Michael S. answered • 14d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
The whole question is really asking you to recognize which two quantities make the Arrhenius equation into a straight line. Start from
k = A e-Ea/RT
and take the natural log of both sides:
ln k = (-Ea/R)(1/T) + ln A
Compare that with y = mx + b. The y variable is ln k, the x variable is 1/T, the slope is -Ea/R, and the intercept is ln A. So you plot ln k against 1/T, never k against T.
The two points, as (1/T, ln k):
T = 325 K: 1/T = 3.08 × 10-3 K-1, ln k = ln(0.400) = -0.916
T = 525 K: 1/T = 1.90 × 10-3 K-1, ln k = ln(0.841) = -0.173
Rise, run, slope:
rise = -0.173 - (-0.916) = 0.743
run = 1.90 × 10-3 - 3.08 × 10-3 = -1.17 × 10-3 K-1
slope = 0.743 / (-1.17 × 10-3) = -634 K
Then, since slope = -Ea/R:
Ea = -(slope)(R) = (634 K)(8.314 J/mol·K) = 5.27 × 103 J/mol ≈ 5.27 kJ/mol
Two things worth carrying forward. First, the instruction about three significant figures is not boilerplate. The run is a difference between two small and fairly close numbers, so rounding 1/325 to 3.1 × 10-3 early shifts the slope by several kelvin and the activation energy with it. Keep the reciprocals to at least three figures and round only at the end. This is the same failure mode as subtracting two similar temperatures, just hiding in the x axis.
Second, read the answer physically. An activation energy near 5 kJ/mol is remarkably small. You can see it directly in the data: raising the temperature by 200 K only moved k from 0.400 to 0.841, barely a doubling. Contrast that with the familiar rule of thumb that reaction rates roughly double for every 10°C near room temperature, which corresponds to an Ea in the neighborhood of 50 kJ/mol, ten times larger. A small activation energy means a low barrier and therefore a rate that is nearly indifferent to temperature, which is exactly the shallow slope you just calculated.
