Sheridan K.

asked • 02/15/24

Consider the reaction data(keep three sig figs)

Table:

T(K) K(s^-1)

225 0.381

875 0.857

What two points should be plotted to graphically determine the activation energy of this reaction?

x1= y1=

x2= y2=


Determine the rise, run, and slope of the line formed by these points


What is the activation energy of this reaction?


Cindy L.

You could use Arrhenius equation to solve this question. lnk=lnA−Ea/RT y1=k1 = ln(0.381) ; y2=k2=ln(0.857); x1=T1=225; x2=T2=875; Gas constant, R, is 8.31 J K-1 mol-1. Then, you have Ea.
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06/03/24

1 Expert Answer

By:

Patrick M. answered • 06/10/24

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Table:

T(K) K(s^-1)

225 0.381

875 0.857


An Arrhenius Plot graphs ln k (y-axis) versus 1/Temperature (x-axis). The resultant slope is equal to negative activation energy, -Ea, divided by the gas constant R (8.31 J/K*mol) (slope = -Ea/R)


What two points should be plotted to graphically determine the activation energy of this reaction?

x1= 1/255 y1= ln(0.381)

x2= 1/875 y2= ln(0.857)


Rise = y2 - y1 = ln(0.857) - ln(0.381) = 0.811 (unitless)

Run = x2 - x1 = (1/875) - (1/255) = -0.00278 (1/K)


Slope = Rise/Run = 0.811 / -0.00278 (1/K) = -292 K



Ea = -Slope * R


Ea = - (-292 K * 8.31 J/K*mol)


Ea = 2430 J/mol





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