Question

Consider the gas-phase reaction between nitric oxide and bromine at 273 ∘C
2NO(g)+Br2(g)→2NOBr(g).
The following data for the initial rate of appearance of NOBr were obtained:

Experiment	[NO](M)	[Br2](M)	Initial Rate of Appearance of NOBr(M/s)
1	0.10	0.20	24
2	0.25	0.20	150
3	0.10	0.50	60
4	0.35	0.50	735

What is the rate of disappearance of Br2 when [NO]= 8.1×10⁻² M and [Br2]= 0.22 M ?

Express your answer using two significant figures.

Answer 1

The rate equation of this reaction can be written as:

Where k is the rate constant and n and m are the reaction orders. We can take experiments 1 and 2 (they have the same concentration of bromine, which cancel out) and calculate:

So, we have

Which means that n = 2.

Analogously, If we take experiments 1 and 3, we have the same concentration of NO, so we can use this:

So now we have:

So m = 1.

Knowing n and m, we can calculate the constant k using experiment 4 and the rate equation:

So,

We can now calculate the rate of appearence of NOBr with the given concentrations:

BUT we have to bear in mind that per each 2 moles of NOBr that dissappear, one mole of Br₂ is consumed, so its rate of disappearance will be half this value, and with a negative value, since bromine is a reactant.: -8.7 M/s