A first-order reaction is 25% complete in 30 minutes. How much time will it take for the reaction to be 75% complete?
145 min
120 min
To solve the problem, we use the first-order reaction kinetics formula and the concept of half-life.
1. Use the first-order reaction formula:
$ \ln \frac{[A]_0}{[A]} = kt $
2. Given data:
- 25% complete in 30 minutes means 75% remains:
$ \frac{[A]}{[A]_0} = 0.75 $
- We want to find time $t$ when 75% is complete, meaning 25% remains:
$ \frac{[A]}{[A]_0} = 0.25 $
3. Calculate the rate constant $k$:
$ k = \frac{1}{t} \ln \frac{[A]_0}{[A]} = \frac{1}{30} \ln \frac{1}{0.75} = \frac{1}{30} \ln \frac{4}{3} $
4. Calculate time for 75% completion:
$ t = \frac{1}{k} \ln \frac{1}{0.25} = \frac{1}{k} \ln 4 $
5. Substitute $k$:
$ t = 30 \times \frac{\ln 4}{\ln \frac{4}{3}} $
6. Approximate values:
$ \ln 4 \approx 1.386 $
$ \ln \frac{4}{3} \approx 0.2877 $
$ t \approx 30 \times \frac{1.386}{0.2877} = 30 \times 4.82 = 144.6 \, \text{minutes} $
7. Interpretation:
The time for 75% completion is about 145 minutes.
Final Answer:
The time required for 75% completion is approximately $ {145\, \text{minutes}} $.
For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?