[ 1) 8,b · 2019-09-20 · t = 19 year solve for k = 20 t-1, and substitute k into the integrated...
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.[1) 8,b Using units of ~As N'': <
(a) Convert from mg/L to mole/L: i [No-1- 70 rngNOs X l,molo~ X lgNO~ -11 X 10-a moleNO& O,S' rb thh
• 3 - L 62gN a 1000mg . Os - · . L 1-} ii [NH+] -100 mgl'ffl4 x lmoJ11Nll, X 1,NH.& -5 6 X 10-s moleNHf J.r ' .4 - -i;- l811Nll4 lOO0mgNH~ - ' L
iii [NH l - 0 5 mgrHs X 1roolo§ X lgNH~ - 2 9 X 10-6 molLNH3 • 3 - · 17gNHa 1000Jng 7·11 - •
iv. Total moles/L = 1.1 x 10-3 + 5.6 x 10-3 + 2.9 x 10-6 = 6.7 x 10-3 ~
(b) Convert from mole/L to m':L°N :Note that one mole of ND;, Nllt, and NHa each contain <me mole of NJ
i. [NO-] = 70 ~ X lipolllNOa· X 1 m~ X . ...l!l!L = 16 !Y1i 3 L 02g•N0s lbtO 3 linoleN -i;-
ii. [NHt] = 100 ~ X 1~go~ ~ 1~~~~~4 X 1~~:N =_78~
iii [NB)= 0 5 ~ X lrno Nils X 11110 ~ X ~ = 0 41 !!!Ill: • . 8 • L 17~NH.1 imol.111'1 finoloN' • -i;-
iv. (N03] + (NHt] + fNHs] = 16 "'f N + 78 ~ + 0.41 "'f N = 94 !Df N
(c) Convert from mole/L to~:
, i. 1.1 X 10-s moJo~O'i := 1.1 X 10-a rnolqN ii. 5.6 >< 10-3 moteLNnt = 5.6 x 10-8 mn];N
iii. 2.9·x 10-5 moltHs = 2.9 X 10-6 mo~N
iv. ·fNOsJ + (NHt] + (NHs] = 1.1 x 10-3 + 5.6 x 10-s ~ 2.9 x 10-11 =·6.7 x 10-s morN
(d) Perts (a). and (c) are equivalent b auae there is 1 mole of N per mole of N03, 1 mole + . of N per mole of NH.,1 and 1 mole f N per mole of NH4 • ~ I tJ
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@ B f''p
L = fbC, o-+...,· L f-:- n- r:i aJl.(..
JL11 -= [5>2-Ja [o;Jre--i)
_____ .- ----·
[oJ c~f) s lcP-"'iVL j
/ I.
3. {B,h) [VC] - 1!!!g -0 05 g (aq) - 60mL - ·
L
MWvc = 62.5 ~
[VCJ(aq) = 0.05 f X 1r~~ = 8 X 10-4 mte I rt
Ktt = 26.8 Lsatm/mole
(a)
Write the equilibrium relationship of aqueous and gaseous VC-make sure you write the equation in such a way that the units of Ktt work out.
= 26.8 Leatm/mole
Calculate the concentration of VC (moles/L) in the air from Henry's constant:
VC(g) = Ktt x VC(aq) = 26.8x8.0x 10-4 = 0.0214atm
p 0.021 atm [VC]g = VC 106 = -- X 106 = 21, 400 ppm,
X 1 atm
Ptot
21,400 ppm, > > 10 ppm, /
BAD NEWS!
(b) The students should have put the open ( or unsealed bottle) in the fume hood as quickly as possible and told everyone to leave the lab immediately. (;Xf,a.,A- ~ fr..i
4. CS (0) H2S04 is a strong acid:
(a) To calculate moles of [H+], calculate number of moles of H2S04 and multiply by 2:
lo mg H2S04 x 1 g 1 moles 2 mole H+ 4 mole + --- X --- X ---- = 2 X 10- -- H
L 1000 mg 98 g mole H2S04 L
pH= -log[H+] = -log(2 x 10-4) = 3.7 lf r
(b) Normality of H2S04 solution equal the moles per Liter of Ht, which equals 2 x 10-4 ~ ;l.,J:-1 L
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~ I ti Acld-Baee Equillbrlwn for HOC! in drinking wat«-, ~ 0 r 'Calculate &action of BOCl \hat Js -in the dlsaoclateod form:
plra = 7.5 K.- 10-T~ Sbloe it'• drbiJdng water, ~ume pH - 7. g+ = 10-7 .
K. [H+l(QOJ-J - 10-'l',II 3,,.t,- ,.., (HOOi) - I
Rearrange equ/ltion, end substitute wluea for K. and a+:
(001-1 = 10-o.a >< (HOOl] = 0'.316(HO01)
~b I
_ (BOOl] =- [HOOi 1 Q/h aactionDiss~~ted -:- fHOOlJ + [OCl-J . 001 + 0.316(HOCI • l + O;:fl~ == ~
1
(~}' ~ ,ts E,qm,,,9 50 mg/L 9f HCO5 aa: • eq/L:
..: mg lmole l eq lg -4 · [HCOa] = 60-L x -61 x -1 -1 x 1000 == 8.2 x 10 eq/L • ·g Ill O mg
• mole/L Since 1 eq of HOOa1 = 1 mole of HCOi1,
• mg/L as OaCOa: (HC01] = 8.2 x 10""' mole/L kcf rf:S
[HOOjj] = 8.2 x 10-'eq/L x !iOg oq Oa x lO 10mg = '1.mg/LuO..00.
Calculate alkalinity in mg/L 88 CaC08:
[HCO-J • 111 ~ X l mole lg X l (!(II = 1.8 X 10-a eqL of alkalinity I L Gig X lOOOtn mo . .
~ mg lmole lg 2e J -4 eq (CVs ] = 11 y x 60 g x 1000 mg x ~ - 15, 1 x 10 1 of alkallnity
Approximate Alkalinity (eq/L) = 1.8 x 10-a + 5.7 >< 10-• = 2.4 x 10-8 ! ~k Approxlme.t e Alkalinity (eq/L) = 2., >< 10-8 f X 100,n!CO;, x lleq = 119 IJ:I u CaOOa, ~
First Order Reaction:
(a) 90% of A destroyed, SQ 10% of initial concentration remains.
0.1.Ao = Ao X e-0.1'
ln(Y:.!) .... 1 t = =230d- - G.o)d ay -1 , ~-
(b) Same•as part (a) except that 99% destroyed, BO 1% temainmg:
-~) 3 J.. t = o.o• = 400 days ,~
(c) Same as pa.rt (a) except tha~ 00.9% destroyed, BO 0.1% remaining:
_ t.la.ooO ~Ah t= o,o\ =690days -=,-~
(Q) - . ,~,~ "'6il Spill nineteen yea.rs ago:
Co = 400 '¥ C alter 19y = 400 i;' i. 'D:y a zero order rate equation:
J dC -=-k dt
After integration:
C=Co-kt where Co = A00 ~ 0=20 !!!ft g
- t = 19 year
Solve for k = 20 t-1, and substitute k into the integrated zero rate eque.ti'on above to obtain
C = Co - 20y8&1'-l x 20year = 0
Answer: Yes the engineer ie correct if the degradation rate is zero order. i,-li ii. To find the "worst-case ecenario," calcula.te the concentration of the pollutant after
twenty yea.re using a first order and second order rate equation.
First Order:
Solve fork:
· ln.52. · ·,ln~ k= -~ = - = 0.l6y-1 t · 19y
. Use k and solve for the time it will take to for C = 1 1e:, assuming first.order kinetics:
1n 0. }n...L t = -~ = - 400 = 38y k 0.16yea.r-\ ·
Second Order:
Co C=l+C
Rearrange and solve fork:
k = l:_i = (t - i&i) ~ = 0.0025 kg t 19y mg xy
Uae k and solve for the time it "7ill t• for q = 1 f, aasum~ second-order kinetics:
I - I (-!--..L)k t = L.]i!_ = = 1 400 I< IJIII = 39!)y
k 0.003~
In conclusion the "worst-case scena.rio" is second order, in which case, it would take ~99 y for the pollutant to degrade. However, first order is more likely,, ,3 f +s
(D -~C ~ ~ - -
I I I de -}Ldb - J{_ rrf-, - - C 2.
- c.. t r -·2. -l•f' /; ) C JL - Co 0
-J:: ~- - -~t' tJ - f -J C,.;,
-[t-t] - -}l-t- 1 tl-J - I - .J_ -t- ~t - - G G,
_L.._ - + {01'-t (_
Co
\ C -=- Co
)+ (o }..t ! ~ rh
0