Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2 Ce+4 → TI+3 + 2 Ce+3 The elementary steps, in the presence of Mn(II), are: Step 1: Mn+2+ Ce+4 Mn+3 + Ce+3 Step 2: Mn+3 + Ce+4 Mn+4 + Ce+3 Step 3: TI+ + Mn+4 → Mn+2 +TI+3 Identify the catalyst, intermediates and the rate-determining step if the rate law is rate = = K[Ce+4][Mn+2]. 16

Chemistry: Matter and Change
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Chapter16: Reaction Rates
Section16.4: Instantaneous Reaction Rates And Reaction Mechanisms
Problem 38SSC
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Thallium (1) is oxidized by cerium (IV) as follows:
TI+ + 2 Ce+4 → TI+3 + 2 Ce+3
The elementary steps, in the presence of Mn(II), are:
Step 1: Mn+2+ Ce+4
Mn+3 + Ce+3
Step 2: Mn+3 + Ce+4
Mn+4 + Ce+3
Step 3: TI+ + Mn+4 → Mn+2 +TI+3
Identify the catalyst, intermediates and the rate-determining step if the rate
law is rate = = K[Ce+4][Mn+2].
16
Transcribed Image Text:Thallium (1) is oxidized by cerium (IV) as follows: TI+ + 2 Ce+4 → TI+3 + 2 Ce+3 The elementary steps, in the presence of Mn(II), are: Step 1: Mn+2+ Ce+4 Mn+3 + Ce+3 Step 2: Mn+3 + Ce+4 Mn+4 + Ce+3 Step 3: TI+ + Mn+4 → Mn+2 +TI+3 Identify the catalyst, intermediates and the rate-determining step if the rate law is rate = = K[Ce+4][Mn+2]. 16
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