If you were going to graphically determine the enthalpy The equilibrium constant

Question

If you were going to graphically determine the enthalpy The equilibrium constant, K, of a certain first order reaction was measured at two temperatures, T r, for this reaction, what points would you plot? The data is shown in this table Number Number T(K) K 275 3.11 875 7.55 avoid rounding errors, use three significant figures in the x values and four significant figures in point 1 Number Number e y values. point 2 Determine the rise, run, and slope of the line formed by these points rise run slope Number Number Number What is the standard enthalpy of this reaction? Number J/mol

Explanation / Answer

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

K = 3.11

ln K = ln (3.11) = 1.13462

T = 275.0

1/T = 0.00364

point 1 is:

x = 0.00364

y = 1.13462

K = 7.55

ln K = ln (7.55) = 2.02155

T = 875.0

1/T = 0.00114

point 2 is:

x = 0.00114

y = 2.02155

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rise = (y1-y2) = 1.13462 - 2.02155 = -0.88692

rise = (X1-X2) = 0.00364 - 0.00114 = 0.00249

slope = rise/run

= -0.88692/0.00249

= -355.69381

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

use

slope = - delta H/R

= -355.69381 = -delta H/8.314

delta H = 2957 J/mol

Answer: 2957 J/mol

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