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anonymous
 3 years ago
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?
anonymous
 3 years ago
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?

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0rate = k [A]^2 just plug in 2.7*10^(5) = k*.1^2 ^rate ^.1 molles/liter maybe and then solve for k :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0umhm so wait... would it be the same as 2.7*10^3M^1s^1?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well unites should be M*s^1 so ( moles/liter)^2* k's units= M*s^1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0umhm alright. Thank you!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0remember 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
 3 years ago
Best ResponseYou've already chosen the best response.0Soo, I was right? :D woho! Thanks alot electrokid!
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