Question

Consider the reaction: 3A + B + 2 C -> D + E where rate = k[A]2 [C]2 .A reaction was performed where [B]o = [C]o = 0.600 M and [A]o = 2.20 x 10-5 M. A plot of 1/[A] vs time (min) gave a plot with a straight line relationship and after 7.00 min, [A] = 2.4 x 10-6 M. What was the initial rate of the reaction described above?

Answer 1

Step-by-step explanation:

As the given reaction is
`3A + B + 2C \rightarrow D + E`

Also, rate = k
`[A]^(2)[C]^(2)`

So, rate is zero order with respect to B and it is second order with respect to both A and C.

It is given that [A] =
`2.4 * 10^(-6) M` and
`[A]_(0) = [tex]2.20 * 10^(-5) M`

For second order rate law, equation is as follows.

`(1)/([A]) = (1)/([A]_(0)) + kt`

`(1)/(2.4 * 10^(-6)) = (1)/(2.20 * 10^(-5)) + k * 7 min`

k =
`0.412 * 10^(5) M^(-1) sec^(-1)`

As, the initial rate = k
`[A]^(2)_(o) [C]^(2)_(o)`

= 0.412 \times 10^{5} M^{-1} sec^{-1} \times (2.20 \times 10^{-5} M)^{2} \times (0.6 M)^{2}[/tex]

=
`0.717 * 10^(-5) M^(3) sec^(-1)`

Thus, we can conclude that the initial rate of the reaction described above is
`0.717 * 10^(-5) M^(3) sec^(-1)`.