If the activation energy of a reaction is 80.9 kJ mol , the fraction of molecules at 700 K, having e — Chemical Kinetics Chemistry Question
Question
If the activation energy of a reaction is 80.9 kJ mol , the fraction of molecules at 700 K, having enough energy to react to form products is e . The value of x is ______. (Rounded off to nearest integer) [Use R = 8.31 J K mol ] -1 -x -1 -1
💡 Solution & Explanation
**Step 1: Identify the formula** The fraction of molecules with sufficient activation energy follows the Arrhenius distribution: $$\text{Fraction} = e^{-E_a/RT}$$ Given that this fraction equals e^(-x), we need to find x. **Step 2: Set up the equation** $$e^{-x} = e^{-E_a/RT}$$ Therefore: $$x = \frac{E_a}{RT}$$ **Step 3: Convert activation energy to consistent units** E_a = 80.9 kJ/mol = 80,900 J/mol **Step 4: Identify given values** - E_a = 80,900 J/mol - R = 8.31 J K⁻¹ mol⁻¹ - T = 700 K **Step 5: Calculate x** $$x = \frac{80,900}{8.31 \times 700}$$ $$x = \frac{80,900}{5,817}$$ $$x = 13.91$$ **Step 6: Round to nearest integer** x ≈ 14.00 Therefore, the answer is 14.00.