Respuesta :
Answer :
(1) The initial rate when [A] is halved and [B] is tripled is, 0.288 M/s
(2) The initial rate when [A] is tripled and [B] is halved is, 0.096 M/s
Explanation:
The given rate law expression is:
[tex]Rate=k[A][B]^2[/tex]
Now we have to determine the initial rate when [A] is halved and [B] is tripled.
The new rate law expression will be:
[tex]Rate=k\times (\frac{[A]}{2})\times (3\times [B])^2[/tex]
[tex]Rate=k\times (\frac{[A]}{2})\times 9\times [B]^2[/tex]
[tex]Rate=k\times (\frac{9}{2})\times [A]\times [B]^2[/tex]
Given:
Initial rate = 0.0640 M/s
As, Initial rate = [tex]k[A][B]^2[/tex] = 0.0640 M/s
Thus,
[tex]Rate=(\frac{9}{2})\times 0.0640M/s[/tex]
[tex]Rate=0.288M/s[/tex]
Now we have to determine the initial rate when [A] is tripled and [B] is halved.
The new rate law expression will be:
[tex]Rate=k\times (\frac{[B]}{2})^2\times (3\times [A])[/tex]
[tex]Rate=k\times (\frac{[B]^2}{4})\times 3\times [A][/tex]
[tex]Rate=k\times (\frac{3}{4})\times [A]\times [B]^2[/tex]
Given:
Initial rate = 0.0640 M/s
As, Initial rate = [tex]k[A][B]^2[/tex] = 0.0640 M/s
Thus,
[tex]Rate=(\frac{3}{4})\times 0.0640M/s[/tex]
[tex]Rate=0.096M/s[/tex]
The rate when [A] is halved and [B] is tripled rate and when [A] is tripled and [B] is halved is mathematically given as
Rate=0.288M/s
Rate2=0.096m/s
What are the rate when [A] is tripled and [B] is halved and vice versa?
Question Parameter(s):
ate=k[A][B]2^2
Has an initial rate of 0.0640 M / s.
Generally, the equation for the Chemical Reaction is mathematically given as
A + B ⟶ C + D
Hence, The rate law is
rate=k[A][B]2Y^2
The rate for [A] is halved and [B] is tripled is
[tex]Rate=k* (\frac{[A]}{2})* (3*[B])^2\\\\Rate=k* (\frac{9}{2})*[A]* [B]^2\\\\[/tex]
Hence
Rate =k[A][B]2Y^2
Rate=9/2* 0.0640
Rate=0.288M/s
In conclusion, when [A] is tripled and [B] is halved.
[tex]Rate=k*(\frac{3}{4})* [A]* [B]^2[/tex]
Hence
Rate2=3/4* 0.0640
Rate2=0.096m/s
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