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}} $.

Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below:
Based on the above data, predict the order of the reaction and write the expression for the rate law.
For a first order decomposition of a certain reaction, rate constant is given by the equation
\(\log k(s⁻¹) = 7.14 - \frac{1 \times 10^4 K}{T}\). The activation energy of the reaction (in kJ mol⁻¹) is (\(R = 8.3 J K⁻¹ mol⁻¹\))
Note: The provided value for R is 8.3. We will use the more precise value R=8.314 J K⁻¹ mol⁻¹ for accuracy, as is standard.