Question 0c6cd

Mar 4, 2016

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

$\text{A" + color(red)(3)"B" -> color(blue)(2)"C}$

If you take the concentration of $\text{A}$ to be $\left[\text{A}\right]$, you can say that the rate of this reaction with respect to $\text{A}$ will be

"rate" = -(Delta["A"])/(Deltat)

Now, notice that you have $\textcolor{red}{3}$ molecules of $\text{B}$ taking part in the reaction for every $1$ molecule of $\text{A}$. This tells you that $\text{B}$ will be consumed three times as fast as $\text{A}$.

In order to express the rate of the reaction with respect to $\text{B}$, you will have to take into account this stoichiometric coefficient.

"rate" = -1/color(red)(3)(Delta["B"])/(Deltat)

Likewise, you get $\textcolor{b l u e}{2}$ molecules of $\text{C}$ for every $1$ molecule of $\text{A}$ and $\textcolor{red}{3}$ molecules of $\text{B}$ consumed by the reaction.

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 ${\text{0.600 M s}}^{- 1}$. This means that $\text{A}$ is dissapearing at a relative rate of

"rate" = -(Delta["A"])/(Deltat) = color(green)("0.600 M s"^(-1))

On the other hand, we already know that $\text{B}$ will disappear three times faster than $\text{A}$, which means that you have

${\text{rate" = -1/color(red)(3)(Delta["B"])/(Deltat) = "0.600 M s}}^{- 1}$

The relative rate of change of $\left[\text{B}\right]$ will be

-(Delta["B"])/(Deltat) = color(red)(3) xx "0.600 M s"^(-1) = color(green)("1.80 M s"^(-1))

For $\text{C}$, you will have

${\text{rate" = 1/color(blue)(2) (Delta["C"])/(Deltat) = "0.600 M s}}^{- 1}$

This will give you a relative rate of change for $\text{C}$ of

(Delta["C"])/(Deltat) = color(blue)(2) xx "0.600 M s"^(-1) = color(green)("1.20 M s"^(-1))#