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

Feb 10, 2016

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

#### Explanation:

The problem

$\text{A" stackrelcolor(blue)(k_1color(white)(m))(→) "B}$

$\text{A" stackrelcolor(blue)(k_2color(white)(m))(→) "C}$

Derive the overall rate law and the relative amounts of $\text{B}$ and $\text{C}$.

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 $\left[\text{A}\right]$

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

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

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

$\ln {\left[\text{A"]_"A₀"^"A} = - {k}_{3} t\right]}_{0}^{t}$

$\ln {\left[\text{A"] – ln["A}\right]}_{0} = - {k}_{3} t$

$\ln {\text{[A]/"[A]}}_{0} = - {k}_{3} t$

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

$\left[{\text{A"] = ["A}}_{0}\right] {e}^{- {k}_{3} t}$

Integrated rate law for $\left[\text{B}\right]$

(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)

${\left[\text{B"] = k_1/k_3["A}\right]}_{0} \left(1 - {e}^{- {k}_{3} t}\right)$

Integrated rate law for C

Similarly,

${\left[\text{C"] = k_2/k_3["A}\right]}_{0} \left(1 - {e}^{- {k}_{3} t}\right)$

Product ratio

${\left[\text{B"] = k_1/k_3["A}\right]}_{0} \left(1 - {e}^{- {k}_{3} t}\right)$

${\left[\text{C"] = k_2/k_3["A}\right]}_{0} \left(1 - {e}^{- {k}_{3} t}\right)$

"[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))))))

$\text{[B]"/"[C]} = {k}_{1} / {k}_{2}$