Could you derive the rate law for two competing first order reactions and the formula for the product ratio, please?

1 Answer
Feb 10, 2016

Answer:

The rate law is #"rate" = -(k_1+ k_2)["A"]#

Explanation:

The problem

#"A" stackrelcolor(blue)(k_1color(white)(m))(→) "B"#

#"A" stackrelcolor(blue)(k_2color(white)(m))(→) "C"#

Derive the overall rate law and the relative amounts of #"B"# and #"C"#.

The differential rate law

#(d["B"])/dt = k_1["A"]#

#(d["C"])/dt = k_2["A"]#

#-(d["A"])/dt = k_1["A"] + k_2["A"] = (k_1 + k_2)["A"]#

Let #k_3 = k_1 +k_2#

Then

#"rate" = -(d["A"])/dt = k_3["A"]#

Integrated rate law for #["A"]#

#(d["A"])/dt = -k_3["A"]#

#(d["A"])/"[A]" = -k_3dt#

#int_("A₀")^"A" (d["A"])/"[A]" = -int_0^tk_3dt#

#ln["A"]_"A₀"^"A" = -k_3t]_0^t#

#ln["A"] – ln["A"]_0 = -k_3t#

#ln"[A]/"[A]"_0 = -k_3t#

#"[A]"/["A"]_0 = e^(-k_3t)#

#["A"] = ["A"_0]e^(-k_3t)#

Integrated rate law for #["B"]#

#(d["B"])/dt = k_1["A"] = k_1["A"]_0e^(-k_3t)#

#int_0^"B" d["B"] = int_0^t k_1["A"]_0e^(-k_3t)dt#

#["B"] = k_1/k_3["A"]_0e^(-k_3t)]_0^t = -k_1/k_3["A"]_0(e^(-k_3t) –e^0) = -k_1/k_3["A"]_0(e^(-k_3t) –1) #

#["B"] = k_1/k_3["A"]_0(1 -e^(-k_3t)) #

Integrated rate law for C

Similarly,

#["C"] = k_2/k_3["A"]_0(1 -e^(-k_3t)) #

Product ratio

#["B"] = k_1/k_3["A"]_0(1 -e^(-k_3t)) #

#["C"] = k_2/k_3["A"]_0(1 -e^(-k_3t)) #

#"[B]"/"[C]" = (k_1/color(red)(cancel(color(black)(k_3)))color(red)(cancel(color(black)(["A"]_0(1 -e^(-k_3t))))))/ (k_2/color(red)(cancel(color(black)(k_3)))color(red)(cancel(color(black)(["A"]_0(1 -e^(-k_3t))))))#

#"[B]"/"[C]" = k_1/k_2#