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Author Question: Tabulated below are initial rate data for the reaction 2Fe(CN)63 + 2I 2Fe(CN)64 + ... (Read 79 times)

bobypop

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Question 1

Tabulated below are initial rate data for the reaction:
              2Fe(CN)63– + 2I→ 2Fe(CN)64– + I2
 
Run
[Fe(CN)63–]0
[I]0
[Fe(CN)64–]0
[I2]0
Rate (M/s)
1
0.01
0.01
0.01
0.01
1 × 10–5
2
0.01
0.02
0.01
0.01
2 × 10–5
3
0.02
0.02
0.01
0.01
8 × 10–5
4
0.02
0.02
0.02
0.01
8 × 10–5
5
0.02
0.02
0.02
0.02
8 × 10–5
What is the value of k?

107M–5 s–1
103M–3 s–1
10 M–2 s–1
50 M–2 s–1
None of these are correct.

Question 2

Tabulated below are initial rate data for the reaction
              2Fe(CN)63– + 2I→ 2Fe(CN)64– + I2
 
 
 
 
 
Initial
Run
[Fe(CN)63–]0
[I]0
[Fe(CN)64–]0
[I2]0
Rate (M/s)
1
0.01
0.01
0.01
0.01
1 × 10–5
2
0.01
0.02
0.01
0.01
2 × 10–5
3
0.02
0.02
0.01
0.01
8 × 10–5
4
0.02
0.02
0.02
0.01
8 × 10–5
5
0.02
0.02
0.02
0.02
8 × 10–5
The experimental rate law is:


= k[Fe(CN)63–]2[I]2[Fe(CN)64–]2[I2]
= k[Fe(CN)63–]2[I][Fe(CN)64–][I2]
= k[Fe(CN)63–)]2[I]
= k[Fe(CN)63–][I]2

= k[Fe(CN)63–][I] [Fe(CN)64–]


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Marked as best answer by bobypop on Mar 21, 2021

princessflame2016

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bobypop

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Reply 2 on: Mar 21, 2021
:D TYSM


Dominic

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Reply 3 on: Yesterday
Wow, this really help

 

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