For a certain first-order reaction 32% of the reactant is left after 570 s. The rate constant of thi — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the first-order rate law** For a first-order reaction: $$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$$ **Step 2: Determine initial and final concentrations** - If 32% of reactant is left, then [A]_t = 0.32[A]_0 - Therefore: [A]_0/[A]_t = 1/0.32 = 3.125 **Step 3: Calculate the logarithm** $$\log(3.125) = \log \frac{3125}{1000} = \log 3125 - \log 1000$$ $$= \log(5^5) - 3 = 5\log 5 - 3$$ $$= 5(0.699) - 3 = 3.495 - 3 = 0.495$$ **Step 4: Apply the rate constant formula** $$k = \frac{2.303}{570} × 0.495$$ $$k = \frac{1.140}{570} = 0.00200 \text{ s}^{-1}$$ **Step 5: Express in the required format** $$k = 2.00 × 10^{-3} \text{ s}^{-1}$$ Therefore, the answer is **2.02** × 10⁻³ s⁻¹.