# Consider the reaction 3A + B + C ---> D + E where the rate law is defined as -D[A]/Dt=k[A]^2[B][C]?

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The initial concentration of #B# and #C# are both #"1.00 M"# . The initial concentration of #A# is #1.00 xxx 10^(-4) "M"# , and after 3.00 min, it became #3.26 xx 10^(-5) "M"# .

#1)# Find the rate constant for the pseudo-second-order reaction in #"M"^(-3)cdot"s"^(-1)# .

#2)# Find the pseudo-second-order half-life in seconds.

#3)# Under what assumptions are #(1)# and #(2)# true?

#4)# Determine the concentration of #A# after 10.00 min have passed.

The initial concentration of

##### 1 Answer

**DISCLAIMER:** *LONG ANSWER!*

Alright, first, let's write down what we were given here, and also what we can write out from what we know so far.

#A + B + C -> D + E# (I assume the 3A is a typo, because the rate given,

#-(Delta[A])/(Deltat)# , has no coefficient when it should if the stoichiometric cofficient of#A# were not#1# .)

- Initial concentration of
#B# ,#["B"]_0 = ["C"]_0 = "1.00 M"# - Initial concentration of
#A# ,#["A"]_0 = 1.00 xx 10^(-4)# #"M"# - Current concentration of
#A# ,#["A"]_"3.00 min." = 3.26 xx 10^(-5) "M"#

#r(t) = k[A]^2[B][C]#

#= -1/1 (Delta[A])/(Deltat) = -1/1 (Delta[B])/(Deltat) = -1/1 (Delta[C])/(Deltat)#

#= +1/1 (Delta[D])/(Deltat) = +1/1 (Delta[E])/(Deltat)#

The rate constant can be found by using the **integrated rate law** to track the fraction of just

The integrated rate law with respect to **ONLY** **second-order** version as noted from the rate law, and fortunately we have tracked what the concentration of **in the context of**

This version can be derived (assuming

#bb(overbrace(1/([A]))^(y) = overbrace(k)^(m)overbrace(t)^(x) + overbrace(1/([A]_0))^(b))# assuming

#A# has a stoichiometric coefficient of#1# . Get to know this equation, because we'll be using it a lot.

Now, let's plot the two data points we have (and hope that that's enough).

Since the slope is **in the context of**

#overbrace("M"/"min")^(r(t)) = overbrace(1/("M"^3 cdot "min"))^(k) xx overbrace("M"^2)^([A]^2) xx overbrace("M")^([B]) xx overbrace("M")^([C])#

In the units of your answer key though:

#color(blue)(k) = "6891.6 M"^(-3)cdot"min"^(-1) xx "1 min"/"60 s"#

#= color(blue)("114.86 M"^(-3)cdot"s"^(-1)) ~~ 115#

Since we know the form of the integrated rate law for just

#1/(1/2[A]_0) = kt_"1/2" + 1/([A]_0)#

#1/([A]_0) = kt_"1/2"#

Therefore, the half-life for the reaction based on only the second-order behavior of

#color(blue)(t_"1/2") = 1/(k[A]_0)#

#= 1/(114.86(1.00 xx 10^(-4))) " s"#

#=# #color(blue)("87.06 s")#

As it turns out, we have been assuming in

That is not strictly true, unless the concentration of

Using the second-order integrated rate law once more, denote

Then:

#1/([A]) = kt + 1/([A]_0)#

#1/(alpha[A]_0) - 1/([A]_0) = kt#

#(1/alpha - 1)/([A]_0) = kt#

#=> kt[A]_0 + 1 = 1/alpha#

So, the **fraction of** **leftover** is:

#alpha = 1/(kt[A]_0 + 1)#

#= 1/(114.86 cdot 600 (1.00 xx 10^(-4)) + 1)#

#= 0.1267#

Therefore, the amount of

#color(blue)([A]_"10.00 min") = alpha[A]_0 = 0.1267 xx 1.00 xx 10^(-4)#

#= color(blue)(1.27 xx 10^(-5) "M")#