Respuesta :
Considering the reaction:
C₂H₄Br₂ + 3 KI ----> C₂H₄ + 2 KBr+ KI₃
[tex]Rate of disappearance of C_{2}H_{4}Br_{2} =\frac{1}{3}\times Rate of disappearance of KI[/tex]
[tex]2.0 \times 10^{-5} =\frac{1}{3}\times Rate of disappearance of KI[/tex]
[tex]Rate of disappearance of KI=6.0 \times 10^{-5}[/tex]
[tex]So, Rate of disappearance of KI is 6.0 \times 10^{-5} ms^{-1}[/tex]
Answer: The value of the rate disappearace of KI is [tex]6.0\times 10^{-5}[/tex]
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
[tex]C_2H_4Br_2+3KI\rightarrow C_2H_4+2KBr+KI_3[/tex]
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.[tex]Rate=-\frac{1d[C_2H_4Br_2]}{dt}=-\frac{1d[KI]}{3dt}=[/tex]
Given: [tex]\frac{d[C_2H_4Br_2]}{dt}]=2.0\times 10^{-5}[/tex]
Putting in the values we get:
Rateof disappearace of KI=[tex]3\times 2.0\times 10^{-5}=6.0\times 10^{-5}[/tex]
Thus the value of the rate disappearace of KI is [tex]6.0\times 10^{-5}[/tex]
