Define the rate of reaction of HBr in the following reaction. How is this related to the rate of formation of Br2?
4 HBr(g) + O2(g) --> 2Br2(g) + 2 H2O(g)
Answer (1)
Rate of reaction for HBr:rate = [tex]-\frac{1}{4} \frac{d[HBr]}{dt}[/tex]negative since HBr is a reactant (consumed)1/4 from the balanced equation (4 moles of HBr)Rate of formation for Br₂rate = [tex]\frac{1}{2} \frac{d[Br2]}{dt}[/tex]positive since Br₂ is a product (formed)1/2 from the balanced equation (2 moles Br₂)Equate the rates to get the relationship:[tex]-\frac{1}{4} \frac{d[HBr]}{dt} = \frac{1}{2} \frac{d[Br2]}{dt}[/tex]multiply both sides by 4[tex]- \frac{d[HBr]}{dt} = 2\frac{d[Br2]}{dt}[/tex]therefore: rate of consumption of HBr is twice the rate of formation of Br₂
