# For a reaction #A -> B#, why do we put a negative sign in front of the rate of disappearance of #A#?

##### 2 Answers

#### Answer:

Here's what's going on here.

#### Explanation:

As you know, the **rate of a chemical reaction** is a measure of how the concentration of the reactants or of the products changes over time.

Now, we define the change in concentration as the difference between the value at a given time

#Delta["concentration"] = "concentration at t"_2 - "concentration at t"_1#

In other words, we always subtract the value we had at a final state from the value we had at an initial state.

Similarly, we define the change in time as the difference between

This means that the rate of a reaction can be expressed as

#"rate" = (Delta"concentration")/(Delta"time")#

Because a chemical reaction *consumes* **reactants** and *produces* **products**, you can say that the rate of the reaction will always come out *positive* for the products since you will have more products as the reaction progresses, i.e. as we move from

#"rate" = ("concentration at t"_2 - "concentration at t"_1)/(t_2 - t_1) color(white)( (color(blue)(larr " > 0 for products")/color(red)(larr" always > 0")#

#"rate" > 0 -># for products

On the other hand, the rate of the reaction will come out *negative* for the reactants because their concentration will **decrease** as the reaction progresses.

#"rate" = ("concentration at t"_2 - "concentration at t"_1)/(t_2 - t_1) color(white)( (color(blue)(larr " < 0 for reactants")/color(red)(larr" always > 0")#

#"rate" < 0 -># for reactants

Now, *by convention*, we express the rate of a reaction as a **positive value**. This means that in order to express the rate in terms of the **rate of disappearance** of the reactants and still get a positive value, we must add a minus sign to the expression of the rate.

#"rate" = underbrace(-overbrace((Delta"concentration")/(Delta"time"))^(color(purple)("negative for reactants")))_(color(purple)("positive value"))#

So, for example, a reaction

#"A " -> " B"#

will have the rate of reaction expressed in terms of the **appearance** of

#"rate" = (Delta[B])/(Deltat)#

and in terms of the **disappearance** of

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

This means that the rate of the reaction is **equal** for all the chemical species that take part in the reaction because

#overbrace("rate")^(color(purple)(> 0)) = overbrace((Delta[B])/(Deltat))^(color(purple)(>0)) = underbrace(- overbrace((Delta["A"])/(Deltat))^(color(purple)(<0)))_(color(purple)(>0))#

Okay, here's another approach to explaining this (which is an entirely valid question!).

The **rate** of a chemical reaction can be expressed using the **rate law**, as a function of time,

#r(t) = -(Delta[A])/(Deltat) = +(Delta[B])/(Deltat)# ,

for the reaction

#A -> B# .

When we write rate laws, we usually describe the **initial rate** here, which *by default* (without explicitly saying "with respect to . . . ") is **with respect to the product** with a stoichiometric coefficient of

Since what we want to describe is therefore the *forward* reaction, we specify that with a **positive** sign on

However, reactants (such as *consumed* in a forward reaction, so the final concentration of *less than* the initial concentration of

#(Delta[A])/(Deltat) -= ("final conc. of A - initial conc. of A")/("final time - initial time") < 0# ,

and as a result, we tack on a **negative sign for reactants** so that

#color(blue)(ul(r(t) = underbrace(-overbrace((Delta[A])/(Deltat))^"negative")_"positive" > 0))# ,

**matching** the sign for the initial rate of the *forward* reaction. (This is why you only see the negative sign in front of the rate of change in concentration over time for the *reactants*.)

On the other hand, product *produced*, so the change in concentration is *positive*, and the rate of reaction with respect to **still positive**:

#color(blue)(ul(r(t) = underbrace(+overbrace((Delta[B])/(Deltat))^"positive")_"positive" > 0)#

And this **positive** value for the initial rate of the overall reaction, **products** with stoichometric coefficients of

**WHAT ABOUT REALLY SHORT TIME INTERVALS?**

The only change between this and **instantaneous** reaction rate is that we have

#lim_(Deltat -> 0) (pm(Delta[A])/(Deltat)) = pm(d[A])/(dt)#

*The sign conventions stay exactly the same.*

Positive for products, negative for reactants, and positive if considering the so-called "rate of reaction" word-for-word.