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#
