It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant get — Chemical Kinetics Chemistry Question
Question
It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be _______ kJ mol . [nearest integer] (Given In10 = 2.3, R = 8.3 J K mol , log 2 = 0.30) –1 –1 -1
💡 Solution & Explanation
**Step 1: Use the Arrhenius equation for two temperatures** When rate constant doubles with a 9 K temperature rise: $$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ Given: k₂ = 2k₁, so k₂/k₁ = 2 **Step 2: Calculate ln(2)** $$\ln(2) = 2.303 \times \log(2) = 2.303 \times 0.30 = 0.691$$ **Step 3: Set up temperature values** - T₁ = 300 K - T₂ = 309 K (rise of 9 K) **Step 4: Calculate the temperature term** $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{309}$$ $$= \frac{309 - 300}{300 \times 309} = \frac{9}{92,700} = 9.71 \times 10^{-5} \text{ K}^{-1}$$ **Step 5: Solve for activation energy** $$0.691 = \frac{E_a}{8.3} \times 9.71 \times 10^{-5}$$ $$E_a = \frac{0.691 \times 8.3}{9.71 \times 10^{-5}} = \frac{5.736}{9.71 \times 10^{-5}}$$ $$E_a = 59,092 \text{ J/mol} = 59.1 \text{ kJ/mol}$$ Therefore, the answer is **59.00** kJ/mol.