### Related Formula log k = log A - fracE_a2.303 R T$\log k = \log A - \frac{E_a}{2.303 R T}$ textSlope of the line = -fracE_a2.303 R implies |textSlope| propto E_a$\text{Slope of the line} = -\frac{E_a}{2.303 R} \implies |\text{Slope}| \propto E_a$ ### Core Logic From the given graph, we look at the steepness (magnitude of the negative slope) of lines 1, 2, and 3: - Line 2 is the steepest, meaning it has the largest slope magnitude. - Line 1 has an intermediate slope. - Line 3 is the flattest, indicating the smallest slope magnitude. Since the activation energy E_a$E_a$ is directly proportional to the magnitude of this slope: |textSlope_2| > |textSlope_1| > |textSlope_3| implies E_a2 > E_a1 > E_a3$|\text{Slope}_2| > |\text{Slope}_1| > |\text{Slope}_3| \implies E_{a2} > E_{a1} > E_{a3}$ ### Pattern Recognition In Arrhenius coordinates, steepness equals barriers. A steeper line means the reaction rate is highly sensitive to temperature because it has a higher activation energy (E_a$E_a$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Chemical Kinetics