Question

Half-life of a first order reaction is \(10\) minutes. What is the rate of reaction after \(20\) minutes, if the initial concentration is \(10^{-2}\ \mathrm{M}\)?

• A. \(1.73\times10^{-4}\ \mathrm{M\,min^{-1}}\)
• B. \(1.73\times10^{-2}\ \mathrm{M\,min^{-1}}\)
• C. \(3.46\times10^{-4}\ \mathrm{M\,min^{-1}}\)
• D. \(4.19\times10^{-5}\ \mathrm{M\,min^{-1}}\) \bigskip

Hint:

For a first-order reaction: \[ \boxed{t_{1/2}=\frac{0.693}{k}} \] After each half-life, the concentration becomes half of its previous value.

Answer 1

The Correct Option is A

Concept: First-order: \(t_{1/2} = 0.693/k\). After n half-lives, [A] = [A]₀/2ⁿ. Rate = k[A].

Step 1: \(k = 0.693/10 = 0.0693\) min⁻¹. 20 min = 2 half-lives, [A] = 10⁻²/4 = 2.5×10⁻³ M. Rate = 0.0693 × 2.5×10⁻³ = 1.73×10⁻⁴ M min⁻¹.

Step 2: Write the final answer. \(\boxed{1.73\times10^{-4}\ \mathrm{M\,min^{-1}}}\)