Question

Consider the following proposed mechanisms for Reaction (1): Mechanism A (one step): H2O2 + 2I- +...

Consider the following proposed mechanisms for Reaction (1): Mechanism A (one step): H2O2 + 2I- + 2H3O+ à I2 + 4H2O Reaction (1) Mechanism B (three steps) H2O2 + I- à OH- + HOI (slow) H3O+ + OH- ßà 2H2O (fast) HOI + H3O+ + I- à I2 + H2O (fast) a) Show that when you sum the three elementary steps, the net result of Mechanism B is Reaction (1): b) What would be the expected rate law for mechanism B, assuming that the rate of Reaction (1) is limited by the slow step 1?

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Answer #1

Mechanism A:

H2O2 + 2 I- + 2 H3O+ ---------> I2 + 4 H2O (Reaction 1)

Mechanism B:

H2O2 + I- --------> OH- + HOI (slow) ……(1)

H3O+ + OH- --------> 2 H2O (fast) ……(2)

HOI + H3O+ + I- --------> I2 + 2H2O (fast) …..(3)

1) Add equations (1)-(3) to get

H2O2 + I- + H3O+ + OH- + HOI + H3O+ + I- ---------> OH- + HOI + 2 H2O + I2 + 2 H2O

Cancel out the common terms to obtain

H2O2 + 2 I- + 2 H3O+ --------> I2 + 4 H2O

This is the same as reaction 1 above.

2) The slowest step of mechanism B determines the rate law for the reaction 1 in mechanism A. The rate law for mechanism B is determined by the slowest step of the reaction, i.e, (1). The rate law for step (1) of mechanism B is given as

Rate = k*[H2O2][I-]

where k = second order rate constant.

We must assume that all the steps (1)-(3) are elementary steps.

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