## Chemistry and Chemical Reactivity (9th Edition)

Plotting the data, of $ln[N_2O]$ vs t, we get the linear regression: $ln[N_2O]=-0.0128\times t-2.3038$ with a $r^2=0.999$, so it's definitely a first-order process. This regression leads to $k=0.0128\ min^{-1}$ The rate is given by: $r=k[N_2O]=0.0128\ min^{-1}\times 0.035\ mol/L=4.48\times10^{-4}\ mol/L.min$