Question

A first order reaction is 87.5% complete at the end of 30 minutes. The half-life of the reaction is ............... minute(s). 
 

Hint:

For first-order reactions, the half-life is constant and is independent of the initial concentration. Use the formula \( t_{1/2} = \frac{0.693}{k} \) to calculate it.

Answer 1

Correct Answer: 10.2 - 9.8

Step 1: Understanding the relationship for first-order reactions. 
For a first-order reaction, the relationship between the concentration of reactant at time \( t \) and the half-life \( t_{1/2} \) is given by the equation: \[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \] where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant.

Step 2: Using the information given. 
At the end of 30 minutes, the reaction is 87.5% complete. This means only 12.5% of the original reactant remains. Thus, \( [A] = 0.125[A]_0 \).

Step 3: Using the formula for first-order reactions. 
For first-order reactions, the fraction remaining after time \( t \) is given by: \[ \frac{[A]}{[A]_0} = e^{-kt} \] Taking the natural logarithm: \[ \ln(0.125) = -kt \] Solving for \( k \): \[ \ln(0.125) = -2.079 \Rightarrow k = \frac{2.079}{30} = 0.0693 \, \text{min}^{-1} \]

Step 4: Calculating the half-life. 
The half-life \( t_{1/2} \) for a first-order reaction is given by: \[ t_{1/2} = \frac{0.693}{k} \] Substitute the value of \( k \): \[ t_{1/2} = \frac{0.693}{0.0693} = 10 \, \text{minutes} \]

Step 5: Conclusion. 
Thus, the half-life of the reaction is 10 minutes, and the correct answer is  10.