):
\[ \frac{t_{75\%}}{t_{50\%}} = \frac{\frac{2.303}{k} \log 4}{\frac{2.303}{k} \log 2} \]
The \( \frac{2.303}{k} \) cancels out:
\[ = \frac{\log 4}{\log 2} \]

### Step 6: Simplify the Logarithm
Using the property of logarithms, we know that:
\[ \log 4 = \log (2^2) = 2 \log 2 \]
Thus:
\[ \frac{t_{75\%}}{t_{50\%}} = \frac{2 \log 2}{\log 2} = 2 \]

### Final Answer
The ratio \( \frac{t_{75\%}}{t_{50\%}} \) is \( 2 \).

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