Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constan

Question

Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 x 10 3 in HNo, mL M KS and 2.00 mL of water, She finds that in the M Fe (NO) 10-5 M equilibrium mixture the concentration of FesCN is 2.o x Find K for the reaction Fe3+ (aq) sCN aq Fe(SCN) +(aq) Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3.) moles 3 moles S Fe How many moles of FescN are in the mixture at equilibrium? What is the volume of the equi- Step 2 librium mixture? (Use Eq. 3.) mui moles FeSC N2+ How many moles of Fe" and SCN are used up in making the FescN moles 3 moles S Fe Step 3 How many moles of Fes and SCN remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps 1 and 2.) 3+ moles SCN moles Fe Step 4 t are the concentrations of Fe SCN and FescN?+ at equilibrium? (Use Eq. 3 and the results of Step 2 and Step 3.) M, FeSCN2+] M M: ISCN Step 5 What is the value of Ke for the reaction? (Use Eq. 2 and the results of Step 4.) (continued on following page)

Explanation / Answer

Answer to Q1)

Step 1

Number of moles of Fe3+ initially present = Concentration of Fe(NO3)3 solution X Volume of Fe(NO3)3 solution

= 2 X 10-3 M X 5 X 10-3 L

= 1 X 10-5 mol

Number of moles of SCN- initially present = Concentration of KSCN solution X Volume of KSCN solution

= 2 X 10-3 M X 3 X 10-3 L

= 6 X 10-6 mol

Step 2

The reaction is given as:

Fe3+ + SCN- <==> FeSCN2+

The total volume of the mixed solution = Volume of Fe(NO3)3 solution + Volume of KSCN solution + Volume of water

= 5 mL + 3 mL + 2 mL = 10 mL

Number of moles of FeSCN2+ at equilibrium = Concentration of FeSCN2+ solution X total volume of the mixed solution

= 7 X 10-5 M X 10 X 10-3 L

= 7 X 10-7mol

At equilibrium, the number of moles of Fe3+ and SCN– used up = Number of moles of FeSCN2+ formed = 7 X 10-7mol

Step 3

At equilibrium, the number of moles of Fe3+ and SCN– can thus be expressed as

Change in number of moles of Fe3+ = Number of moles of Fe3+ initially present – Number of moles of FeSCN2+ at equilibrium

= 1 X 10-5 mol - 7 X 10-7mol

= 9.3 X 10-6 mol

Change in number of moles of SCN– = Number of moles of SCN– initially present – Number of moles of FeSCN2+ at equilibrium

= 6 X 10-6 mol - 7 X 10-7mol

= 5.3 X 10-6 mol

Step 4

[Fe3+]e = Change in number of moles of Fe3+ / Total volume (L)

= 9.3 X 10-6 mol / 1 X 10-2 L

= 9.3 X 10-4 M

[SCN-]e = Change in number of moles of SCN– / Total volume (L)

= 5.3 X 10-6 mol / 1 X 10-2 L

= 5.3 X 10-4 M

[FeSCN2+]e = number of moles of FeSCN2+/ Total volume (L)

= 7 X 10-7mol / 1 X 10-2 L

= 7 X 10-5 M

Step 5

Kc = [Fe(SCN)2+]e / [Fe3+]e•[SCN-]e

= 7 X 10-5 / 9.3 X 10-4 X 5.3 X10-4

= 0.142 X103

= 142

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