Question

reaction: A + 2 B yields products
rate data:
A: 0.10 0.20 0.30
B: 0.10 0.20 0.20
Rate(M/min) 0.010 0.40 0.060
what is rate law
a. k [B]^2
b. k [A] [B]
c. k [A]^2
d. k [A] [B]^2

Answer 1

are you sure you wrote the data correctly?

Answer 2

yes but it is vertical but it is easier to write it horizontal

Answer 3

Rate(M/min) 0.010 0.\(\color{red}0\)40 0.060

so it's not this ^

Answer 4

o sorry no it is 0.040 I didn't c that error my bad

Answer 5

0.04 = k[0.2]^n [0.2]^m
0.06 = k[0.3]^n [0.2]^m

dividing these two equation gives you a simplified one of

0.667 = 0.667^n

solving for n gives you "1", which is the exponent of the reactant A in the rate law "Rate = k[A]^1[B]^m", which means the reactant A follows a 1st order process

Answer 6

Now we can just choose one Rate law and plug in stuff to find the order for the reactant B

Answer 7

Let's choose the first experiment (column 1)

Answer 8

and let's divide it with experiment 2 (column 2)

Answer 9

So,

0.01 = k[0.1]^1[0.1]^m
0.04 = k[0.2]^1[0.2]^m

Dividing the two equations give you a simplified one of

0.25 = 0.5 x 0.5^m
0.5 = 0.5^m
so solving for m gives us 1, which is the exponent for reactant B which means B also follows 1 order kinetics

Answer 10

Helloooooo anyone there?

Answer 11

hi sorry yea

Answer 12

So are you good with this?

Answer 13

ye i am just not sure how I would do this without a calculator. It seems like this would take alot of time to do for a timed test.

Answer 14

You don't need a calculator. You can make this easier to do on paper by converting these decimals to fractions and dividing or multiplying these fractions

Answer 15

Ya these type of problems can be a little time consuming but if you remember the pattern of how to solve problems like these, it could take like less than a minute

Answer 16

with a calculator i guess

Answer 17

ok I will try to find more problems and c. thanks for the help

Answer 18

np