Question

Fe3(aq) FESCN (aq) SCN (aq) In this experiment, we'll be examining the temperature-dependence of the rate constant for the reaction above. 1. Using the UV-Vis instrument, you measured a room temperature absorbance for FESCN2 at 477 nm of 0.815. What is its concentration in molarity (M)? Please show all work. The molar absorptivity for FeSCN2 is 4,258 м"сm1 a. Answer: b. You calculated the [FeSCN2] above from the observed room temperature absorbance. Which one of the following statements about the calculated [FeSCN2] is correct? i. The number calculated above in part b is the equilibrium [FeSCN2']eq. i. The number calculated above in part b is the initial [FeSCN2]in iii. The number calculated above in part b is the initial [FeNO3]in iv. None of the above 2. In this experiment, you will mix 5.00 mL of 0.0030 M Fe(NO3)3 and 3.00 mL of 0.0050 M KSCN, and you will dilute the solution to a total volume of 25.00 mL What is the initial concentration of Fe(NO3)3 in molarity (M)? Show your work below. a. Answer: What is the initial concentration of KSCN in molarity (M)? Show your work below b.

Answer 1

1)a)we know A= €Cl , A= absorbance of the solution, €= molar absorptivity constant , C = concentration and l= length of the cell . Here, A= 0.815 and €= 4258 lit mol-1 cm-1 , l= 1 . So, C = A/(€×l) = 0.815/4258 = 1.914×10^-4 (M) . So, the concentration of the solution = 1.914×10^-4(M). b) As the above reaction is in equilibrium at room temperature, and the absorbance is measured at room temperature so option 1 is correct. .. 2) a)b) Here, the concentration of Fe(NO3)3 = 0.0030(M), Volume= 5 ml , Total volume of the solution= 25 ml ,. so concentration in mixture= (5×0.0030 )/25 = 6×10^-4(M) . Similarly, concentration of KSCN = 0.0050(M), volume of solution= 3ml , Now in mixture concentration= 3×0.0050 / 25 = 6 ×10^-4(M). But , if we measure the concentration before dilution then total volume of solution = 8 ml. So, the concentration of Fe(NO3)3 = 5×0.0030/8 = 1.875×10^-3(M). Similarly, concentration of KSCN = 3×0.0050/8 = 1.875×10^-3(M).