-0.058 -0.060 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 1/T (1/K) Substance AG (kJ mol-¹) 1₂(aq) I (aq) Is (aq) 16.37 -51.57 -51.4 Calculate AG for this reaction at 298 K. (DO NOT read this value off the graph. Use the data given to calculate a more accurate value.) AG - kJ
-0.058 -0.060 0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037 1/T (1/K) Substance AG (kJ mol-¹) 1₂(aq) I (aq) Is (aq) 16.37 -51.57 -51.4 Calculate AG for this reaction at 298 K. (DO NOT read this value off the graph. Use the data given to calculate a more accurate value.) AG - kJ
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Transcribed Image Text:Iodine is not very soluble in water, but it dissolves readily in a solution containing iodide ions by the following reaction:
I₂ (aq) + I¯ (aq) → 13(aq)
The following graph shows the results of a study of the temperature dependence of AG for this reaction. (The solid line is a best fit to the actual data points.) Notice that the quantity on the y axis is AG /T, not just AG. Additional
data relevant to this reaction are also given in the table that follows the graph.
AGOT (kJ/K)
-0.050
-0.052
-0.054
-0.056
-0.058
-0.060
0.0030 0.0031 0.0032 0.0033 0.0034 0.0035 0.0036 0.0037
1/T (1/K)
Substance AG (kJ mol-¹)
1₂ (aq)
16.37
I (aq)
13 (aq)
-51.57
-51.4
a Calculate AG for this reaction at 298 K. (DO NOT read this value off the graph. Use the data given to calculate a more accurate value.)
AG =
kJ
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