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The equilibrium constant, K, of a certain first order reaction was measured at two temperatures, T....

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

T(K) 325 = 3.38K

T(K) 625 = 6.07K

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

What is the standard enthalpy of this reaction in J/mol?

If you were going to graphically determine the enthalpy, ΔH°, for this reaction, what points would you plot?

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Answer #1

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

K = 3.38

ln K = ln (3.38) = 1.21788

T = 325.0

1/T = 0.00308

point 1 is:

x = 0.00308

y = 1.21788

K = 6.07

ln K = ln (6.07) = 1.80336

T = 625.0

1/T = 0.0016

point 2 is:

x = 0.0016

y = 1.80336

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

rise = (y1-y2) = 1.21788 - 1.80336 = -0.58548

rise = (X1-X2) = 0.00308 - 0.0016 = 0.00148

slope = rise/run

= -0.58548/0.00148

= -396.42071

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

use

slope = - Ho/R

-396.42071 = -Ho/8.314

Ho = 3296 J/mol

We will plot point 1 and point 2 mentioned above

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