Question

Write and the integrated rate laws hor zeroth-first- second-order rate laws.

Answer 1

- 0th:
`C_A=C_(A0)-kt`

- 1st:
`C_A=C_(A0)exp(-kt)`

- 2nd:
`(1)/(C_A)=kt+(1)/(C_(A0))`

Step-by-step explanation:

Hello,

For the ideal reaction A→B:

- Zeroth order rate law: in this case, we assume that the concentration of the reactants is not included in the rate law, therefore the integrated rate law is:

`(dC_A)/(dt)=-k\\ \int\limits^(C_A)_{C_(A0)} {} \ dC_A= \int\limits^(t)_(0) {-k} \ dt\\C_A-C_(A0)=-kt\\C_A=C_(A0)-kt`

- First order rate law: in this case, we assume that the concentration of the reactant is included lineally in the rate law, therefore the integrated rate law is:

`(dC_A)/(dt)=-kC_A\\ \int\limits^(C_A)_{C_(A0)} {(1)/(C_A) } \ dC_A= \int\limits^(t)_(0) {-k} \ dt\\ln((C_(A))/(C_(A0)) )=-kt\\C_A=C_(A0)exp(-kt)`

- Second order rate law: in this case, we assume that the concentration of the reactant is squared in the rate law, therefore the integrated rate law is

`(dC_A)/(dt)=-kC_A^(2) \\ \int\limits^(C_A)_{C_(A0)} {(1)/(C_A^(2) ) } \ dC_A= \int\limits^(t)_(0) {-k} \ dt\\-(1)/(C_A)+(1)/(C_(A0))=-kt\\(1)/(C_A)=kt+(1)/(C_(A0))`

Best regards.

Answer 2

Step-by-step explanation:

The integrated rate law for the zeroth order reaction is:

`[A]=-kt+[A]_0`

The integrated rate law for the first order reaction is:

`[A]=[A]_0e^(-kt)`

The integrated rate law for the second order reaction is:

`(1)/([A])=kt+(1)/([A]_0)`

Where,

`[A]` is the active concentration of A at time t

`[A]_0` is the active initial concentration of A

t is the time

k is the rate constant