Question

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

Answer 1

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:

$\frac{1}{[A]}=kt+\frac{1}{[A]_0}$

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

Answer 2

Answer:

- 0th: $C_A=C_{A0}-kt$

- 1st: $C_A=C_{A0}exp(-kt)$

- 2nd: $\frac{1}{C_A}=kt+\frac{1}{C_{A0}}$

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:

$\frac{dC_A}{dt}=-k\\ \int\limits^{C_A}_{C_{A0}} {} \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\C_A-C_{A0}=-kt\\C_A=C_{A0}-kt$

- 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:

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

- 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

$\frac{dC_A}{dt}=-kC_A^{2} \\ \int\limits^{C_A}_{C_{A0}} {\frac{1}{C_A^{2} } } \ dC_A= \int\limits^{t}_{0} {-k} \ dt\\-\frac{1}{C_A}+\frac{1}{C_{A0}}=-kt\\\frac{1}{C_A}=kt+\frac{1}{C_{A0}}$

Best regards.