Question

In a first order reaction, the concentration of reactant decreases from \(400\ \text{moles lit}^{-1}\) to \(50\ \text{moles lit}^{-1}\) in \(7.5\times10^{3}\ \text{s}\). The rate constant of the reaction is (approximately):

• A. $1 \times 10^{-2}\text{ s}^{-1}$
• B. $2.5 \times 10^{-3}\text{ s}^{-1}$
• C. $1 \times 10^{-5}\text{ s}^{-1}$
• D. $2.77 \times 10^{-4}\text{ s}^{-1}$

Hint:

For first-order reactions, instead of using the log formula, check if the concentration ratio $\frac{[A]_0}{[A]_t}$ is a power of 2 ($2^N$). If yes, then the elapsed time is simply $N \times t_{1/2}$, which allows you to find $t_{1/2}$ and $k$ within seconds!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to calculate the rate constant ($k$) for a first-order reaction where the reactant concentration falls from an initial concentration of $400 \text{ M}$ to $50 \text{ M}$ in a given time interval of $7.5 \times 10^3\text{ s}$.


Step 2: Key Formula or Approach:
We can use the half-life concept for a first-order reaction to solve this quickly:
A concentration decrease of a first-order reactant over successive half-lives ($t_{1/2}$) follows:
\[ [A]_0 \xrightarrow{t_{1/2}} \frac{[A]_0}{2} \xrightarrow{t_{1/2}} \frac{[A]_0}{4} \xrightarrow{t_{1/2}} \frac{[A]_0}{8} \]
Once the half-life is found, the rate constant is given by:
\[ k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{t_{1/2}} \]


Step 3: Detailed Explanation:
Let us observe the concentration reduction:
- Initial concentration, $[A]_0 = 400 \text{ mol L}^{-1}$
- Final concentration, $[A]_t = 50 \text{ mol L}^{-1}$
Let us determine the number of half-lives ($N$) required for this change:
\[ 400 \xrightarrow{\text{1st } t_{1/2}} 200 \xrightarrow{\text{2nd } t_{1/2}} 100 \xrightarrow{\text{3rd } t_{1/2}} 50 \]
This means exactly $3$ half-lives have passed in the given time interval:
\[ 3 \times t_{1/2} = 7.5 \times 10^3\text{ s} \]
\[ t_{1/2} = 2.5 \times 10^3\text{ s} \]
Now, let us calculate the rate constant ($k$):
\[ k = \frac{0.693}{t_{1/2}} = \frac{0.693}{2.5 \times 10^3\text{ s}} \]
\[ k = 0.2772 \times 10^{-3}\text{ s}^{-1} = 2.772 \times 10^{-4}\text{ s}^{-1} \]
This matches option (D).


Step 4: Final Answer:
The correct option is (D).