Question

For a first-order reaction, the concentration of a reactant decreases from 0.8 M to 0.4 M in 20 minutes. What is the half-life of the reaction? (Use $\ln 2 = 0.693$)

• A. 20 min
• B. 10 min
• C. 40 min
• D. 15 min

Hint:

For first-order reactions, use the half-life formula \( t_{1/2} = \frac{\ln 2}{k} \), where \( k \) is found from concentration changes over time.

Answer 1

The Correct Option is A

Given: \[ [\text{A}]_0 = 0.8 \, \text{M}, \quad [\text{A}] = 0.4 \, \text{M}, \quad t = 20 \, \text{min}, \quad \ln 2 = 0.693 \] Step 1: Formula for First-Order Reaction
For a first-order reaction: \[ k = \frac{1}{t} \ln \left( \frac{[\text{A}]_0}{[\text{A}]} \right) \] The half-life is: \[ t_{1/2} = \frac{\ln 2}{k} \] Step 2: Calculate Rate Constant
Substitute the values: \[ k = \frac{1}{20} \ln \left( \frac{0.8}{0.4} \right) = \frac{1}{20} \ln 2 = \frac{0.693}{20} \] Step 3: Calculate Half-Life
\[ t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{\frac{0.693}{20}} = 20 \, \text{min} \] Thus, the half-life is: \[ \boxed{20 \, \text{min}} \]