Question

The reaction A + B ⟶ C + D rate = k [ A ] [ B ] 2 A+B⟶C+Drate=k[A][B]2 has an initial rate of 0.0640 M / s. 0.0640 M/s. What will the initial rate be if [ A ] [A] is halved and [ B ] [B] is tripled?What will the initial rate be if [A] is tripled and [B] is halved? answer ________ M/s

Answer 1

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:

$Rate=k[A][B]^2$

Now we have to determine the initial rate when [A] is halved and [B] is tripled.

The new rate law expression will be:

$Rate=k\times (\frac{[A]}{2})\times (3\times [B])^2$

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

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

Given:

Initial rate = 0.0640 M/s

As, Initial rate = $k[A][B]^2$ = 0.0640 M/s

Thus,

$Rate=(\frac{9}{2})\times 0.0640M/s$

$Rate=0.288M/s$

Now we have to determine the initial rate when [A] is tripled and [B] is halved.

The new rate law expression will be:

$Rate=k\times (\frac{[B]}{2})^2\times (3\times [A])$

$Rate=k\times (\frac{[B]^2}{4})\times 3\times [A]$

$Rate=k\times (\frac{3}{4})\times [A]\times [B]^2$

Given:

Initial rate = 0.0640 M/s

As, Initial rate = $k[A][B]^2$ = 0.0640 M/s

Thus,

$Rate=(\frac{3}{4})\times 0.0640M/s$

$Rate=0.096M/s$