What is the rate law for the following mechanism in terms of the overall rate constant k? Step 1: #A + B leftrightarrow C# (fast) Step 2: # B + C -> D# (slow)

Question

127337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-16T01:42:18

The rate law is #"rate" = k["A"]["B"]^2#.

The reaction is

#"A + B"color(white)(l) ⇌ "C"# (fast)
#"B + C" color(white)(l)stackrelcolor(blue)(k_2color(white)(m))(→) "D"# (slow)

The rate of the reaction is governed by the slow step.

This gives the rate equation

#"rate" = k["B"]["C"]#

The problem is that #"C"# isn't one of the reactants.

We must express #"[C]"# in terms of #"[A]"# and #"[B]"#.

We can do that because the first step is an equilibrium.

#K = "[C]"/"[A][B]"#

∴ #["C"] = K["A"]["B"]#

Substituting into the original rate expression, we get

#"rate" = k_2["B"] × K["A"]["B"] = k_2K["A"]["B"]^2#

Let #k = k_2K#, and the rate law becomes

#"rate" = k["A"]["B"]^2#