Jan 2024 Update: 3 Rates of Chemical Reactions I: A Clock Reaction A. Preliminary Experiments 1. What are the colors of the solutions

Question

3 Rates of Chemical Reactions I: A Clock Reaction A. Preliminary Experiments 1. What are the colors of the solutions containing the following ions? K 2. The color of the starch-I, complex is blue-black colorless or color less 3 B. Kinetics Experiment Solution 1. Initial (S. OS 1-0.050 M; initial [l =0.050 M. Time experiment started Time (s) benveen Total moles of Aliquor no appearances of color Cumulative times (8) $0, consumed 1 188 2.0x to 2 190 338 4.0x16 200 578 6.0x104 4 215 793 80x100+ 235 1023 10.10- 255 12.83 12x10 7 284. 1557 14*10-4 Solution 2. Initial _02-1=0.10 M. Inicial [r]=0.050 m. Time experiment started 2 Time (8) between Total moles of Aliquor no. appearances of color Cumulative rimtes is) So, consumed 1 9 91 2010- 2 95 186 4,0 10+ 97 283 60X10- 382 8.0x10 482 10. 10- 100 582 12x 10- 102 684 Solution 3. Initial (5.0, 1-0.050 Bf, initial TO $ 0,2-consumed Cumulative times (8) 24% 2.0x10-4 4.0x10 6.0x10- 8.0x10-4 10.x 10- 12x10+ 14x104 523 680 True (s) beheer Aliquor no appearances of color 1 2 126 3 384 136 143 153 160 840 170 1010 Solution 4. Initiat S 0,2-)-0,10 M; initial [1] =0,025 M. Time experiment started Total moles of Time (s) between Aliquot no. appearances of color Cumulative tintes (5) 1 192 192 2 198 390 3 203 593 6.0x10 209 802 8,0x10-4 5 209 jon 6 218 1229 7 221 1450 14*10-4 5,0,?- consumed 2.0x10-4 4.0x10- + 10.x10- 12x10+

Asked By MysticMist72 at 30 Jan 2024

Answered By Expert

Stewart

Expert · 4.3k answers · 4k people helped

Step 1/4

There are 4 solutions given for which we have to calculate k.

For solution 1, we need to find rate= `{1}/{2}{[S_{2}O_8]}/{t}`

Explanation:

rate = k [S 2 O 8 2- ] x [I - ] y

Step 2/4

Putting the values of [S 2 O 8 2- ]= 2 ` x` `10^{-4}` for aliquot no.1 For no. 1: rate= `{1}/{2}{2 xx 10^-4}/{188}=5.138 xx 10^-7 mol //sec`For no. 2: rate= `{1}/{2}{4 xx 10^-4}/{378}=5.29 xx 10^-7 mol //sec`For no. 3; rate= `{1}/{2}{6 xx 10^-4}/{578}=5.19 xx 10^-7 mol //sec`For no. 4: rate= `{1}/{2}{8 xx 10^-4}/{793}=5.04 xx 10^-7 mol //sec`For no. 5: rate= `{1}/{2}{10 xx 10^-4}/{1023}=4.88 xx 10^-7 mol //sec`For no. 6: rate= `{1}/{2}{12 xx 10^-4}/{1283}=4.67 xx 10^-7 mol //sec`For no. 7: rate= `{1}/{2}{14 xx 10^-4}/{1557}=4.49 xx 10^-7 mol //sec`

Explanation:

Putting the values of cumulative time and number of moles of [S 2 O 8 2- ] for each aliquot number.

Step 3/4

We have seven different readings for the experiment so we need to calculate the mean.

the mean rate= `4.98 xx 10^-7 mol//sec` .

Explanation:

We have seven different readings for the experiment so we need to calculate the mean. To calculate the mean rate = `{rate 1+rate 2+rate 3+rate 4+rate 5+rate 6 +rate 7}/{7}` = `4.98 xx 10^-7 mol//sec`

Step 4/4

[S 2 O 8 2- ]=0.050

[I -1 ]=0.050

k= `{rate}/{[S2O8 2- ][I-1]}`

For solution 1

k= `{4.98 xx 10^-7}/{{0.050}{0.050}}=1.993 xx 10^-4 mol^{-1} sec^{-1}litre`

Explanation:

The formula of rate= k[S 2 O 8 2- ][I -1 ]

Final Answer

k 1 = 1.993 ` xx 10^-4`

Similarly we will calculate k for all 4 solutions and find the mean k.

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