The decomposition reaction HBr on simple substance proceeds in the vessel volume 2 liters.Initially, the in vessel containing the 0,5 mol/lit of hydrogen

given:
Δt1  = 20 sec
Δt2  = 40 sec
Δt2  = 1 min = 60 sec
с1(НВr) = 0,5 mol/lit
с2(НВr)  = 0,3 mol/lit
с3(НВr)  = 0,1 mol/lit
с4(НВr)  = 0,05 mol/lit
find:
υ1 - ?
υ2 - ?
υ3 - ?
Solution:
υ=(С₁-С₂)/(Δt•V);
υ₁=((0,5-0,3)mol/lit)/(20 sec.)=0,01 mol/lit∙sec.
υ₂=(С₂-С₃)/(Δt₂)=((0,3-0,1)mol/lit)/(40 sec.)=0,05 mol/lit∙sec
υ₃=(С₃-С₄)/(Δt₃)=((0,1-0,05)mol/lit)/(60 sec.)=0,0008 mol/lit•sec.
Answer: υ1 = 0,01 mol/lit • s
υ2 = 0,05 mol/lit • s
υ3 = 0,0008  mol/lit• s