# Given the reaction #A -> B#, if the rate multiplies by #100# when the concentration of #A# is multiplied by #10#, what is the order of the reaction?

##
I know it's 2 but why?

I know it's 2 but why?

##### 1 Answer

Pretend you don't know the answer yet. The **reaction** given is:

#stackrel("Reactant(s)")overbrace(A) -> stackrel("Product(s)")overbrace(B)#

When we write the **rate law**, or the equation that relates the *rate* *rate constant* *concentration*

#\mathbf(r(t) = k stackrel("Reactant(s) Only")overbrace([A]^m) = stackrel("Reactant(s)")overbrace(-1/nu_A(d[A])/(dt)) = stackrel("Product(s)")overbrace(+1/nu_B(d[B])/(dt)))# where

#nu_A# is thestoichiometric coefficientof#A# .

NOTE:Stoichiometric coefficients do not necessarily correspond with the order.

#(d[X])/(dt)# is therate of change in concentrationof#X# . If#X# is a reactant, it is a rate ofdisappearance. If#X# is a product, it is a rate ofappearance.

Now, we know that **constant** no matter what, as long as it's the same reaction. Therefore, when *must change*, because *same*.

But, there's also the **order** of the reactant

The question asks us to **multiply** **by** **tells us how much** **changes**, so let's do that.

#ul(ul(color(red)(x)))*r(t) = k(10[A])^(m) = ul(ul(color(red)(10^m)))*k[A]^m#

But whatever we do to one side, we have to do to the other, so we have to multiply by *dependent* on

#100 = 10^m#

Since *order* of reactant **resultant rate law** has the exponent

#color(blue)(r(t) = k[A]^color(red)(2))#

**NOTE:**

It's a little different when a second reactant comes into play, but in that case, we rig different experiment trials so that the concentration of other reactants stay the same as well.

That way, the influence of changing concentrations gives a clear relationship with the rate.