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Cyanide redox equilibria
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Topic: Cyanide redox equilibria (Read 3489 times)
Alberto
Frequent Contributor
Posts: 14
Cyanide redox equilibria
«
on:
July 12, 2015, 12:49:57 AM »
Hello. I am trying to solve a problem without success. First some background:
Cyanide(CN-) salts (mainly NaCN) are used to dissolve gold.
Unfortunately the process leaves tailings rich in cyanide that are a hazard to the environment and to public health. Cyanide with time degrades to cyanate, and then to ammnonia and carbonate. I want to simulate that process, but first I need to make a database for cyanide that is not present in PHREEQC.
I have thermodynamic tables about cyanide compounds, but what I really have not found are the oxidation states of carbon and nitrogen in cyanide. That oxidation states are needed to see if I must (or not) introduce a new "master species" for cyanide.
For example, the
THERMODDEM database
( [url][/http://thermoddem.brgm.fr]) assumes that the oxidation states are
N(-5)
and
C(+4)
. I however seriously doubt that nitrogen could get 5 extra electrons in its outer shell and carbon be still in its most oxidized state. And with respect to the simulations, nearly all the cyanide I put into solution become cyanate (OCN-), and carbon is not reported in the results.
A more plausible option could be
N(-3)
and
C(+2)
, but I am not sure. And this itroduces the problem of including cyanide in the solution, since the default ("master") species for that oxidation states are ammonia(NH3) and carbon monoxide (CO). As far as I know, CO does not form out of cyanide as would be suggested by the reaction:
CO + NH3 = CN- + H2O + H+
log_k -14.33
delta_h 81.79 kJ
So maybe my guess is wrong.
The other cyano compound, cyanate, decomposes to ammonia and bicarbonate:
OCN- + 2H2O = HCO3- + NH3
log_k 56.05
delta_h -41.85 kJ
So probably the carbon is C(+4) and nitrogen is N(-3).
I attach the failed calculation using the thermoddem database, download it from THERMODDEM (made by the French Geological Survey)
Thank you in advance for your help.
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dlparkhurst
Top Contributor
Posts: 3619
Re: Cyanide redox equilibria
«
Reply #1 on:
July 12, 2015, 03:27:56 AM »
If you define cyanide as CN and cyanate as OCN, that is, as part of the C (carbon) and N (nitrogen) equilibria, I think you will find that neither species will exist at equilibrium under any reasonable chemical environments. In other words, they are metastable and thermodynamically unstable relative to other C and N species.
I think you want to define new master species Cyanide and Cyanate. These species will react only by REACTION or KINETICS. The minteq.dat database has these species defined. So you can either use minteq.dat or extract the Cyanide and Cyanate species and add them to another database.
An example that kinetically transforms Cyanide to Cyanate to C/N species follows. The rates of reaction were arbitrarily assigned.
#DATABASE minteq.dat
SOLUTION 1
temp 25
pH 13 charge
pe -11
redox pe
units mmol/kgw
density 1
Na 1
Cyanide 1
-water 1 # kg
RATES
Cyanide_2_Cyanate
-start
10 k = 1 # per day
20 k = k / 86400 # per sec
30 rate = k * TOT("Cyanide")
40 moles = rate * TIME
50 save moles
-end
Cyanate_2_C_and_N
-start
10 k = 0.5 # per day
20 k = k / 86400 # per sec
30 rate = k * TOT("Cyanate")
40 moles = rate * TIME
50 save moles
-end
END
KINETICS 1
Cyanide_2_Cyanate
-m 1
# CN- + 0.5O2 = OCN-
-formula OCyanide -1 Cyanate 1
Cyanate_2_C_and_N
-m 1
# OCN = OCN
-formula Cyanate -1 OCN 1
-step 864000 in 100
END
USE solution 1
USE kinetics 1
USER_GRAPH 1
-headings TIME Cyanide Cyanate N(-3)
-axis_titles "TIME, IN DAYS" "MOLALITY" ""
-initial_solutions false
-connect_simulations true
-plot_concentration_vs x
-start
10 GRAPH_X TOTAL_TIME / 86400
20 GRAPH_Y TOT("Cyanide"), TOT("Cyanate"), TOT("N(-3)")
-end
-active true
END
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