## Chemistry 9th Edition

a) Average rate of decomposition of $H_2O_2$: $2.31\times 10^{-5}\space \frac{mol}{L\times s}$ Average rate of production of $O_2$: $1.16\times 10^{-5}\space \frac{mol}{L\times s}$ b) Average rate of decomposition of $H_2O_2$: $1.16\times 10^{-5}\space \frac{mol}{L\times s}$ Average rate of production of $O_2$: $5.8\times 10^{-6}\space \frac{mol}{L\times s}$
a) The average rate of decomposition of $H_2O_2$ can be found by taking the negative change in the amount of $H_2O_2$ and dividing that by the change in time. By using the values given to us, we get $\frac{-(0.500\space mol/L-1.000\space mol/L)}{2.16\times10^4\space s-0\space s}=\frac{0.500\space mol}{2.16\times10^4\space s}$ which equals $2.31\times 10^{-5}\space \frac{mol}{L\times s}$ The reaction is $2H_2O_2\rightarrow 2H_2O+O_2$, so for every $2\space mol$ of $H_2O_2$ consumed, $1\space mol$ of $O_2$ is produced, and the rate of production of $O_2$ is $\frac{1}{2}$ times the rate of decomposition of $H_2O_2$. Therefore, the rate of production of $O_2$ is $2.31\times 10^{-5}\space \frac{mol}{L\times s}\times \frac{1}{2}$ or $1.16\times 10^{-5}\space \frac{mol}{L\times s}$. b) The average rate of decomposition of $H_2O_2$ can be found by taking the negative change in the amount of $H_2O_2$ and dividing that by the change in time. By using the values given to us, we get $\frac{-(0.250\space mol/L-0.500\space mol/L)}{4.32\times10^4\space s-2.16\times10^4\space s}=\frac{0.250\space mol}{2.16\times10^4\space s}$ which equals $1.16\times 10^{-5}\space \frac{mol}{L\times s}$. The reaction is $2H_2O_2\rightarrow 2H_2O+O_2$, so for every $2\space mol$ of $H_2O_2$ consumed, $1\space mol$ of $O_2$ is produced, and the rate of production of $O_2$ is $\frac{1}{2}$ times the rate of decomposition of $H_2O_2$. Therefore, the rate of production of $O_2$ is $1.16\times 10^{-5}\space \frac{mol}{L\times s}\times \frac{1}{2}$ or $5.8\times 10^{-6}\space \frac{mol}{L\times s}$.