Elemental sulfur can exist as molecules with the formula S_8. The S_8 molecule is represented by the — Kinetics Chemistry Question
Problem Context
Elemental sulfur can exist as molecules with the formula S_8. The S_8 molecule is represented by the incomplete Lewis diagram above.
[VISUAL]
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3 S_8 + 8 OH^- → 8 S_3^- + 4 HOOH
In an experiment, a student studies the kinetics of the reaction represented above and obtains the data shown in the following table.
| Experiment | Initial [S_8] (M) | Initial [OH^-] (M) | Initial Reaction Rate (M/s) |
|---|---|---|---|
| Trial 1 | 0.100 | 0.0100 | 0.699 |
| Trial 2 | 0.300 | 0.0100 | 2.10 |
| Trial 3 | 0.300 | ? | 4.19 |
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The next day the student conducts trial 4 using the same concentrations of S_8 and OH^- as in trial 1, but the reaction occurs at a much slower rate than the reaction in trial 1. The student observes that the temperature in the lab is lower than it was the day before.
Model Answer
Two
1 point is earned for the correct answer.
Model Answer
109.5 (Acceptable range: 104 <= angle <= 110)
(The experimentally determined angle is 107.8)
1 point is earned for an angle in the acceptable range.
Model Answer
1s2 2s2 2p6 3s2 3p4 OR [Ne]3s2 3p4
1 point is earned for a correct configuration.
Model Answer
Peak X represents electrons in a 1s orbital. A Cl atom has one more proton in its nucleus than does a S atom; therefore, the electrons in Cl are more strongly attracted to the nucleus, and the binding energy of the 1s electrons in the Cl atom is greater than that of the 1s electrons in the S atom.
1 point is earned for identifying the orbital as 1s.
1 point is earned for a valid explanation.
Model Answer
See example of a correct response (dashed peak in the scoring guidelines) which is located to the right of peak Y and has the correct height (representing 4 electrons for sulfur compared to 5 electrons for chlorine).
1 point is earned for a peak that is to the right of peak Y and that has the correct height.
Model Answer
The reaction is first order with respect to S8. Comparing trials 1 and 2, tripling [S8] while keeping [OH‒] constant results in the tripling of the initial reaction rate.
OR
rate2 / rate1 = (k[S8]^x [OH-]^y)_2 / (k[S8]^x [OH-]^y)_1
2.10 / 0.699 = k(0.300)^x (0.0100)^y / k(0.100)^x (0.0100)^y ⇒ 3.00 = 3.00^x ⇒ x = 1
1 point is earned for the correct order and a valid justification.
Model Answer
Comparing trials 2 and 3, [S8] is kept constant and the initial reaction rate doubles. Since the reaction is first order with respect to OH, the concentration of OH in trial 3 must be 2 * 0.0100 M = 0.0200 M.
1 point is earned for the correct concentration of OH and a valid justification.
Model Answer
Explanation 1 (Energy of collisions):
The temperature was lower on the second day so the average kinetic energy of the reactant particles was lower. Therefore, there were fewer collisions between particles with sufficient energy to react.
Explanation 2 (Frequency of collisions):
Since the temperature was lower, the kinetic energy was lower and the average speed of the particles was lower. At the lower speeds, the reactant particles collided less frequently.
1 point is earned for a valid explanation that addresses the effect on energy of collisions.
1 point is earned for a valid explanation that addresses the effect on frequency of collisions.