anonymous
  • anonymous
The rate law for a hypothetical reaction is rate = k [A]^2. When the concentration is 0.10 moles/liter, the rate is 2.7 × 10^-5 M*s^-1. What is the value of k?
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
rate = k [A]^2 just plug in 2.7*10^(-5) = k*.1^2 ^rate ^.1 molles/liter maybe and then solve for k :)
anonymous
  • anonymous
k = 0.27e-2
anonymous
  • anonymous
umhm so wait... would it be the same as 2.7*10^-3M^-1s^-1?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
well unites should be M*s^-1 so ( moles/liter)^2* k's units= M*s^-1
anonymous
  • anonymous
umhm alright. Thank you!
anonymous
  • anonymous
is moles/liter=M?
anonymous
  • anonymous
remember the unit M=mol/L \[ 2.7\times10^{-5}\,{\rm M/s}=k\times \left(0.1\,{\rm M}\right)^2\\ k=\frac{2.7\times10^{-5}}{0.01}\times \frac{{\rm M}}{\rm s\cdot M^2}\\ \boxed{k=2.7\times10^{-3}{\rm M^{-1}\cdot s^{-1}}} \] the reaction is thus, a second order reaction.
anonymous
  • anonymous
Soo, I was right? :D woho! Thanks alot electrokid!
anonymous
  • anonymous
yaya. good job.

Looking for something else?

Not the answer you are looking for? Search for more explanations.