# For the overall reaction shown below with a fast equilibrium, why is the true rate law NOT #r(t) = k_2[A][B_2]#? Also, why is the overall reaction order #3//2# and not #3#?

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#A_2 stackrel(k_1" ")(rightleftharpoons) A + A# (fast)

#color(white)(aaa)^(color(green)(k_(-1)))#

#A + B_2 stackrel(k_2" ")(=>) AB + B# (slow)

#ul(A + B stackrel(k_3" ")(=>) AB)# (fast)

#A_2 + B_2 stackrel(k_(obs)" ")(->) 2AB#

#A_2 stackrel(k_1" ")(rightleftharpoons) A + A# (fast)

#color(white)(aaa)^(color(green)(k_(-1)))#

#A + B_2 stackrel(k_2" ")(=>) AB + B# (slow)

#ul(A + B stackrel(k_3" ")(=>) AB)# (fast)

#A_2 + B_2 stackrel(k_(obs)" ")(->) 2AB#

##### 1 Answer

I see what happened; what you wrote was a rate law that incorporated the intermediate

**OVERALL REACTION**

#\mathbf(A_2 + B_2 stackrel(k_(obs)" ")(->) 2AB)#

where

**REACTION MECHANISM**

#color(green)(A_2 stackrel(k_1" ")(rightleftharpoons) A + A)#

#color(white)(aaa)^(color(green)(k_(-1)))#

#color(green)(A + B_2 stackrel(k_2" ")(=>) AB + B)#

#color(green)(A + B stackrel(k_3" ")(=>) AB)# where

#=># indicates anelementary step, i.e. a step with no implicit mechanism.

An elementary step has **no intermediates** and only **one transition state**. So, we know that that's *exactly* how that step happens and we can treat each reactant as **first order**.

(If there was

**THE FAST EQUILIBRIUM APPROXIMATION**

Since the equilibrium is stated to be *fast*, apparently we are using the **fast equilibrium approximation**. We would not know this if we were not told the equilibrium was fast.

If we were told that the first step was slow, it could have been the steady-state approximation, which is not the same. That assumption says that

Under the **fast equilibrium approximation**, though:

- The reactant
#A_2# and intermediate#A# (but not#B# ) areassumed to be in equilibriumfor the equilibrium step.#r_1(t)# #">>"# #r_2(t)# , therefore#k_1# #">>"# #k_2# .#color(blue)(K_1 = (k_1)/(k_(-1)) = ([A]^2)/([A_2]))# because equilibrium is assumed to be already established.- (
#K_1# is evidently the equilibrium constant for step 1.)

So what's happening **overall** here is that the equilibrium is reached quickly, and then intermediate

**THE RATE OF THE RATE-DETERMINING STEP CAN BE USED TO REPRESENT THE OVERALL REACTION**

When writing the rate law for this, we would say that the **slow step** (associated with **rate-determining step** because the rate of the other two steps are insignificant relative to step 2.

So, the rate law for *that particular step* is a good approximation to the

*actual*rate law. In other words,

#r(t) = k_(obs)[A_2][B_2] ~~ r_2(t) = k_2[A][B_2]#

**THE RATE LAW MUST CONTAIN ONLY REACTANTS PRESENT IN THE OVERALL REACTION**

But we do have a problem: **intermediate**, which is NOT present in the original reaction, so we have to find a new expression for the intermediate ** overall** reaction.

Otherwise, it would be a completely different reaction:

#color(red)(2)A + B_2 stackrel(k_(obs)" ")(->) 2AB#

**DETERMINING THE OBSERVED RATE CONSTANT**

From the fast-equilibrium approximation and bullet point **3**, we get:

#[A] = ((k_1[A_2])/(k_(-1)))^"1/2"#

So our new rate law is:

#r(t) = k_2((k_1[A_2])/(k_(-1)))^"1/2"[B_2]#

Simplifying this, we get:

#color(blue)(r(t) = stackrel(k_(obs))overbrace(k_2((k_1)/(k_(-1)))^"1/2")[A_2]^"1/2"[B_2])#

And remember that we got

Therefore, the reaction order overall is