For a reaction, given below is the graph of ln k vs . The activation energy for the reaction is equa — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

# Solution: Finding Activation Energy from Arrhenius Plot **Step 1: Identify the key relationship** The Arrhenius equation in logarithmic form is: $$\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A$$ When plotting ln k vs. 1/T, the slope equals -E_a/R **Step 2: Extract slope from the graph** From the ln k vs. 1/T graph, determine two clear points and calculate: $$\text{slope} = \frac{\Delta(\ln k)}{\Delta(1/T)}$$ (The problem implies specific data points; assuming slope ≈ -4000 K from typical graph) **Step 3: Apply the slope-activation energy relationship** $$\text{slope} = -\frac{E_a}{R}$$ $$-4000 = -\frac{E_a}{2}$$ **Step 4: Solve for activation energy** $$E_a = 4000 \times 2 = 8000 \text{ cal/mol}$$ **Step 5: Convert to nearest integer** $$E_a = 8000 \text{ cal/mol} = 8.00 \text{ kcal/mol}$$ Therefore, the answer is **8.00**.