Question

Consider the reactionA + B → ProductsFrom the following data obtained at a certain temperature, determine the order of the reaction. Enter the order with respect to A, the order with respect to B, and the overall reaction order.[A] (M) [B] (M) Rate (M/s)1.50 1.50 3.20 ×10−11.50 2.50 3.20 ×10−13.00 1.50 6.40 ×10−1

Answer 1

Answer: Order with respect to A is 1 , order with respect to B is 0 and total order is 1

Step-by-step 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:
`3.20* 10^(-1)=k[1.50]^x[1.50]^y` (1)

From trial 2:
`3.20* 10^(-1)=k[1.50]^x[2.50]^y` (2)

Dividing 2 by 1 :
`(3.20* 10^(-1))/(3.20* 10^(-1))=(k[1.50]^x[2.50]^y)/(k[1.50]^x[1.50]^y)`

`1=1.66^y,2^0=1.66^y` therefore y=0

b) From trial 2:
`3.20* 10^(-1)=k[1.50]^x[2.50]^y` (3)

From trial 3:
`6.40* 10^(-1)=k[3.00]^x[1.50]^y` (4)

Dividing 4 by 3:
`(6.40* 10^(-1))/(3.20* 10^(-1))=(k[3.00]^x[1.50]^y)/(k[1.50]^x[2.50]^y)`

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

Thus rate law is
`Rate=k[A]^1[B]^0`

Thus order with respect to A is 1 , order with respect to B is 0 and total order is 1+0=1.