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Why is rate always positive?

socratic.org - For a reaction A -> B, why do we put a negative sign in front of the rate of ...

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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

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What is the rate order?

chem.libretexts.org - 3.3.3: Reaction Order - Chemistry LibreTexts

The reaction order is the relationship between the concentrations of species and the rate of a reaction.

Introduction

The order of a rate law is the sum of the exponents of its concentration terms. Once the rate law of a reaction has been determined, that same law can be used to understand more fully the composition of the reaction mixture. More specifically, the reaction order is the exponent to which the concentration of that species is raised, and it indicates to what extent the concentration of a species affects the rate of a reaction, as well as which species has the greatest effect. For the N 2 O 5 decomposition with a rate law of k[N 2 O 5 ], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. It can also be said that the reaction is "first order in N 2 O 5 ". For more complicated rate laws, the overall reaction order and the orders with respect to each component are used. As an example, consider the following reaction,

[ A + 3B + 2C

ightarrow ext{products} ]

whose experimental rate law is given by:

[ ext{rate} = k[A][B]^2 ]

This reaction is third-order overall, first-order in A, second-order in B, and zero-order in C. Zero-order means that the rate is independent of the concentration of a particular reactant. Of course, enough C must be present to allow the equilibrium mixture to form.

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