Question

1. In this lab we will be expressing rate as a change in concentration of reactants over time. In order to we need to know the initial concentrations of our reagents after they are combined but before the reaction occurs. This is a dilution process. Recall that the calculation for a dilution is accomplished using the relationship MiV1 = M2V2. Complete the table below by calculating the concentrations of l'(aq) and Fe"(aq) in the initial reaction solutions. The stock solutions used in the trials are: 0.0500 M Fe(NO3)3 () 0.0500 M KI (a) 0.00250 M S.Oz (0) S202 Fe(NO3)2 Total solution volume Trial KI (mL) (mL) (mL) (including HNO3, H20 [1] (M) [Fe?"] (M) 18:03"](M) and starch; mL) 10.0 10.0 10.0 100.25 .00099 0.00499* 2.49x104 10.0 10.0 5.0 100.25 0.004990.00249 2.49x10-4 3 5.0 10.0 10.0 100.25 0.00249 *Provided example: [Fe3+] = (0.0100 L x 0.0500 mol/LY(0.10025 L) = 0.00499 mol/L CT] 0.005 X 0.0500)/0.10025 Che] =0.0 (DO 0.0500)/0.10025 2. Show work for one of the calculations in the I-column below Fill in Data Table 3 (report form) with the [Fe") and [T) you calculated in Prelab Question 1. 3. If the [S203) is initially 2.49x10-M, and the reaction takes 145 seconds for the solution to turn blue, what is the rate of the reaction reported in terms of 12? (See equation 3 in the discussion section of the procedure)

Answer 1

1. The initial concentration of the I^- and Fe³⁺ in the solution is calculated using the relationship for dilution. The calculation is given below:

For [P-J (M) 0 For Trial I. Mi = concentration of k l M₂ = Concentration of I in solution V = Volume of KI N2 = Volume of solution. For dilution MI V = MoV2 0.0500 M x 10.0lme) - M2*100.25 (ml) M2 = 0.0600 x 10.0_Migall 100125 ml = 0.00499 M = 4.99x10-3 M ③ For Terial 2 Mire for dilution. Mili= M2 V2 -0.0500 Mx 10.0(ml) - Mg x 100125 (ml) M2 = 0.0500x10 Mant R 100:25 at M2 = 0.00499 M = 4.99x103M ③ For Trial 3 For dilution, MN. = Ma V2. 0.0 500 na x 5.0(ml)= Mg x 100-25 ml -M2 = 0.0500 * 5.0 More - 100.25 mnt = 0.00249 M = 2.49 x103M Scanned with CamScanner

- For. [FeBt] (M) 0 For Trial 1 Me = Concentration of Fe(NO3), My » Concentration of [Fe3+] in solution Viz Goncalecal volume of Fe(NOG), Vos volume of W ig solution For dilution, MiNi= MeV/2. 0.0500 MX 10 (me) - M2 x 100.25 (me) M2 = 0.0500 x10 Moot 100 25 mt L = 0.00499 M = 4.99x1013 M For Trial 2 . For dilection, Mil, = M2/2 0.0500 MX 5.0 (me) = M2 x 100'25 (ml) - M2 = 0.0500 x 50 m not - 100·25 ml > 0.00249 M = 2:49x103 M ③ For Tercial 3. For dilution, Milli M2N2. 0.0500 M x 10.0 ml = M2 x100.25 ml M₂ = 0.0500x10 Mant 100.25 ml = 0.00499 M -4.99 X10-3M. Scanned with CamScanner

For [S2.032.J(M) 0 For Trial 1 My = concentration of 520² - My = concentration of 520 in solution Vi= Volume of S203² V2 = Volume of solution Far dileefon Mil, = Mala. 0.00250 M x 100 ml - M2. X100-25 ml M2 = 0.00250 X 10.0 manat 100.25 ml = 0.000249 M= 2.849 x104 M * For Tesal 2 Mili= MaV2 0000250 MX 1000 ml = Max 108.25 ml M2 = 0.00250 x 10'0 na bad 100.25 m Mz 0.000249 M = 2.49x109M ③ For Trial 3 Mili= Mala 0.00250M * 100ml - M2 x 100.25 ml M2 = 0.00250x1010 Marl 100'25 ml M2 = 0.000249 M = 2.498104M Scanned with C CamScanner

Trial	KI (mL)	S2O32- (mL)	Fe(NO3)3 (mL)	Total solution volume	[I-] (M)	[Fe3+] (M)	[S2O32-] (M)
1	10.0	10.0	10.0	100.25	4.99 x 10-3	4.99 x 10-3	2.49 x 10-4
2	10.0	10.0	5.0	100.25	4.99 x 10-3	2.49 x 10-3	2.49 x 10-4
3	5.0	10.0	10.0	100.25	2.49 x 10-3	4.99 x 10-3	2.49 x 10-4

3. The rate of the reaction is given as the change in concentration of the reactants with time.

The rate of reaction in terms of S₂O₃^2- is equal to rate of reaction in terms of I^-, as the rate of a reaction is constant for a particular reaction. We can find the rate of reaction when final concentration and initial concentration of the reactants for a time interval is given. The equation below can be used to find the rate of reaction:

Rate of Reaction = [Final concentration of Rectants)_Catial concentration gemsbody - Time o As only suitial concentration of 320 is given Radt of Reactions [Penel come.of spo3 ]-et (249x 104) MST Time [ final conc. of 5203.] - 2.49x1024 Mst 145 Tes Scanned with Senner

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