Question

In a first order reaction, the concentration of the reactant is reduced to 1/8 of the initial concentration in 75 minutes. The \(t_{1/2}\) of the reaction (in minutes) is (\(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60\))

• A. 60.2
• B. 50.2
• C. 25.1
• D. 75.1

Hint:

For first-order reactions: - Integrated rate law: \( \ln([A]_0/[A]_t) = kt \) or \( [A]_t = [A]_0 e^{-kt} \). - Half-life: \( t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k} \). - If concentration reduces to \( (1/2)^n \) of initial, then \( n \) half-lives have passed. Time \( t = n \cdot t_{1/2} \). Here, \( 1/8 = (1/2)^3 \). So, 3 half-lives have passed in 75 minutes. \( 3 \times t_{1/2} = 75 \) minutes. \( t_{1/2} = \frac{75}{3} = 25 \) minutes. This is the quickest way. The discrepancy to 25.1 likely comes from using approximate log values in a multi-step calculation as shown above.

Answer 1

The Correct Option is C

For a first-order reaction, the relationship between concentration and time is given by: \[ \ln\left(\frac{[A]_0}{[A]_t}\right) = kt \text{or} k = \frac{2.
303}{t} \log_{10}\left(\frac{[A]_0}{[A]_t}\right) \] where \( [A]_0 \) is the initial concentration, \( [A]_t \) is the concentration at time \(t\), and \(k\) is the rate constant.
Given that the concentration is reduced to 1/8 of the initial concentration, so \( \frac{[A]_t}{[A]_0} = \frac{1}{8} \), which means \( \frac{[A]_0}{[A]_t} = 8 \).
Time \( t = 75 \) minutes.
\[ k = \frac{2.
303}{75} \log_{10}(8) \] We know \( \log_{10}(8) = \log_{10}(2^3) = 3 \log_{10}(2) \).
Given \( \log_{10} 2 = 0.
30 \).
(Using \( \log_{10} 2 = 0.
3010 \) is more accurate, but problem gives 0.
30) So, \( \log_{10}(8) = 3 \times 0.
30 = 0.
90 \).
\[ k = \frac{2.
303 \times 0.
90}{75} \, \text{min}^{-1} \] The half-life \( t_{1/2} \) for a first-order reaction is given by \( t_{1/2} = \frac{\ln 2}{k} = \frac{0.
693}{k} \).
Or, \( t_{1/2} = \frac{2.
303 \log_{10} 2}{k} \).
\[ t_{1/2} = \frac{2.
303 \times 0.
30}{k} \] Substitute the expression for k: \[ t_{1/2} = \frac{2.
303 \times 0.
30}{\frac{2.
303 \times 0.
90}{75}} = \frac{2.
303 \times 0.
30 \times 75}{2.
303 \times 0.
90} \] Cancel \(2.
303\): \[ t_{1/2} = \frac{0.
30 \times 75}{0.
90} = \frac{0.
30}{0.
90} \times 75 = \frac{1}{3} \times 75 \] \[ t_{1/2} = \frac{75}{3} = 25 \, \text{minutes} \] If we use \( \log_{10} 2 = 0.
30 \), then \( t_{1/2} = 25 \) min.
Option (3) is 25.
1.
This suggests a slightly more precise log value might have been used implicitly by the question setter, or just rounding.
Using \( \ln 2 \approx 0.
693 \): \( k = \frac{\ln 8}{75} = \frac{3 \ln 2}{75} = \frac{\ln 2}{25} \).
\( t_{1/2} = \frac{\ln 2}{k} = \frac{\ln 2}{(\ln 2)/25} = 25 \) minutes.
The result is exactly 25 minutes with either log base.
The 25.
1 option might be a distractor or based on different rounding in an intermediate step if one were to calculate \(k\) first then \(t_{1/2}\).
\(k = \frac{2.
303 \times 0.
90}{75} = \frac{2.
0727}{75} \approx 0.
027636 \).
\(t_{1/2} = \frac{0.
693}{0.
027636} \approx 25.
076 \).
Rounding to one decimal place gives 25.
1 minutes.
So, the calculation of k first and then \(t_{1/2}\) using 0.
693 leads to 25.
1.
This matches option (3).