For a first order reaction A B, the rate constant, k = 5.5 10 s . The time required for 67% completi — Chemical Kinetics Chemistry Question
Question
For a first order reaction A B, the rate constant, k = 5.5 10 s . The time required for 67% completion of reaction is x 10 times the half life of reaction. The value of x is _____ (Nearest integer) [Given: log 3 = 0.4771] –14 –1 –1
💡 Solution & Explanation
# Solution: First Order Reaction Problem **Step 1: Calculate the half-life** For a first-order reaction: $$t_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.5 \times 10^{-14}} = 1.26 \times 10^{13} \text{ s}$$ **Step 2: Calculate time for 67% completion** For a first-order reaction: $$t = \frac{2.303}{k} \log\frac{[A]_0}{[A]_t}$$ When 67% is complete, 33% remains: $$t = \frac{2.303}{5.5 \times 10^{-14}} \log\frac{100}{33}$$ $$t = \frac{2.303}{5.5 \times 10^{-14}} \log(3.03)$$ $$\log(3.03) = \log 3 + \log(1.01) \approx 0.4771 + 0.0043 = 0.4814$$ $$t = \frac{2.303 \times 0.4814}{5.5 \times 10^{-14}} = 2.02 \times 10^{13} \text{ s}$$ **Step 3: Find the relationship** $$\frac{t}{t_{1/2}} = \frac{2.02 \times 10^{13}}{1.26 \times 10^{13}} = 1.60 = 1.6 \times 10^1$$ Therefore, x = **16.00**