@BD.

Just for your own benefit: this is why just calculating k's from single points and averaging is not a good way to do things.

Using your data I calculate k*[CH_{4}] of 0.108 using your "averaging method" and 1.0979 using a basic linear regression with partial least squares. Not too bad, right?

Ok, well change your first point (t = 2) from 0.57 to 0.17. Using the averaging method now I get a value of 0.09328, a drop of about 9% or so. Using the PLS regression, I get a value for k*[CH_{4}] of 1.0917, a deviation of <1%.

Even a simple regression can tolerate an anomalous outlier point, whereas just averaging the values cannot, because it weights each point evenly.