Table 2.

Calculation of the amount of H₂O₂ eliminated (x) at indicated time t.

k	t	[Pyr]0	[H2O2]0	D	E	x	[H2O2]
(µM−1 m−1)	(min)	(µM)	(µM)	(µM)		(µM)	(µM)
0.000142	33.0	150	50	−100	0.6267	23.59	26.41
0.000142	22.0	1,000	50	−950	0.0518	47.53	2.47

*The values for k, t, [Pyr]₀, and [H₂O₂]₀ were entered. The remaining terms are calculated using the following equations: D = [H₂O₂]₀ − [Pyr]₀; E = $e^{ktD}$; $x=\frac{({[\mathrm{Pyr}]}_0{{[\mathrm{H}}_2{\mathrm{O}}_2]}_0-{[\mathrm{Pyr}]}_0{{[\mathrm{H}}_2{\mathrm{O}}_2]}_0\mathrm{E}}{{[\mathrm{Pyr}]}_0-{{[\mathrm{H}}_2{\mathrm{O}}_2]}_0\mathrm{E}}$; and [H₂O₂] = [H₂O₂]₀ − x.