1 or shore seeps ean be the rate- outlined above determined maction order. and c
ID: 518114 • Letter: 1
Question
1 or shore seeps ean be the rate- outlined above determined maction order. and chemical reasonin" determine and 10 No so in the due sro," and initial2 will be shows half as only a small determining the of sro, for and starch rimer smaller than 1 of initial rates thuwe reactions, making the method will be carried out in which the same coneeni Meli aa varying the reaction temperat are ured while or [sao, 1. is the same for each trial and can be instead related to if takes seconds for another trial In each of these trials the volume of the reaction therefore, the quantity Alsao, ch nge to occur in one trial and 20 seconds for the color change to in rate with an initial (with the same quantity of trials) then the later trial measured from the start of 1 can that was exactly haur that of the former. Thus, only Ar need to be the reagents to the time when the color of the reaction solution therefore be re-written as eq. 2, shown below where trial hanges. are the initial Isao n trial respect tively the measured time of reaction, and the the condi repl quantity 1 has been relationship to AIS20, 1 laced by its (eq. 2) trial 1 Rate trial 1 In another trial call it trial Atrial 24terial 1 however, the initial 2 for example) the same equation can be written: identical for concentration ofeither Tor changed relative to trial 1. equations by both trials provided the temperature is the same: therefore, by dividing these two constant) each other one the reaction (either depending on which quantity is held can be determined upon measuring Ar for both trials. 30a,b) shows this manipulation below, where in this case the ITl was identical in both trials Rate trial 1 [S20H trial (eq. 3a) Rate trial 2 24 trial 2 trial 2 Rate trial 1 -A Sao TJ/2Attrial 1 8 o, trial 1. Rate trial 2 -AlS203 /24 2 [S208 lo trial 2 ttrial (eq. 3b) ation 3b shows with knowledge of A, AIS2032], and [S2082 lo for both trials the reaction n can be determined. In a third trial [TTo lsnos is held can be varied relative to the first and second trial n can constant and using a similar manipulation as in eq. 3(a,b) the reaction y one be determined. Finally, with knowledge of both react reaction orders and data priate of the previous trials the reaction specific rate constant k, can be determined with units ith knowledge of a reaction's experimentally determined reaction order it is possible to ualitatively reaction mechanism. Recall a reaction mechanism is a series of or individual, steps that must add up to give the global reaction; in this case RXN l h elementary step's reaction stoichiometry is identical to its reaction order and cularity the step. example, a reaction order of 1 involyf unimolecular, of a reaction order of2 inyolyExplanation / Answer
Q1) The proposed two step mechanism for iodine clock reaction is:
S2O82- + I- ------> SO42- + SO4I-
SO4I-+ I- ------> SO42- + I2
S2O82- + 2I- ------> 2SO42- + I2
Note, iodine and iodide ion react with each other to form triodide ion. This reaction is given as:
I2+ I- <------> I3-
Q2)
Trial
t in sec
[I-]0
[S2O82-]0
K (s mol-2 L2)
Temperature in K
1
55.21
10/25 mL X 0.2 M = 0.08 M
10/25 mL X 0.1 M = 0.04 M
55.21/ (0.08 X 0.04) = 1.72 X 104
292.6
2
99.63
10/25 mL X 0.2 M = 0.08 M
5/25 mL X 0.1 M = 0.02 M
99.63/ (0.08 X 0.02) = 6.23 X 104
292.8
3
99.63
5/25 mL X 0.2 M = 0.04 M
10/25 mL X 0.1 M = 0.04 M
99.63/ (0.04 X 0.04) = 6.23 X 104
292.8
8
11.99
10/25 mL X 0.2 M = 0.08 M
10/25 mL X 0.1 M = 0.04 M
11.99/ (0.08 X 0.04) = 3.75 X 103
291.9
Comparing Trial 1 and 2 at constant [I-] when [S2O82-] is halved time doubles. Therefore the reaction is 1st order with respect to [S2O82-] i.e n = 1
Similarly, comparing Trial 1 and 3 at constant [S2O82-] when [I-] is halved time doubles. Therefore the reaction is 1st order with respect to [I-] i.e m= 1
Overall order of the reaction = m+n =2
Now Rate = k [I-]m[S2O82-]n = k [I-]1[S2O82-]1
k = Rate / [I-]1[S2O82-]1
Thus we use this equation to find the k for trial 8.
Arrhenius equation is given as:
ln(k2/k1) = -Ea/ R (1/T2 – 1/T1)
Using information from trial one and trial 8
ln (3.75 X 103 /1.72 X 104) = -Ea/ 8.314 J mol-1 K-1 (1/291.9 K – 1/292.6K)
ln (0.2180) = -Ea/ 8.314 J mol-1 K-1 (0.7/ 291.9• 292.6)K-1
ln (0.2180) = -Ea/ 8.314 J mol-1 X (0.7/ 291.9• 292.6)
-1.5233 = -Ea/ 8.314 J mol-1 X (0.7/ 291.9• 292.6)
Ea = 1.5233 X 8.314 J mol-1 X (291.9• 292.6) / 0.7
= 1.54 X 106 J mol-1
= 1.54X 103 kJ mol-1
3) Yes. Answer 2 made sense. Here is the reason why.
ln(k2/k1) = -Ea/ R (1/T2 – 1/T1)
Using information from trial one and trial 2
ln (6.23 X 104 /1.72 X 104) = -Ea/ 8.314 J mol-1 K-1 (1/292.8 K – 1/292.6K)
ln (3.622) = -Ea/ 8.314 J mol-1 K-1 (-0.2/ 292.8• 292.6)K-1
1.287 = Ea/ 8.314 J mol-1 X (0.2/ 292.8• 292.6)
Ea = 1.287 X 8.314 J mol-1 X (292.8 •292.6) / 0.2
= 4.58 X 106 J mol-1
= 4.58 X 103 kJ mol-1
Thus the energy of activation was reduced from 4.58 X 103 kJ mol-1 to 1.54X 103 kJ mol-1 by addition of a catalyst.
Trial
t in sec
[I-]0
[S2O82-]0
K (s mol-2 L2)
Temperature in K
1
55.21
10/25 mL X 0.2 M = 0.08 M
10/25 mL X 0.1 M = 0.04 M
55.21/ (0.08 X 0.04) = 1.72 X 104
292.6
2
99.63
10/25 mL X 0.2 M = 0.08 M
5/25 mL X 0.1 M = 0.02 M
99.63/ (0.08 X 0.02) = 6.23 X 104
292.8
3
99.63
5/25 mL X 0.2 M = 0.04 M
10/25 mL X 0.1 M = 0.04 M
99.63/ (0.04 X 0.04) = 6.23 X 104
292.8
8
11.99
10/25 mL X 0.2 M = 0.08 M
10/25 mL X 0.1 M = 0.04 M
11.99/ (0.08 X 0.04) = 3.75 X 103
291.9
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