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

2 Answers
Jul 3, 2017

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 #t_2# and the value at a given time #t_1#, where #t_1 < t_2#.

#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 #t_2# and #t_1#, which implies that #Deltat > 0#.

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 #t_1# to #t_2#.

#"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 #"B"#

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

and in terms of the disappearance of #"A"#

#"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))#

Jul 4, 2017

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 #1#, so from here on out in this answer, that's what #r(t)# represents.

Since what we want to describe is therefore the forward reaction, we specify that with a positive sign on #r(t)#.

However, reactants (such as #A#) always get consumed in a forward reaction, so the final concentration of #A# is less than the initial concentration of #A#, i.e.

#(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 #B# gets produced, so the change in concentration is positive, and the rate of reaction with respect to #B# is 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, #r(t)#, is strictly because, again, #r(t)# is defined to match the sign of the change in concentration over time of products with stoichometric coefficients of #bb1#, since we really focus on the forward reaction.

WHAT ABOUT REALLY SHORT TIME INTERVALS?

The only change between this and instantaneous reaction rate is that we have #Deltat -> 0#, i.e. the definition of a derivative:

#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.