Question

1/T 0.003626 0.003409 0.003438 0.003274 0.003172 0.003092 ln (ksp) -29.6153 -28.093 -29.1602 -28.4125 -27.8986 -28.8031 he...

1/T
0.003626
0.003409
0.003438
0.003274
0.003172
0.003092
ln (ksp)
-29.6153
-28.093
-29.1602
-28.4125
-27.8986
-28.8031

he lab begins by stating that A linear relationship exists when the values of ln Ksp obtained at various temperatures are plotted as a function of the reciprocal temperature. The (negative) slope of the line equals -delta H/R and the y-intercept (where x=0) equals delta S/R (both are standard states). the analysis of data goes on to say: (2) calculate the solubility product of borax at each temperature (3) plot the natural logorithm of the solubility product versus the recipricol temperature of the measurements (4) extract from the plotted data the thermodynamic properties of delta H and delta S for the dissolution of borax.. below is the information found in the experiment For starters, i think I am not plotting this correctly. Also, how would I find the negative slope for delta S /R. Also find delta G

Homework Answers

Answer #1

ksp = exp(lnksp)

values of ksp are as follows

ksp
1.3748E-13
6.3004E-13
2.1671E-13
4.5773E-13
7.6523E-13

3.0972E-13

here 2nd and 6th value of lnksp are bit odd, plotting a curve between lnksp vs 1/T for rest of the points

1/T

lnksp

0.003626

-29.6153

0.003438

-29.1602

0.003274

-28.4125

0.003172

-27.8986

from curve slope = -3799.9 and intercept = -15.937

and

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