For the reaction, A + 2 B > C + 2D, some measurements of the initial rate of rea

ID: 837513 • Letter: F

Question

For the reaction, A + 2 B > C + 2D, some measurements of the initial rate of reaction at varying concentration gave the following data.

From this data, what is the rate law for the reaction? Select one:

a. The rate law is: rate = k[A]2[B]

b. The rate law is: rate = k[A][B]2

c. The rate law is: rate = k[A]2[B]2

d. The rate law is: rate = k[A][B]

e. The rate law is: rate =[A][B]3

i keep getting rate =k[A][B]3 which isnt listed. can someone step me through this so i can see what im doing wrong? Thanks

I .Let us take concentration of A constant :

Then ,let rate = k[A]^a [B]^b

Frome the table, taking second and third rows, we get :

1. 0.000540 = k (150)^a (200)^b

2. 0.001055 = k(150)^a (250)^b

Dividing 2 by 1, we get :

0.001055/0.000540 = (250/200)^b

=> 1.9537 = (5/4)^b

Taking log on both sides :

ln(1.9537)=b ln(5/4)

From here, we get : b=3

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II .Let us take concentration of B constant :

Then ,let rate = k[A]^a [B]^b

Frome the table, taking first and second rows, we get :

1. 0.000540 = k (150)^a (200)^b

2. 0.000360 = k(100)^a (200)^b

Dividing 1 by 2, we get :

0.000540/0.000360 = (150/100)^a

=> 1.5 = (1.5)^a

From here, we get : a=1

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Thus, the rate law is :

Rate= K [A][B]^3

The correct option is e.

I hope this helps. Please rate.

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