Question

In the second-order reaction \( 2A \to B \), the initial concentration of A is 1.0 M and after 30 minutes, the concentration of A is 0.5 M. The rate constant of the reaction is ............ L/mol/h (round off to 2 decimal places)

Hint:

For a second-order reaction, use the rate law \( \frac{1}{[A]} - \frac{1}{[A]_0} = k t \) to solve for the rate constant.

Answer 1

Correct Answer: 1.99

Step 1: Second-order reaction rate law.
For a second-order reaction, the rate law is given by: \[ \frac{1}{[A]} - \frac{1}{[A]_0} = k t \] where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( k \) is the rate constant.
Step 2: Apply the given values.
We are given: \[ [A]_0 = 1.0 \, \text{M}, \quad [A] = 0.5 \, \text{M}, \quad t = 30 \, \text{minutes} = 0.5 \, \text{hours} \] Substitute into the rate law: \[ \frac{1}{0.5} - \frac{1}{1.0} = k \times 0.5 \] Simplify: \[ 2 - 1 = k \times 0.5 \] \[ k = \frac{1}{0.5} = 0.693 \, \text{L/mol/h} \] Step 3: Conclusion.
Thus, the rate constant is \( \boxed{0.693} \, \text{L/mol/h} \).