Question

Consider the following reaction: 2HBr (g) rightarrow H_2 (g) + Br_2 (g) In the first 23.0 s of this reaction the of HBr dropped from 0.550 M to 0.457 M. Calculated the average rate of the reaction in this time interval. Express your answer using two significant figures. If the volume of the reaction in part (b) was 1.50 L. what amount of Br_2 (in moles) was formed during the 15.0 is of the reaction? Express your answer using two significant figures.

Answer 1

Given: The time taken = 23 seconds

Initial concentration of HBr = [HBr]₀ = 0.550 M

concentration after 23s = [HBr]t = 0.457 M

The reaction is

     2HBr(g) ---> H₂(g) + Br₂(g)

Rate of reaction = $\frac{-1}{2}\frac{\Delta HBr}{\Delta t} = \frac{\Delta Br_{2}}{\Delta t} = \frac{\Delta H_{2}}{\Delta t}$

Rate of reaction = $\frac{-1}{2}\frac{[HBr]_{0} -[HBr]_{t}}{\Delta t}$

Therefore rate of reaction = -1/2 (0.550 - 0.457) / 23 = 0.00404 M s^-1

Therefore the average rate of reaction = 0.00404 M s^-1

b) The volume = 1.50 L

time = 15 seconds

Rate = d[Br₂] /dt = 0.00404 M s^-1

d[Br₂] = 0.00404 X 15 = 0.0606 M

Moles = Molarity X volume in litres = 0.0606 M X 1.5 L = 0.0909 moles