Question

In this lab we will be expressing rate as a change in concentration of reactants over time. In order to calculate this, 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 M1V1 = M2V2.

1. In this lab we will be expressing rate as a change in concentration of reactants over time. In order to calculate this 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 I'(aq) and Fe+ (aq) in the initial reaction solutions. The stock solutions used in the trials are: 0.0500 M Fe(NO3)3(aq) | 0.0500 M KI (aq) 0.00250 M S2032 (aq) S2032 Fe(NO3)3 Total solution volume Trial KI (mL) (mL) [Fe3+] (M) [S2O3+] (M) (including HNO3, H20 (mL) and starch; mL) 1 10.0 10.0 10.0 100.25 0.00499* 10.0 10.0 5.0 100.25 5.0 10.0 10.0 100.25 *Provided example: [Fe3+] = (0.0100 L x 0.0500 mol/L)/(0.10025 L) = 0.00499 mol/L

If the [S2O3 2- ] is initially 2.49x10-4 M, and the reaction takes 145 seconds for the solution to turn blue, what is the rate of the reaction reported in terms of I2? (See equation 3 in the discussion section of the procedure)

Answer 1

FE(NO₃)3 7-M) Festm) 5202 (M) Trial KIme (mid ป Total vol. (me) 100.25 10 10 10 10 0.0049 0.0049 0.00025 100.25 0.0049 0.0049 0.000125 100:25 0.0049 0.0049 0.00025 20.00025 Initial concentration of 5₂00 = 2.498104 m The reaction takes 145 seconds time. so the rate of reaction = 3.49810 145 sec = 1.717 X10-6 MIS This rate is in terms of [5₂0g 2-1. 9.49x10-4M on =