Gradebook 2/23/2018 11:59 PM 48.5/1002/22/2018 10:19 PM Print CaloulatorPeriodic

Question

Gradebook 2/23/2018 11:59 PM 48.5/1002/22/2018 10:19 PM Print CaloulatorPeriodic Table Question 24 of 30 Consider this reaction data: If you were going to graphically determine the activation energy this reaction, what points would you plot? A products Number 325 0.384 625 0.655 To avoid rounding errors, use at least Naber Number figures in all values. point 2 Determine the rise, run, and slope of the line formed by these points rise run slope Number Number What is the activation energy of this reaction? Number J/ mol privacy policy terms of use | contact us | help

Explanation / Answer

here graph is drawn between ln K   and 1/T

lnK = y -axis

1/T = x - axis

point 1:   x = 3.077 x 10^-3   ,   y = -0.957

point 2 : x = 1.60 x 10^-3   , y = -0.423

rise =   y2-y1   = 0.534

run = x2 - x1   = - 1.48 x 10^-3

slope = rise / run = - 0.534 / 1.48 x 10^-3 = -361

-Ea / R = slop

Ea = 360.8 x 8.314 = 1761 J / mol = actiavation energy

actiavation energy = 3.00 x 10^3 J/mol

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