The problem
"A" stackrelcolor(blue)(k_1color(white)(m))(→) "B"Ak1m−−→B
"A" stackrelcolor(blue)(k_2color(white)(m))(→) "C"Ak2m−−→C
Derive the overall rate law and the relative amounts of "B"B and "C"C.
The differential rate law
(d["B"])/dt = k_1["A"]d[B]dt=k1[A]
(d["C"])/dt = k_2["A"]d[C]dt=k2[A]
-(d["A"])/dt = k_1["A"] + k_2["A"] = (k_1 + k_2)["A"]−d[A]dt=k1[A]+k2[A]=(k1+k2)[A]
Let k_3 = k_1 +k_2k3=k1+k2
Then
"rate" = -(d["A"])/dt = k_3["A"]rate=−d[A]dt=k3[A]
Integrated rate law for ["A"][A]
(d["A"])/dt = -k_3["A"]d[A]dt=−k3[A]
(d["A"])/"[A]" = -k_3dtd[A][A]=−k3dt
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
