Question

The decomposition of NOBr is studied manometrically because the number of moles of gas changes; it cannot be studied colorimetrically because both NOBr and Br2 are reddish-brown. 2NOBr(g) → 2NO(g) + Br2(g) Use the data below to make the following determinations: (a) the average rate of decomposition of NOBr over the entire experiment. (b) the average rate of decomposition of NOBr between 2.00 and 4.00 seconds. Time (s) [NOBr] (mol/L) 0.00 0.0100 2.00 0.0071 4.00 0.0055 6.00 0.0045 8.00 0.0038 10.00 0.0033 The rates of decomposition of NOBr are

Answer 1

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) = 3.4 x 10⁻⁴ M/s

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) = 4 x 10⁻⁴ M/s