The reaction A + B ⟶ C + D rate = k [ A ] [ B ] 2 has an initial rate of 0.0440 M / s. What will the initial rate be if [ A ] is halved and [ B ] is tripled? initial rate: .00978 M / s What will the initial rate be if [ A ] is tripled and [ B ] is halved? initial rate:

Respuesta :

Answer :

(1) The initial rate when [A] is halved and [B] is tripled is, 0.198 M/s

(2) The initial rate when [A] is tripled and [B] is halved is, 0.0440 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 = 0.0440 M/s

As, Initial rate = [tex]k[A][B]^2[/tex] = 0.0440 M/s

Thus,

[tex]Rate=(\frac{9}{2})\times 0.0440M/s[/tex]

[tex]Rate=0.198M/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})\times (3\times [A])^2[/tex]

[tex]Rate=k\times (\frac{[B]}{2})\times 9\times [A]^2[/tex]

[tex]Rate=k\times (\frac{9}{2})\times [A]\times [B]^2[/tex]

Given:

Initial rate = 0.00978 M/s

As, Initial rate = [tex]k[A][B]^2[/tex] = 0.00978 M/s

Thus,

[tex]Rate=(\frac{9}{2})\times 0.00978M/s[/tex]

[tex]Rate=0.0440M/s[/tex]