Answer
See the answer below.
Work Step by Step
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:
$k=9224.4\ L/mol.h$
