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The equilibrium constant, K, of a certain first order reaction was measured at two temperatures, T. The data is shown in this

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Answer #1

We need to plot lnK on y axis and 1/T on x axis

K = 3.04

ln K = ln (3.04) = 1.11186

T = 275.0

1/T = 0.00364

point 1 is:

x = 0.00364

y = 1.11186

K = 5.57

ln K = ln (5.57) = 1.7174

T = 675.0

1/T = 0.00148

point 2 is:

x = 0.00148

y = 1.7174

----------------------------------------------------------------------------

rise = (y1-y2) = 1.11186 - 1.7174 = -0.60554

rise = (X1-X2) = 0.00364 - 0.00148 = 0.00215

slope = rise/run

= -0.60554/0.00215

= -281.00726

----------------------------------------------------------------------------

use

slope = - Ho/R

-281.00726 = -Ho/8.314

Ho = 2336 J/mol

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