# Find the rate constant for parallel or competitive reactions?

##
#a)# #k_"obs" = k_1+k_2+...+k_n#

#b)# #k_"obs" = nk_i#

#c)# #k_"obs" = k^n#

##### 1 Answer

I get

*DISCLAIMER: LONG ANSWER!*

Okay, so for the 1st order reaction, we have:

#A stackrel(k_1" ")(->) A_1#

#A stackrel(k_2" ")(->) A_2#

#" "vdots#

#A stackrel(k_n" ")(->) A_n#

We can say that the same starting

Also, we are partitioning the same reactant

Next, we can write the rate law for each parallel competitive **subreaction**:

#r_1(t) = k_1[A] = (d[A_1])/(dt)# (Eq. 1.1)

#r_2(t) = k_2[A] = (d[A_2])/(dt)# (Eq. 1.2)

#" "vdots#

#r_n(t) = k_n[A] = (d[A_n])/(dt)# (Eq. 1.n)where:

- The
reaction rateis#r_i# . These may or may not be different across subreactions.- The
rate constantis#k_i# . These may or may not be different across subreactions.- The
partitioned concentrationis#[A]_i# . These are presumed to be the same across each subreaction.- The
rate of appearanceof#A_i# with respect to timeis#(d[A_i])/(dt)# . These may or may not be different across subreactions.

The full rate law would be

#\mathbf(r(t) = k[A] = -(d[A])/(dt)),# (Eq. 2)

and *total initial* concentration.

Now, ultimately, since each subreaction is presumed to start at the same time, and the sum of all the rates of appearance of products is equal to the rate of disappearance of

#-(d[A])/(dt) = (d[A_1])/(dt) + (d[A_2])/(dt) + cdots + (d[A_n])/(dt),# (Eq. 3)

and consequently, from **Eqs. 1.1 - 1.n** and **Eq. 3**, we get:

#r(t) = -(d[A])/(dt) = r_1(t) + r_2(t) + cdots + r_n(t)# .(Eq. 4)

Now, adding up **Eqs. 1.1 - 1.n** and utilizing **Eq. 4**:

#r(t) = r_1(t) + r_2(t) + cdots + r_n(t)#

#= k_1[A] + k_2[A] + cdots + k_n[A]#

#= (k_1 + k_2 + cdots + k_n)[A]#

Therefore, comparing back to **Eq. 2**:

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

#= color(blue)((k_1 + k_2 + cdots + k_n) [A])#

Comparing back to the options:

#\mathbf(a)# **matches this result**.#b# would imply that#n[A] = [A]# , but we already assumed that#n ne 1# .#c# would change the units of#k# compared to each#k_i# , which doesn't make sense. It would falsely imply the units of#k# to be#"s"^(-n)# instead of#"s"^(-1)# , for a*1st*-order reaction.