Question

A first-order reaction is 25% complete in 30 minutes. How much time will it take for the reaction to be 75% complete?

• A. 90 min
• B. 60 min
• C. 120 min
• D. 150 min

Hint:

A useful shortcut for first-order reactions: The time for 75% completion (\(t_{75\%}\)) is exactly two half-lives (\(2 \times t_{1/2}\)). However, since 25% is not a half-life, you must use the ratio of logs as shown above.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a first-order reaction, the time required for a certain fraction of the reactant to be consumed is independent of the initial concentration. We can use the integrated rate law to find the rate constant $k$, and then use it to find the time required for any other given completion percentage.
Step 2: Key Formula or Approach:
The integrated rate law for a first-order reaction is: \[ k = \frac{2.303}{t} \log \left( \frac{a}{a - x} \right) \] where $a$ is the initial concentration (assumed 100%) and $x$ is the amount reacted.
Step 3: Detailed Explanation:
Case 1: 25% complete
$t_1 = 30$ min, $a = 100$, $x = 25 \implies a - x = 75$.
\[ k = \frac{2.303}{30} \log \left( \frac{100}{75} \right) = \frac{2.303}{30} \log \left( \frac{4}{3} \right) \] Using logarithmic approximations ($\log 4 \approx 0.602$, $\log 3 \approx 0.477$): \[ \log \left( \frac{4}{3} \right) = \log 4 - \log 3 \approx 0.602 - 0.477 = 0.125 \] So, $k = \frac{2.303 \times 0.125}{30} \text{ min}^{-1}$.
Case 2: 75% complete
Let the time required be $t_2$. Here, $x = 75 \implies a - x = 25$.
\[ t_2 = \frac{2.303}{k} \log \left( \frac{100}{25} \right) = \frac{2.303}{k} \log(4) \] Substitute the expression for $k$ from Case 1: \[ t_2 = \frac{2.303}{\frac{2.303 \times \log(4/3)}{30}} \log(4) \] \[ t_2 = 30 \times \frac{\log(4)}{\log(4/3)} \] Using the values $\log 4 \approx 0.602$ and $\log(4/3) \approx 0.125$: \[ t_2 \approx 30 \times \frac{0.602}{0.125} = 30 \times 4.816 \approx 144.5 \text{ min} \] Alternatively, using the standard approximations $\log 4 = 0.60$ and $\log 3 = 0.48$: \[ \log(4/3) = 0.60 - 0.48 = 0.12 \] \[ t_2 = 30 \times \frac{0.60}{0.12} = 30 \times 5 = 150 \text{ min} \] Given the options, 150 min is the intended standard answer using typical approximations.
Step 4: Final Answer:
The reaction will take approximately 150 minutes to be 75% complete. The correct option is (D).