## Chemistry 9th Edition

Zero order: $[A]_t=-kt+[A]_0$ First order: $ln[A]_t=-kt+ln[A]_0$ Second order: $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$
Zero order: We are given that the rate equals $\frac{-d[A]}{dt}=k[A]^0$; therefore, we take the equation $\frac{-d[A]}{dt}=k[A]^0$ and integrate both sides from 0 to t to get $\int_{[A]_0}^{[A]_t}d[A]=-\int_0^tkdt$. Simplifying this integral gives $[A]_t-[A]_0=-kt$ which can be rewritten as $[A]_t=-kt+[A]_0$. First order: We are given that the rate equals $\frac{-d[A]}{dt}=k[A]^1$; therefore, we take the equation $\frac{-d[A]}{dt}=k[A]$ and integrate both sides from 0 to t to get $\int_{[A]_0}^{[A]_t}\frac{d[A]}{[A]}=-\int_0^tkdt$. Simplifying this integral gives $ln[A]_t-ln[A]_0=-kt$ which can be rewritten as $ln[A]_t=-kt+ln[A]_0$. Second order: We are given that the rate equals $\frac{-d[A]}{dt}=k[A]^2$; therefore, we take the equation $\frac{-d[A]}{dt}=k[A]^2$ and integrate both sides from 0 to t to get $\int_{[A]_0}^{[A]_t}\frac{d[A]}{[A]^2}=-\int_0^tkdt$. Simplifying this integral gives $-\frac{1}{[A]_t}+\frac{1}{[A]_0}=-kt$ which can be rewritten as $\frac{1}{[A]_t}=kt+\frac{1}{[A]_0}$.