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The reaction has an initial rate of 0.0300 M/s.


A+B YIELDS C+D rate=k[A][B]^2


What will the initial rate be if [A] is halved and [B] is tripled? (answer) _________ M/s


What will the initial rate be if [A] is tripled and [B] is halved? answer ________ M/s

Respuesta :

CintiaSalazar

Answer:

The initial rate if [A] is halved and [B] is tripled is 0.135 [tex]\frac{m}{s}[/tex]

The initial rate if [A] is tripled and [B] is halved is 0.0225 [tex]\frac{m}{s}[/tex]

Explanation:

You have:

  • A + B → C + D
  • rate=k*[A]*[B]² and being the initial rate= 0.0300 m/s → 0.0300 m/s=k*[A]*[B]²

If [A] is halved and [B] is tripled, the new concentrations are:

  • [A]=[A]÷2
  • [B]=[B]*3

Replacing these new concentrations in the velocity expression:

[tex]rate=k*\frac{[A]}{2} *(3*[B])^{2}[/tex]

[tex]rate =k*\frac{[A]}{2} *9*[B]^{2}[/tex]

[tex]rate=\frac{9}{2}* k*[A] *[B]^{2}[/tex]

Replacing k*[A]*[B]² with 0.0300 m/s:

[tex]rate=\frac{9}2} *0.0300 \frac{m}{s}[/tex]

[tex]rate=0.135 \frac{m}{s}[/tex]

The initial rate if [A] is halved and [B] is tripled is 0.135 [tex]\frac{m}{s}[/tex]

If [A] is tripled and [B] is halved, the new concentrations are:

  • [A]=[A]*3
  • [B]=[B]÷2

Replacing these new concentrations in the velocity expression:

[tex]rate=k*[A]*3 *(\frac{[B]}{2} )^{2}[/tex]

[tex]rate=k*[A]*3*\frac{[B]^{2}}{4}[/tex]

[tex]rate =\frac{3}{4}* k*[A] *[B]^{2}[/tex]

Replacing k*[A]*[B]² with 0.0300 m/s:

[tex]rate =\frac{3}{4}* 0.0300 \frac{m}{s}[/tex]

[tex]rate=0.0225 \frac{m}{s}[/tex]

The initial rate if [A] is tripled and [B] is halved is 0.0225 [tex]\frac{m}{s}[/tex]

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