Question #0c6cd
1 Answer
Here's what I got.
Explanation:
The rate of a reaction will essentially tell you how the concentration of a chemical species that takes part in the reaction changes over a period of time.
#color(blue)("rate of reaction" = (Delta["chemical species"])/(Deltat))#
Now, the concentration of a reactant will always decrease as the reaction proceeds. On the other hand, the concentration of a product will always increase as the reaction proceeds.
Remember, the rate of the reaction can be written from the perspective of each chemical species that takes part in the reaction. For reactants, you will have
#"rate" = - (Delta["reactant"])/(Deltat)#
Here the minus sign is used because the concentration of a reactant decreases with time.
For products, you will have
#"rate" = (Delta["product"])/(Deltat)#
This means that you can write
#color(blue)("rate" = - (Delta["reactant"])/(Deltat) = (Delta["product"])/(Deltat))#
Moreover, these concentrations will change in a way that is reflective of the stoichiometric coefficients of the chemical species in question.
In your case, the balanced chemical equation looks like this
#"A" + color(red)(3)"B" -> color(blue)(2)"C"#
If you take the concentration of
#"rate" = -(Delta["A"])/(Deltat)#
Now, notice that you have
In order to express the rate of the reaction with respect to
#"rate" = -1/color(red)(3)(Delta["B"])/(Deltat)#
Likewise, you get
You can thus say that
#"rate" = 1/color(blue)(2)(Delta["C"])/(Deltat)#
Since the rate of the reaction is the same regardless of which chemical species you write it for, you will have
#"rate" = -(Delta["A"])/(Deltat) = -1/color(red)(3)(Delta["B"])/(Deltat) = 1/color(blue)(2)(Delta["C"])/(Deltat)#
The problem tells you that the rate of the reaction is equal to
#"rate" = -(Delta["A"])/(Deltat) = color(green)("0.600 M s"^(-1))#
On the other hand, we already know that
#"rate" = -1/color(red)(3)(Delta["B"])/(Deltat) = "0.600 M s"^(-1)#
The relative rate of change of
#-(Delta["B"])/(Deltat) = color(red)(3) xx "0.600 M s"^(-1) = color(green)("1.80 M s"^(-1))#
For
#"rate" = 1/color(blue)(2) (Delta["C"])/(Deltat) = "0.600 M s"^(-1)#
This will give you a relative rate of change for
#(Delta["C"])/(Deltat) = color(blue)(2) xx "0.600 M s"^(-1) = color(green)("1.20 M s"^(-1))#