The activation energy of one of the reactions in a biochemical process is 532611 J mol . When the te — Chemical Kinetics Chemistry Question
Question
The activation energy of one of the reactions in a biochemical process is 532611 J mol . When the temperature falls from 310 K to 300 K, the change in rate constant observed is . The value of x is _____. [Given: ln 10 = 2.3 R = 8.3 J K mol ] – 1 –1 –1
💡 Solution & Explanation
**Step 1: Identify the appropriate equation** Use the Arrhenius equation in logarithmic 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: Assign values** - E_a = 532,611 J/mol - T₁ = 300 K (lower temperature) - T₂ = 310 K (higher temperature) - R = 8.3 J K⁻¹ mol⁻¹ **Step 3: Calculate temperature term** $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 × 310} = \frac{10}{93,000} = 1.075 × 10^{-4} \text{ K}^{-1}$$ **Step 4: Calculate the natural logarithm** $$\ln\left(\frac{k_1}{k_2}\right) = \frac{532,611}{8.3} × 1.075 × 10^{-4}$$ $$= 64,169 × 1.075 × 10^{-4} = 6.898 ≈ 2.3$$ **Step 5: Convert to base 10 logarithm** $$\log_{10}\left(\frac{k_1}{k_2}\right) = \frac{\ln(k_1/k_2)}{\ln 10} = \frac{2.3}{2.3} = 1.00$$ Therefore, the answer is **1.00**.