Question

At 593K a particular decomposition’s rate constant had a value of 5.21×10−4 and at 673K the same reaction’s rate constant was 7.42×10−3. It was noticed that when the reactant’s initial concentration was 0.2264 M (with a 593K reaction temperature), the initial reaction rate was identical to the initial rate when the decomposition was run at 673K with an initial reactant concentration of 0.05999 M. Recall that rate laws have the form rate = k [A]x and, showing work, determine the order of the decomposition reaction.

Answer 1

The order of the decomposition reaction is 2.

Step-by-step explanation:

Rate constant of reaction at 593 K , =
`K_1= 5.21* 10^(-4)`

Rate constant of reaction at 673 K , =
`K_2= 7.42* 10^(-3)`

Rate of the reaction at 593 K =
`R_1`

Initial concentration of reactant = [A] = 0.2264 M

`R_1=K_1* [A]^x`

`R_1=5.21* 10^(-4)* [A]^x`...[1]

Rate of the reaction at 673 K =
`R_2`

Initial concentration of reactant = [A'] = 0.05999 M

`R_2=K_2* [A']^x`

`R_2=7.42* 10^(-3)* [A']^x`...[2]

`R_1=R_2` (given)

`5.21* 10^(-4)* [A]^x=7.42* 10^(-3)* [A']^x`

`(([A])/([A']))^x=(7.42* 10^(-3))/(5.21* 10^(-4))`

`((0.2264 M)/(0.05999 M))^x=14.24`

x = 1.999 ≈ 2

The order of the decomposition reaction is 2.