# Write the rate law for this reaction, and give the numerical value of the rate constant?

##

What will the observed initial rate be if the initial concentration

[Y]=0.400 and [Z]=0.300?

What will the observed initial rate be if the initial concentration

[Y]=0.400 and [Z]=0.300?

##### 1 Answer

The **rate law** in general just relates the rate

#r(t) = k[Y]^m[Z]^n# where

#m# is the order of reactant#Y# and#n# is the order of reactant#Z# . We do not know#m# or#n# yet, so we must find those to finish writing the rate law.

**RATE LAW ORDERS, AND RATE LAW**

To make our lives easier, let us set ** experiments 1 and 2** and the change in the initial

#r_(i,1)(t) = k[Y]_(i,1)^m[Z]_(i,1)^n#

#r_(i,2)(t) = k[Y]_(i,2)^m[Z]_(i,2)^n#

But since

#(r_(i,2)(t))/(r_(i,1)(t)) = (cancel(k)[Y]_(i,2)^mcancel([Z]_(i,2)^n))/(cancel(k)[Y]_(i,1)^mcancel([Z]_(i,1)^n))#

#(r_(i,2)(t))/(r_(i,1)(t)) = ([Y]_(i,2)^m)/[Y]_(i,1)^m#

#(1.6xx10^(-4))/(4.0xx10^(-5)) = ("0.200 M"/"0.100 M")^m#

If you do the math, you'd get a comparison:

#4 = 2^m#

Thus, **second order**, or the reaction is "second order with respect to

For ** experiments 1 and 3**, such that

#(r_(i,3)(t))/(r_(i,1)(t)) = ([Z]_(i,3)^n)/[Z]_(i,1)^n#

#(8.0xx10^(-5))/(4.0xx10^(-5)) = ("0.200 M"/"0.100 M")^m#

If you do the math, you'd get a comparison:

#2 = 2^m#

Thus, **first order**, or the reaction is "first order with respect to

Therefore, the **overall rate law** is:

#color(blue)(r(t) = k[Y]^2[Z])#

**RATE CONSTANT**

The **rate constant** is *specific to the reaction*, not to the experiment. That means we can pick *any* rate from *any* trial, and find the rate constant for the entire reaction.

#r_(i,1)(t) = k_1[Y]_(i,1)^2[Z]_(i,1)#

#4.0xx10^(-5) = k_1("0.100 M")^2("0.100 M")#

#=> color(blue)(k_1) = (4.0xx10^(-5) "M/s")/(("0.100 M")^2("0.100 M"))#

#= color(blue)("0.04 M"^(-2)cdot"s"^(-1))#

To prove that

#r_(i,2)(t) = k_2[Y]_(i,2)^2[Z]_(i,2)#

#1.6xx10^(-4) = k_2("0.200 M")^2("0.100 M")#

#=> color(blue)(k_2) = (1.6xx10^(-4) "M/s")/(("0.200 M")^2("0.100 M"))#

#= color(blue)("0.04 M"^(-2)cdot"s"^(-1))#

Thus, **the rate constant is the same for any trial we choose**, since it is merely the same reaction at a different test run.

At this point you should be able to finish the problem yourself. Once you have the rate constant, you can use your given concentrations in the last part of the problem to find the initial rate of the reaction that corresponds to those new concentrations.