## Chemistry and Chemical Reactivity (9th Edition)

Plotting the data, of $ln[NH_3]$ vs t, we get the linear regression: $ln[NH_3]=-0.00586\times t-14.05$ with a $r^2=0.999$ For the $1/[NH_3]$ vs t: $1/[NH_3]=9224.4\times t+1.0\times10^6$ with a $r^2=1$ Since this regression leads to a slightly higher $r^2$, the process is more likely to be of second-order so: $k=9224.4\ L/mol.h$