For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 5 — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

**Step 1: Recall the first-order rate equation** For a first-order reaction: $$t = \frac{2.303}{k} \log\frac{[A]_0}{[A]_t}$$ where t is time, k is the rate constant, [A]₀ is initial concentration, and [A]_t is concentration at time t. **Step 2: Calculate time for 50% completion (t₁/₂)** When 50% is complete, [A]_t = 0.5[A]₀ $$t_{1/2} = \frac{2.303}{k} \log\frac{[A]_0}{0.5[A]_0} = \frac{2.303}{k} \log(2)$$ **Step 3: Calculate time for 75% completion (t₇₅)** When 75% is complete, [A]_t = 0.25[A]₀ $$t_{75} = \frac{2.303}{k} \log\frac{[A]_0}{0.25[A]_0} = \frac{2.303}{k} \log(4)$$ **Step 4: Find the ratio** $$\frac{t_{75}}{t_{1/2}} = \frac{\frac{2.303}{k} \log(4)}{\frac{2.303}{k} \log(2)} = \frac{\log(4)}{\log(2)}$$ **Step 5: Simplify using logarithm properties** $$\frac{\log(4)}{\log(2)} = \frac{\log(2^2)}{\log(2)} = \frac{2\log(2)}{\log(2)} = 2$$ Therefore, the answer is **2.00.**