Question

\(A \rightarrow B\) is a first order reaction. What is the concentration of reactant A after 28.8 seconds if the initial concentration of A is 0.1 M and rate constant \(k = 0.0693\,s^{-1}\)? (Given \(e = 2.718\))

• A. \(0.1/e\) M
• B. \(0.1/e^2\) M
• C. \(0.1/e^3\) M
• D. \(0.1/e^4\) M

Hint:

For first-order reactions, always remember: \([A] = [A]_0 e^{-kt}\), so \(kt\) directly gives the power of \(e\).

Answer 1

The Correct Option is B

Step 1: Understanding first order reaction integrated rate law.
For a first order reaction, the concentration of reactant at time \(t\) is given by: \[ [A] = [A]_0 e^{-kt} \] where \([A]_0\) is initial concentration, \(k\) is rate constant, and \(t\) is time.

Step 2: Substituting given values.
We are given: \[ [A]_0 = 0.1\,M,\quad k = 0.0693\,s^{-1},\quad t = 28.8\,s \]

Step 3: Calculating exponent term.
\[ kt = 0.0693 \times 28.8 \] \[ kt = 1.994 \approx 2 \] So, \[ e^{-kt} \approx e^{-2} \]

Step 4: Calculating concentration.
\[ [A] = 0.1 \times e^{-2} = \frac{0.1}{e^2} \]

Step 5: Final interpretation.
The concentration decreases exponentially with time for a first-order reaction, and here it reduces to \(1/e^2\) times the initial value.
Final Answer: \[ \boxed{\frac{0.1}{e^2}\,M} \]