Question

A first-order reaction has rate constant \(1 \times 10^{-2} \, \text{s}^{-1}\). What time will it take for 20 g of reactant to reduce to 5 g?

• A. 346.5 s
• B. 238.6 s
• C. 138.6 s
• D. 693.0 s

Hint:

For first-order reactions, use the integrated rate law to calculate the time taken for a specific concentration to be reached.

Answer 1

The Correct Option is C

Step 1: Using the first-order reaction equation.
The integrated rate law for a first-order reaction is: \[ \ln \left(\frac{[A_0]}{[A_t]}\right) = kt \] Where: - \( [A_0] = 20 \, \text{g} \) (initial concentration) - \( [A_t] = 5 \, \text{g} \) (concentration at time \(t\)) - \( k = 1 \times 10^{-2} \, \text{s}^{-1} \) (rate constant)
Step 2: Calculation.
Substitute the values into the rate law: \[ \ln \left(\frac{20}{5}\right) = 1 \times 10^{-2} \times t \] \[ \ln(4) = 1 \times 10^{-2} \times t \] \[ 1.386 = 1 \times 10^{-2} \times t \] Solving for \(t\): \[ t = \frac{1.386}{1 \times 10^{-2}} = 138.6 \, \text{s} \]
Step 3: Conclusion.
The time required is 138.6 s.