Answer:
[tex]r = k [A]^{2}[B]^{2}[/tex]
Explanation:
A + B + C ⟶ D
[tex]\text{The rate law is } r = k [A]^{m}[B]^{n}[C]^{o}[/tex]
Our problem is to determine the values of m, n, and o.
We use the method of initial rates to determine the order of reaction with respect to a component.
(a) Order with respect to A
We must find a pair of experiments in which [A] changes, but [B] and C do not.
They would be Experiments 1 and 2.
[B] and [C] are constant, so only [A] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{2}}{r_{1}} & = & \dfrac{ k[A]_{2}^{m}}{ k[A]_{1}^{m}}\\\\\dfrac{2.50\times 10^{-2}}{6.25\times 10^{-3}} & = & \dfrac{0.100^{m}}{0.0500^{m}}\\\\4.00 & = & 2.00^{m}\\m & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to A}[/tex]
(b) Order with respect to B
We must find a pair of experiments in which [B] changes, but [A] and [C] do not. There are none.
They would be Experiments 2 and 3.
[A] and [C] are constant, so only [B] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{3}}{r_{2}} & = & \dfrac{ k[B]_{3}^{n}}{ k[B]_{2}^{n}}\\\\\dfrac{1.00\times 10^{-1}}{2.50\times 10^{-2}} & = & \dfrac{0.100^{n}}{0.0500^{n}}\\\\4.00 & = & 2.00^{n}\\n & = & \mathbf{2}\\\end{array}\\\text{The reaction is 2nd order with respect to B}[/tex]
(c) Order with respect to C
We must find a pair of experiments in which [C] changes, but [A] and [B] do not.
They would be Experiments 1 and 4.
[A] and [B] are constant, so only [C] is changing.
[tex]\begin{array}{rcl}\dfrac{r_{4}}{r_{1}} & = & \dfrac{ k[C]_{4}^{o}}{ k[C]_{1}^{o}}\\\\\dfrac{6.25\times 10^{-3}}{6.25\times 10^{-3}} & = & \dfrac{0.0200^{o}}{0.0100^{o}}\\\\1.00 & = & 2.00^{o}\\o & = & \mathbf{0}\\\end{array}\\\text{The reaction is zero order with respect to C.}\\\text{The rate law is } r = k [A]^{2}[B]^{2}[/tex]
