What is the rate law equation for the following reaction (given the experimental data below)?

1 Answer
Write your answer here...
Start with a one sentence answer
Then teach the underlying concepts
Don't copy without citing sources
preview
?

Answer

Write a one sentence answer...

Answer:

Explanation

Explain in detail...

Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

3
Mar 9, 2018

Answer:

The rate law is #r = k["A"]^2#.

Explanation:

The rate law is

#r = k"[A]"^m#

where

#rcolor(white)(l) =# the reaction rate
#k color(white)(l)=# the rate constant
#m =# an integer whose value we must determine

Consider Trial Runs 1 and 2

#r_2/r_1 = (color(red)(cancel(color(black)(k)))(0.8 color(red)(cancel(color(black)("mol/L")))) ^m)/(color(red)(cancel(color(black)(k)))(0.4 color(red)(cancel(color(black)("mol/L"))))^m) = (0.8/0.4)^m = (8.0 color(red)(cancel(color(black)("mol·L"^"-1""s"^"-1"))))/ (2.0 color(red)(cancel(color(black)("mol·L"^"-1""s"^"-1"))))#

#2^m =4.0#

Thus, doubling the concentration quadruples the rate.

The reaction is second order in #["A"]#.

If you use logarithms, you get

#color(blue)(mlog2 = log4.0)#

#color(blue)(m = log4.0/log2 = 2.0 ≈ 2)#

The reaction is second order in #"[A]"#.

The rate law is #r = k["A"]^2#.

Was this helpful? Let the contributor know!
1500
Impact of this question
30 views around the world
You can reuse this answer
Creative Commons License