Question

Find the rate constant of a first order reaction A \(\rightarrow\) B reaction where \(\text{t}_{1/2} = 10 \text{ minutes}\) at \(300 \text{ K}\) is (Given log 2 = 0.3)

• A. \(1.15 \times 10^{-1} \text{ s}^{-1}\)
• B. \(1.15 \times 10^{-3} \text{ s}^{-1}\)
• C. \(2.15 \times 10^{-1} \text{ s}^{-1}\)
• D. \(2.15 \times 10^{-3} \text{ s}^{-1}\)

Hint:

Always glance at the units of the options before starting calculations in chemical kinetics. Failing to convert time units (minutes to seconds) is a classic trap.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first-order reaction, the half-life (\(t_{1/2}\)) is a constant value that does not depend on the initial concentration of the reactant. It is directly related to the rate constant (\(k\)) of the reaction. Because the options are provided in units of seconds inverse (\(\text{s}^{-1}\)), a unit conversion from minutes to seconds is required.
Step 2: Key Formula or Approach:
The mathematical relationship between the rate constant (\(k\)) and half-life (\(t_{1/2}\)) for a first-order reaction is:
\[ t_{1/2} = \frac{\ln 2}{k} = \frac{2.303 \log 2}{k} \]
Rearranging to solve for \(k\):
\[ k = \frac{2.303 \log 2}{t_{1/2}} \]
Step 3: Detailed Explanation:
The given values are:
\(t_{1/2} = 10 \text{ minutes}\)
\(\log 2 = 0.3\)
First, convert the half-life from minutes to seconds to match the units in the options:
\[ t_{1/2} = 10 \text{ minutes} \times 60 \text{ seconds/minute} = 600 \text{ seconds} \]
Now, substitute the values into the rearranged formula:
\[ k = \frac{2.303 \times 0.3}{600} \]
Multiply the numerator:
\[ 2.303 \times 0.3 = 0.6909 \]
Now perform the division:
\[ k = \frac{0.6909}{600} \]
To make the calculation easier, express the numbers in scientific notation:
\[ k = \frac{6.909 \times 10^{-1}}{6 \times 10^2} \]
\[ k = \left(\frac{6.909}{6}\right) \times 10^{-1 - 2} \]
\[ k = 1.1515 \times 10^{-3} \text{ s}^{-1} \]
Rounding to three significant figures, we get \(1.15 \times 10^{-3} \text{ s}^{-1}\).
Step 4: Final Answer:
The rate constant is approximately \(1.15 \times 10^{-3} \text{ s}^{-1}\).