If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentrat — Chemical Kinetics Chemistry Question
Question
If formation of compound [B] follows the first order of kinetics and after 70 minutes the concentration of [A] was found to be half of its initial concentration. Then the rate constant of the reaction is x × 10 s . The value of x is______. (Nearest Integer) –6 –1
💡 Solution & Explanation
**Step 1: Identify the reaction order and given information** - First-order reaction - Half-life (t₁/₂) = 70 minutes - Need to find rate constant k in units of s⁻¹ **Step 2: Apply the half-life formula for first-order reactions** For a first-order reaction: $$t_{1/2} = \frac{0.693}{k}$$ **Step 3: Solve for k in min⁻¹** $$k = \frac{0.693}{t_{1/2}} = \frac{0.693}{70 \text{ min}}$$ $$k = 0.0099 \text{ min}^{-1}$$ **Step 4: Convert k from min⁻¹ to s⁻¹** $$k = \frac{0.0099 \text{ min}^{-1}}{60 \text{ s/min}}$$ $$k = \frac{0.0099}{60} \text{ s}^{-1}$$ $$k = 1.65 × 10^{-4} \text{ s}^{-1}$$ **Step 5: Express in the form x × 10⁻⁶ s⁻¹** $$1.65 × 10^{-4} = 165 × 10^{-6}$$ Therefore, **x = 165**.