Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. X + Y yields products

Answer 1

you haven't provide the experimental data. Usually is a table like this

experiment [X], M [Y], M rate = -d[X]/dt, M h-1
1 0.5 0.5 1.2
2 1 0.5 4.8
3 2 1 38.4

Answer 2

sorry ill put that now'

Answer 3

Trial [X] [Y] Rate
1 0.20 M 0.15 M 2.4 × 10-2 M/min
2 0.20 M 0.30 M 4.8 × 10-2 M/min
3 0.40 M 0.30 M 19.2 × 10-2 M/min

Answer 4

@cuanchi

Answer 5

take the two first esperiments
Trial [X] [Y] Rate
1 0.20 M 0.15 M 2.4 × 10-2 M/min
2 0.20 M 0.30 M 4.8 × 10-2 M/min

the concentration of X is the same and the concentration of Y increase twice, the rate also increase twice
then the order for Y will be =1

Rate = k [X]^m [Y]^n

n= 1

then you take the Trials 2 and 3
2 0.20 M 0.30 M 4.8 × 10-2 M/min
3 0.40 M 0.30 M 19.2 × 10-2 M/min

here the concentration od Y is constant and the concentration of X change to double and the rate increase 4 times then the order for X is....?????

Answer 6

i don't really understand this lesson

Answer 7

so is it 2

Answer 8

m=2???

Answer 9

@cuanchi what do i do after since i have n=1 m=2?

Answer 10

write the rate law

Rate = k [X]^m [Y]^n

replace the m and n for the values

Answer 11

also you can calculate the value of k

Answer 12

so k[A]^2[B]^1

Answer 13

then i use one of the trials above right

Answer 14

yes

Answer 15

so [.20]^2[.15]^2

Answer 16

[.15]^1

Answer 17

2.4 × 10-2 M/min=k[.20]^2[.15]^1

Answer 18

what do i do now? @cuanchi

Answer 19

k=2.4 ?/m^3s^3

Answer 20

i mean M^3 m^3

Answer 21

i think i got it is it k=2.4 4/M^3 m^3

Answer 22

@sweetburger

Answer 23

isolate the k

Answer 24

i dont know what to do after that to be honest

Answer 25

k= Rate / ([X]^m [Y]^n)

Answer 26

so its not 1?

Answer 27

Oh i was looking at this " [.20]^2[.15]^2"

Answer 28

If you determined its 1. Then you are correct.

Answer 29

ok

Answer 30

k= 2.4 × 10-2 M/min 4/M^3 m^2?

Answer 31

NO

Answer 32

k= Rate / ([X]^m [Y]^n)

k= 2.4 × 10-2 M/min / ((0.20 M ^1) (0.15 M ^2))

Answer 33

\[k =\frac{ 2.4 \times 10^{-2}M/\min }{ (0.2M)^{1}\times(0.15M)^{2} } \]

Answer 34

k=.0027

Answer 35

to double check you can replace this value in the rate law and calculate the rate with the concentrations of the Trial 2

Rate = k [X]^m [Y]^n

2 0.20 M 0.30 M 4.8 × 10-2 M/min

and see if you get 4.8 × 10-2 M/min

Answer 36

same thing

Answer 37

thx