**Answer:**

a) The average rate of decomposition of NOBr over the entire reaction is 3.4 x 10⁻⁴ M/s

b) The average rate of decomposition of NOBr between 2.00 and 4.00 seconds is 4 x 10⁻⁴ M/s

**Explanation:**

The average rate of decomposition of a reactant can be written as follows:

v = -1/a Δ[A] / Δt

Where

v: average rate of decomposition of A

a: stoichiometric coefficient of reactant A

Δ[A]: variation of the concentration of reactant A

(final concentration of A - initial concentration of A)

Δt: variation of time (final time - initial time)

a) The average rate of the decomposition of NOBr over the entire experiment can be calculated as follows, using this data:

Initial concentration of NOBr: 0.0100 M

Final concentration of NOBr: 0.0033 M

Initial time: 0.00 s

Final time: 10.00 s

Stoichiometric coefficient of NOBr: 2

Then:

v = -1/2 Δ[NOBr] / Δt

v = -1/2 (0.0033 M - 0.0100 M) / (10.00 s - 0.00 s) = <u>3.4 x 10⁻⁴ M/s</u>

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b) In the same way, the rate of decomposition can be calculated between the 2.00 and 4.00 seconds:

v = -1/2 (0.0055 M - 0.0071 M) / (4.00 s - 2.00 s) = <u>4 x 10⁻⁴ M/s</u>