The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant — Chemical Kinetics Chemistry Question
Question
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 × 10 s and the activation energy E = 11.488 kJ mol , the rate constant at 200 K is ______ × 10 s . (Round off to the Nearest Integer). (Given: R = 8.314 J mol K ) -3 -1 a -1 -5 -1 -1 -1
💡 Solution & Explanation
**Step 1: Identify the applicable formula** Use the Arrhenius equation in two-temperature form: $$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ **Step 2: List given values** - k₁ = 1.0 × 10⁻³ s⁻¹ at T₁ = 300 K - T₂ = 200 K (find k₂) - Eₐ = 11,488 J mol⁻¹ - R = 8.314 J mol⁻¹ K⁻¹ **Step 3: Calculate temperature term** $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{200} = 0.003333 - 0.005 = -0.001667 \text{ K}^{-1}$$ **Step 4: Calculate the exponent** $$\frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{11,488}{8.314} × (-0.001667) = 1382.1 × (-0.001667) = -2.305$$ **Step 5: Solve for k₂** $$\ln\left(\frac{k_2}{k_1}\right) = -2.305$$ $$\frac{k_2}{k_1} = e^{-2.305} = 0.0997$$ $$k_2 = 1.0 × 10^{-3} × 0.0997 = 9.97 × 10^{-5} \text{ s}^{-1}$$ **Step 6: Express in required form** $$k_2 = 9.97 × 10^{-5} ≈