Question

Practice 08: Reactions With Multiple Pathways Key In the presence of O ferrous ion, Fe is oxidized to ferrie lon, Fet. The rate law for this reaction depends on the pH, following one rate law when the pH is greater than 3.6 and a different rate law when the pH IN below 3.5. Part 1: pH > 3.5 The following initial rates (defined is greater than 3.5. ***/dt) were measured ting solutions buffered to pH levels [Fe2+(M) 1.00 x 10-4 2.00 x 10-4 1.00 x 10-4 2.00 x 10-4 pH 5.0 5.0 4.0 5.0 Po, (atm) 0.20 0.20 0.20 0.30 Rate (M/day) 24 x 10-6 48 x 10-6 2.4 x 10- 72 x 10-6 Use this data to determine the reactions orders and the rate constant (with appropriate units) for the rate law # kı[Fe2+1"[H,0**P Note that the rate law is written in terms of H,0+, not pH. Answer. The data for Part I is an example of the initial rate method for finding a rate law. The key to working with data of this sort is to isolate pairs of experiments in which the concentration of only one species is changing • To find the reaction order for po" we compare experiments 1 and 2, finding that doubling its concentration results in a doubling of the rate, thus, the rate is first-order in Fe2+. • To find the reaction order for O, we compare experiments 2 and 4, finding that increasing the partial pressure by 50% leads to a 50% increase in the rate thus, the rate is first-order in O • To find the reaction order for H,0" we compare experiments 1 and 3, finding that increasing its concentration by a factor of 10 causes the rate to decrease by a factor of 100; thus, the rate is inverse second-order in H,0°. Yes, sometimes a species can slow a reaction down. At this point we know that the rate law is * = 1,[Fe2+][H30+1Po. To find the nate constant we substitute concentrations, partial pressures, and rates from each experiment into the mte law and solve for . Doing this for each experiment produces the same result of 1.2 x 10 M atmd Pay particular attention to the units and convince yourself that they are correct. Part II: pH <3.5 Interestingly, at pH levels less than 3.5, the concentration of Hot no longer affects the reaction's rate and the rate law is * = kx[F.**] Po, The data below were collected using p udo-order conditions in which the partial pressure of O, is hekil constant at 0.20 atm. Under these conditions, the rate law is ka[Fe2+1

the two photos above are information need so solve the question below

Answer 1

Answert:- in the above in foumation we get to know the values of Ki and Kzgie Ki= 1x 10""' m2 cutm ' ' Kg: 102x10~4d-' almy Answer to Here given when ki [Fe2+] [H2O ] Po, = Kg [Fe2+] PO2 – 0 To find pH at condition -(1) So, Ki [Fe 2+ ][H3 0+1"? Poz- Ky[ fe2t] Poz Kz 30 Ki [H30*7*2Kg [M90*1= KAO Taking log on both sides we get rog [H30ty? = log ( k ) - Log(430") = log (ua xe ) | .2 XIO

Answer:- m the above on foumation we get to know the values of ki and Kagie KI = 1.2x10'' M12 atm'& Kz: 102x104d' army answer lo Here given when Ki [feat] [H3O+] Po, = Ky [Fe2+] PO2 – 0 To find pH at condition -(0) 50g Ko [Fe2+ ][H3077" Pozo K3[Fe2t] Poz Ki [M30*1**Ky [H90**? = Kg Taking log on both sides we get dog [ Hy oty?? log () a log[190"] log (OS X 12") 1.2 X10 - 2 (-PH)= log(13) [p4-609 [Hgot] åpH = log (10) & Pý - L. pm - Ya = 3.5 PH p*= 3-5% on. = 3.5 When the pH is equal to 3.5 then the in is one pathway to determine the date law fose the feat to Fe3+, Answer 2. Given intial concentration of Feat is 1.0x1074 M Po2 = 0.2 atm Recte law - ņ= ki [Fe2+ ][490+]** PO2 + K3 [Fe 2+] P02 R = Kn [feat] [H30 *jPOL log [R] = log ki + log [fe?"] + loq [30*tog (es) log [n] = log (1.2 X18') + log (1.0*10*) - 2.6091430*] + log (0:2) HYTTETTU log (R) - Log (1-2) + 09 (10") + Log(co4) + 2 pH + log (0-2) = 0.079 - 11 - 4 +2PH +60.69) OOCR) = 2PM - 15.61 comparing with y=mx + c [straight line 7 equation ] pot fi 128 45 16 17 § 9 10 .112.113.14 log [R] [1361-11.61-9.614-7061|-5-61-361|-7-610139|a.394.39639 839|10:39| 12:39 ine Steeaight line quaph tells that with the increase in pH the log coate) also increases. So this shows the oxidation will be more. → Nog [R] T 2 3 4 5 6 7 8 9 10 11 12 13 14 15 PHY -12 15 Scanned with CamScanner