# Question d6c6a

Jan 22, 2016

"rate" = -(d["N"_2])/(dt) = -1/3(d["H"_2])/(dt) = 1/2(d["NH"_3])/(dt)

#### Explanation:

First thing first, make sure that you understand what the rate of a reaction actually represents.

As you know, a chemical reaction implies the consumption of the reactants and the formation of the products.

The rate of a reaction will simply tell you the change in the concentrations of the reactants or the change in the concentrations of the products as the reaction proceeds towards equilibrium.

So, in essence, the rate of a reaction tells you how the concentrations of the reactants (or of the products) will change in time.

$\textcolor{b l u e}{\text{rate" = "change in concentration"/"change in time}}$

Now, take a look at the balanced chemical equation for this reaction

${\text{N"_text(2(g]) + color(red)(3)"H"_text(2(g]) -> color(blue)(2)"NH}}_{\textrm{3 \left(g\right]}}$

Notice that you have a $1 : \textcolor{red}{3}$ mole ratio between nitrogen gas and hydrogen gas. This means that for every $1$ mole of nitrogen gas that takes part in the reaction, you will consume $\textcolor{red}{3}$ moles of hydrogen gas.

Likewise, you have a $1 : \textcolor{b l u e}{2}$ mole ratio between nitrogen gas and ammonia. This means that for every $1$ mole of nitrogen gas consumed by the reaction, $\textcolor{b l u e}{2}$ moles of ammonia will be produced.

Now, let's say that you want to express the rate of the reaction in terms of the time derivative of the concentration of nitrogen gas, ${\text{N}}_{2}$. You can say that

"rate" = - (d["N"_2])/(dt)

The minus sign designates the fact that the concentration is decreasing with time.

It is very important to realize that the rate of the reaction is equal for all chemical species that are taking part in that reaction.

To write the rate of the reaction in terms of the concentration of hydrogen gas, you need to take into account the aforementioned mole ratio that exists between the reactants.

You could say that

"rate" = -(d["H"_2])/(dt)

However, that would not be correct, because you need to have

"rate" = -(d["N"_2])/(dt) = -(d["H"_2])/(dt)

But since it you will consume $\textcolor{red}{3}$ moles of hydrogen per mole of nitrogen, you will need to have

"rate" = -(d["N"_2])/(dt) = -1/color(red)(3)(d["H"_2])/(dt)

Finally, express the rate of the reaction in terms of the concentration of ammonia

"rate" = (d["NH"_3])/(dt)

You do not use a minus sign here because this concentration is increasing as the reaction progresses.

Once again, you need

"rate" = -(d["N"_2])/(dt) = -1/color(red)(3)(d["H"_2])/(dt) = (d["NH"_3])/(dt)

But since you get $\textcolor{b l u e}{2}$ moles of ammonia per mole of nitrogen gas, you will need to have

"rate" = -(d["N"_2])/(dt) = -1/color(red)(3)(d["H"_2])/(dt) = 1/color(blue)(2)(d["NH"_3])/(dt)#

And there you have it - the rate of the reaction expressed in terms of the time derivatives of the concentrations.