Respuesta :
The decomposition of HBr takes place according to the reaction below,
2HBr(g) = H2(g)+ Br2(g)
a) Rate of reaction is the measure of the change in concentration of the reactants or the change in concentration of the product per unit time(seconds).
Therefore;
Rate = -1/2(Δ(HBr)/Δt = Δ(H₂)/Δt =Δ(Br₂)/Δt,
where ΔHBr is the change in concentration of HBr (reactant), ΔH₂ is the change in hydrogen concentration , ΔBr₂ is the change in bromine concentration and Δt is the change in time.
b) Average rate of reaction after 25 sec will be,
rate = -1/2Δ(HBr)/Δt
= -1/2 (0.512 M - 0.6 M)/(25 s-0 s)
= -1/2 (-0.088)/25
= 0.00176 M/s
2HBr(g) = H2(g)+ Br2(g)
a) Rate of reaction is the measure of the change in concentration of the reactants or the change in concentration of the product per unit time(seconds).
Therefore;
Rate = -1/2(Δ(HBr)/Δt = Δ(H₂)/Δt =Δ(Br₂)/Δt,
where ΔHBr is the change in concentration of HBr (reactant), ΔH₂ is the change in hydrogen concentration , ΔBr₂ is the change in bromine concentration and Δt is the change in time.
b) Average rate of reaction after 25 sec will be,
rate = -1/2Δ(HBr)/Δt
= -1/2 (0.512 M - 0.6 M)/(25 s-0 s)
= -1/2 (-0.088)/25
= 0.00176 M/s
The rate of reaction in terms of concentration change has been given by
[tex]\rm \dfrac{1}{2}\;\dfrac{\Delta HBr}{\Delta time}[/tex] = [tex]\rm \dfrac{\Delta H_2}{\Delta time}[/tex] = [tex]\rm \dfrac{\Delta Br_2}{\Delta time}[/tex].
The rate of reaction when the concentration of HBr has been dropped to 0.512 M has been 1.76 [tex]\rm \bold{\times\;10^-^3}[/tex] M/s.
(a) The rate of reaction can be defined as the change in concentration of the reactant and product per unit time.
The rate of reaction in terms of reactant change can be given by:
Rate = n [tex]\rm \dfrac{\Delta concentration}{\Delta time}[/tex]
where, n is the coefficient, [tex]\rm \Delta reactant[/tex] has been the change in reactants concentration, and [tex]\rm \Delta time[/tex] has been the change in the time.
For the given reaction, the rate for change in each reactant and product has been:
Rate = [tex]\rm \dfrac{1}{2}\;\dfrac{\Delta HBr}{\Delta time}[/tex] = [tex]\rm \dfrac{\Delta H_2}{\Delta time}[/tex] = [tex]\rm \dfrac{\Delta Br_2}{\Delta time}[/tex]
(b) The change in the concentration of HBr has been:
[tex]\rm \Delta HBr[/tex] = 0.60 M - 0.512 M
[tex]\rm \Delta HBr[/tex] = 0.088 M
[tex]\rm \Delta time[/tex] = 25 sec.
The rate of reaction in terms of HBr concentration has been:
Rate = [tex]\rm \dfrac{1}{2}\;\dfrac{\Delta HBr}{\Delta time}[/tex]
Rate = [tex]\rm \dfrac{1}{2}\;\dfrac{0.088}{25}[/tex]
Rate = 0.00176 M/s
Rate = 1.76 [tex]\rm \bold{\times\;10^-^3}[/tex] M/s.
The rate of reaction when the concentration of HBr has been dropped to 0.512 M has been 1.76 [tex]\rm \bold{\times\;10^-^3}[/tex] M/s.
For more information about the rate of reaction, refer to the link:
https://brainly.com/question/8592296
