Question

The reaction A→B has been experimentally determined to be second order. The initial rate is 0.0100M/s at an initial concentration of A of 0.300M .


What is the initial rate at [A]=0.900M ?

Answer 1

Have you written the rate law?

Answer 2

no

Answer 3

that's step number 1.

Answer 4

rate=k[A]^2

Answer 5

good stuff, now find \(k\) with the values initially given in the question

Answer 6

.0100=k[.900]^2 is this the correct setup

Answer 7

close, the rate of 0.01 M/s is for \([A]=0.300~M\)

Answer 8

we trippled the .300 to go to .90 so would i tripple the intial rate to get .300

Answer 9

that would work if the reaction was linear but it's not, it's second order.

Answer 10

so how would i find the right answer

Answer 11

first find \(k\)

Answer 12

first i find k then that would allow me to plus that in for the next equation

so

.0100=k[.300]^2
k=.1111 or 1/9

then

rate=1/9[.900]^2
.09=rate

Answer 13

perfect

Answer 14

Yes thank you so much

Answer 15

no problem !