Plot log([I 3 - ] / t) versus log[ S 2 O 8 2– ] 0 from data in Trials 1, 2, and
ID: 530342 • Letter: P
Question
Plot log([I3-] / t) versus log[ S2O82–]0 from data in Trials 1, 2, and 3. Click the box underneath the graph to show the trendline. It will automatically calculate your slope and intercept. Record them below.
Determine the value of q from the graph. Explain your answer.
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a b c d e f g h i Trial t [I3-] / t (M/s) log([I3-] / t) V 0.2 M KI added (mL) [I–]0(M) log[I–]0 V of 0.2 M (NH4)S2O8(mL) [S2O8]0(M) log[S2O8]0 1 19 .0116 -1.94 25 .077 -1.11 25 .077 -1.11 2 40 .0058 -2.24 25 .077 -1.11 12.5 .038 -1.42 3 85 .0144 -1.84 25 .077 -1.11 6.25 0.19 -1.72 4 41 .0055 -1.25 12.5 .038 -1.42 25 .077 -1.11 (Explanation / Answer
The reaction is endothermic (absorbs heat), because the slope is negative.
From Van't Hoff plot for an endothermic reaction:
slope = (H / R )
H = slope * R
H = -0.158 * 0.08314kJ/mol
H = 0.013 kJ/mol