Reso 4/14/2018 11:59 PM 66.7/100 PrintCalculator 9 Gradebook Question 4 of 6 Map

Question

Reso 4/14/2018 11:59 PM 66.7/100 PrintCalculator 9 Gradebook Question 4 of 6 Map Consider this reaction data: If you were going to graphically determine the activation energy this reaction, what points would you plot? A products T(K) k(s) 275 0.381 625 0.778 point 1: To avoid rounding errors, use at least figures in all values point 2: n Determine the rise, run, and slope of the line formed by these points. rise run slope Number What is the activation energy of this reaction? J/mol Previous Give Up& View Solution Check Answer Next Exit

Explanation / Answer

The graph will be between ln(K) versus 1/T

Point1: x=1/275 = 0.003636, y = ln(0.381) = -0.96495

Point2: x=1/625 = 0.0016, y = ln(0.778) = -0.25102

rise = y2 - y1 = -0.25102 + 0.96495 = 0.71393

nun = x2 - x1 = 0.0016 - 0.003636= -0.002036

Slope = rise/nun = 0.71393/-0.002036 = -350.653

Activation energy = Slope * (-R) = 350.653 * 8.314 = 2915.33 J/mol

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