Supplementary table 2
Mass balance for liquid phase
Day 0
Batch A
TNT
=
0.122
mmoles/L
Day 1
Day 3
TNT = 0.000 mmoles/L;
TNT
ADNTs
ADNTs = 0.000 mmoles/L;
ADNTs = 0.0 mmoles/L;
ADNTs
DANTs = 0.130 mmoles/L.
DANTs
mmoles/L;
0.044 mmoles/L.
 (TNT+ADNTs+DANTs) =
mmoles/L.
0.044 mmoles/L.
(TNT+ADNTs+DANTs)
0.13 mmoles/L
 (TNT+ADNTs+DANTs)
 (TNT+ADNTs+DANTs)
= 0.09 mmoles/L*
= 0.06 mmoles/L*
TNT
TNT
mmoles/L;
=
0.018
DANTs
=
Day 7
=
0.000
mmoles/L;
= 0.06 mmoles/L*
Batch B
TNT
=
mmoles/L
0.126
ADNTs = 0.093 mmoles/L;
ADNTs
DANTs = 0.033 mmoles/L. 
mmoles/L;
(TNT+ADNTs+DANTs)
0.053
0.011
DANTs
=
mmoles/L.
(TNT+ADNTs+DANTs)
= 0.13 mmoles/L
*The mismatch
0.13 mmoles/L
=
=
mmoles/L;
=
ADNTs
mmoles/L;
mmoles/L;
0.0
TNT
0.121
0.000
=
TNT = 0.000 mmoles/L;
=
=
TNT
Day 14
0.0
0.089
mmoles/L;
=
0.065
TNT
=
=
0.0
mmoles/L;
=
DANTs
0.0
ADNTs
0.018
=
mmoles/L;
=
DANTs
=
mmoles/L;
mmoles/L.

0.071 mmoles/L.
0.023
DANTs
=
(TNT+ADNTs+DANTs) =
 (TNT+ADNTs+DANTs)
0.12 mmoles/L
= 0.1 mmoles/L*
between the inlet molar concentration of TNT and the generated metabolites on Day 1 (Batch A)
is most likely due to the presence of
hydroxylaminodinitrotoluenes (HADNTs) in the reaction media, which are the first detectable metabolites of reduction of TNT. The mismatch between the inlet molar
concentration of TNT and the generated metabolites on Day 7 (Batch A) and Day 14 (Batch A and Batch B) most likely indicates the formation of TAT or other amino
products (Rieger and Knackmuss, 1995).
Mass balance for solid phase
Batch D
Batch E
Batch G
Batch F
Batch C
Given that the amount of the solution in
On Day 0 after the mixture in the
On Day 0 after the mixture in
Using the same
Using the same
the
TNT
bottle was hand-shaken in the
the bottle was hand-shaken in
aprroach as was
aprroach as
concentration in it was 28.15 mg/L, i.e. the
batches 11.75 mg/L in average
the
used for batches
was used for
solution contained 2.0 mg of TNT [0.07 (L)
remained in the solution, which
average
* 28.2 (mg/L) = 2.0 (mg)]. On Day 0 after
means that 0.8 mg [0.07 (L) *
solution, which
the mixture in the bottle was hand-shaken
11.75 (mg/L) = 0.80 (mg)] were
in the batches 16.90 mg/L in average
bottles
was
70
mL,
the
batches
12.2
mg/L
in
the
D, E and G, the
batches D, E
means
that
calculated
and G, the
0.86 mg [0.07 (L) * 12.2 (mg/L)
extraction yield
calculated
adsorbed on pine bark. Since
= 0.85 (mg)] were adsorbed on
of TNT on Day 0
extraction yield
remained in the solution, which means
the average TNT concentration
pine bark. Since the average
for Batch F was
of TNT on Day
that 15.4 mg/L [28.2 (mg/L) – 16.9 (mg/L)
in 800 µL of acetonitrile on Day0
TNT concentration in 800 µL of
96 %.
0 for Batch C
= 11.3 (mg/L)] or 0.8 mg [0.07 (L) * 11.3
was 175.3 mg/L, 0.14 mg of
acetonitrile on Day0 was 162.0
(mg/L) = 0.8 (mg)] were adsorbed on pine
TNT was extracted from 0.5 g of
mg/L, 0.13 mg of TNT was
bark.
pine bark (wet weight) [175.3
extracted from 0.5 g of pine
concentration in 800 µL of acetonitrile on
(mg/L) * 8* 10-4
= 0.140
bark (wet weight) [162.0 (mg/L)
Day0 was 164.2 mg/L, 0.131 mg of TNT
(mg)]. The total mass of wet
* 8* 10-4 (L) = 0.13 (mg)]. The
was extracted from 0.5 g of pine bark (wet
pine bark in batch C after it’s
total mass of wet pine bark in
content was centrifuged was 3.0
batch C after it’s content was
0.131 (mg)]. The total mass of wet pine
g.
the
centrifuged was 3.2 g. Thus,
bark in batch C after it’s content was
simultaneously adsorbed TNT
0.83 mg of the simultaneously
centrifuged was 3.1 g. Thus, 0.79 mg of
could be extracted from 3.0 g of
adsorbed
the simultaneously adsorbed TNT could
wet PB with ACN. The extraction
extracted from 3.2 g of wet PB
be extracted from 3.1 g of wet PB with
yield of TNT was 0.84 (g) / 0.80
with ACN. The extraction yield
ACN. The extraction yield of TNT was
(g) * 100 (%) = 105 (%).
of TNT was 0.83 (g) / 0.86 (g)
Since
the
average
weight) [164.2 (mg/L) * 8* 10
-4
TNT
(L)
0.79 (g) / 0.8 (g) *100 (%) = 98 (%).
=
Thus,
0.84
(L)
mg
of
remained
in
TNT
*100 (%) = 97 (%).
could
be
was 94 %.