Question

In an enzyme catalyzed first-order reaction, the substrate conversion follows an exponential pattern such that 80% of the substrate is converted in 10 minutes. The first-order rate constant (in min\(^{-1}\)) of the reaction, rounded off to THREE decimal places, is ............

Hint:

To find the rate constant for a first-order reaction, use the equation \( \frac{[A_t]}{[A_0]} = e^{-kt} \) and solve for \( k \).

Answer 1

Correct Answer: 0.16 - 0.17

Step 1: Understanding the first-order reaction equation.
For a first-order reaction, the amount of substrate remaining after time \( t \) is given by: \[ \frac{[A_t]}{[A_0]} = e^{-kt}, \] where \( [A_t] \) is the concentration at time \( t \), \( [A_0] \) is the initial concentration, and \( k \) is the rate constant.

Step 2: Using the information given.
Given that 80% of the substrate is converted in 10 minutes, 20% remains, so: \[ \frac{[A_t]}{[A_0]} = 0.20. \] Substitute into the equation: \[ 0.20 = e^{-k \times 10}. \]

Step 3: Solving for \( k \).
Taking the natural logarithm of both sides: \[ \ln(0.20) = -10k, \] \[ k = - \frac{\ln(0.20)}{10}. \] Using \( \ln(0.20) \approx -1.609 \): \[ k = \frac{1.609}{10} = 0.161. \]

Step 4: Conclusion.
The rate constant \( k \) is \( \boxed{0.161} \, \text{min}^{-1} \).