Question

The initial rates method can be used to determine the rate law for a reaction. Using the data for the reaction shown below, what is the rate law expression for the reaction?

A+B→C

Trial123[A](molL)0.0650.0650.078[B](molL)0.0650.0780.065−Δ[A]Δt(molL s)2.29311875×10−72.29311875×10−73.9625092×10−7

• Write the rate law expression using two significant figures for the rate constant.
• Use brackets for any needed concentrations ([A], [B], or [C]).
• Simplify the expression as much as possible. Do NOT use exponents of 1 or 0.

Answer 1

ANSWER:

Reaction:

A + B +0

Data:

Trial	[A] (mol/L)	[B] (mol/L)	-Δ[A]/Δt (M/s)
1	0.065	0.065	2.29311875x10-7
2	0.065	0.078	2.29311875x10-7
3	0.078	0.065	3.9625092x10-7

According to initial rate method, the rate law for this reaction is

v = k ABY

where k = rate constant and x,y = reaction order of A and B

To find the reaction order of A and B, we need to compare the changes in the reaction rate (-Δ[A]/Δt) when the reactant concentrations are changed:

• Reaction order of B
  • We need to keep constant [A] and change [B] => we use the data of trial 1 and 2
  • In this case the change in [B] does not modify the reaction rate:

V1 = 2.29311875r10-7 M/s = um

• This means that reaction rate does not depend of concentration of B. Then, B is zero order (y = 0)

v = k[A]" [B]° = k[A]"

• Reaction order of A
  • ​​​​​​​We need to keep constant [B] and change [A] => we use the data of trial 2 and 3
  • In this case, the change in [A] modifies the reaction rate:

V1 = 2.29311875.c10-7 M/s

U3 = 3.9625092x10-7 M/S

• This means that reaction rate depends of concentration of A. And the order of A is

​​​​​​​
V1 = k[Ali
and   
V3 = k[A

then

Vi k[Ali U3k[A] Ali [A]

$\frac{v_{1}}{v_{3}}=\left (\frac{[A]_{1}}{[A]_{3}} \right )^{x}$

100(%) – 2 109 (

(問) @!

x = (2.29311875r 10-7 M/S log ( 3.9625092r10-7 M/s) - log (6.065 M)

x= 3

• The reaction is of third order respect A

Then, the rate law of this reaction is

v=k[A]

Now, we need to find the value of the constant rate (k). Using the rate law, we get

k = [A13

• using the data of trial 1

k= 2.29311875.r10-7 M/s (0.065 M3

k = 8.35x10-4 M2.s

Finally, the rate law is

​​​​​​​

v = 8.4x10-4- [A] 3 M2.s