Question 4 Given the following mechanism: H₂O₂ (aq) + I (aq) → H₂O (1) + OI (aq) (slow) H₂O₂ (aq) + 01- (aq) → H₂0 (1) + O₂ (9) + I (aq) (fast) The rate law consistent with this mechanism is [Select] The intermediate in this reaction is [Select) A plot of [Select] This ✓ [Select] V [H202] vs time k vs time In [H202] vs time 1/[H202] vs time and the catalyst is [Select] for this reaction would be linear. te constant is determined to be 7.4 x 104 M*sec¹ (the exponent on M is x to avoid giving awa

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question 4
Given the following mechanism:
H₂O₂ (aq) + (aq) → H₂O (1) + OI (aq) (slow)
H₂O₂ (aq) + OI- (aq) → H₂O (1) + O₂ (9) + I (aq) (fast)
The rate law consistent with this mechanism is [Select]
The intermediate in this reaction is [Select]
A plot of [Select]
This
of
✓ [Select]
[H202] vs time
k vs time
In [H202] vs time
1/[H202] vs time
and the catalyst is [Select]
for this reaction would be linear.
te constant is determined to be 7.4 x 104 M*sec¹ (the exponent on M is x to avoid giving away
Transcribed Image Text:Question 4 Given the following mechanism: H₂O₂ (aq) + (aq) → H₂O (1) + OI (aq) (slow) H₂O₂ (aq) + OI- (aq) → H₂O (1) + O₂ (9) + I (aq) (fast) The rate law consistent with this mechanism is [Select] The intermediate in this reaction is [Select] A plot of [Select] This of ✓ [Select] [H202] vs time k vs time In [H202] vs time 1/[H202] vs time and the catalyst is [Select] for this reaction would be linear. te constant is determined to be 7.4 x 104 M*sec¹ (the exponent on M is x to avoid giving away
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