Question

Determine the rate law, including the values of the orders and rate law constant, for the following reaction using the experimental data provided. (4 points)

A + B yieldsproducts

Trial

[A]

[B]

Rate

1 0.10 M 0.20 M 1.2 × 10-2 M/min
2 0.10 M 0.40 M 4.8 × 10-2 M/min
3 0.20 M 0.40 M 9.6 × 10-2 M/min

Answer 1

Answer: Rate law=
`k[A]^1[B]^2`, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is
`3L^2mol^(-2)s^(-1)`

Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

`Rate=k[A]^x[B]^y`

k= rate constant

x = order with respect to A

y = order with respect to A

n = x+y = Total order

a) From trial 1:
`1.2* 10^(-2)=k[0.10]^x[0.20]^y` (1)

From trial 2:
`4.8* 10^(-2)=k[0.10]^x[0.40]^y` (2)

Dividing 2 by 1 :
`(4.8* 10^(-2))/(1.2* 10^(-2))=(k[0.10]^x[0.40]^y)/(k[0.10]^x[0.20]^y)`

`4=2^y,2^2=2^y` therefore y=2.

b) From trial 2:
`4.8* 10^(-2)=k[0.10]^x[0.40]^y` (3)

From trial 3:
`9.6* 10^(-2)=k[0.20]^x[0.40]^y` (4)

Dividing 4 by 3:
`(9.6* 10^(-2))/(4.8* 10^(-2))=(k[0.20]^x[0.40]^y)/(k[0.10]^x[0.40]^y)`

`2=2^x,2=2^1`, x=1

Thus rate law is
`Rate=k[A]^1[B]^2`

Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.

c) For calculating k:

Using trial 1:
`1.2* 10^(-2)=k[0.10]^1[0.20]^2`

`k=3 L^2mol^(-2)s^(-1)`.