Der Partner Experiment 12 Advance Study Assignment: Determination of the Equilib

Question

Der Partner Experiment 12 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction 1. A student mixes 5.00 mL 2.00 × 10 , M Fe(NO)s with 5.00 mL 2.00 × 10 s M KSCNShe finds that in the equilibrium mixture the concentration of FeSCN2, is 1.40 × 10-M. Find K for the reaction Fe' (aq) + SCN (ag) FeSCN (ag). Step 1 Find the number of moles Fe and SCN initially present. (Use Eq. 3) 0.01"10. , 200*10%-o.oo o L. 0-01 ylo-3 fromemolesFe-; 001:10.holesKN Step 2 How many moles of FeSCN are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3.) 140 10- 8 and SCN are used up in making the FeSCN? moles FeSCN2 How many moles of Fe moles Fe'" ; moles SCN. · Step 3 How many moles of Fe and SCN remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps 1 and 2.) moles Fe', moles SCN Step 4 What are the concentrations of Fe, SCN, and FeSCN2 at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3 and the results of Step 3.) mL Step 5 What is the value of K, for the reaction? (Use Eq, 2 and the results of Step 4.) 103

Explanation / Answer

Experiment 12.

Fe3+(aq) + SCN-(aq) <==> FeSCN^2+(aq)

Step 1.

initial [Fe3+] = 2 x 10^-3 M x 5 ml = 0.01 mmol

initial [SCN-] = 2 x 10^-3 M x 5 ml = 0.01 mmol

Step 2. moles of [FeSCN^2+] at equilibrium = 1.4 x 10^-4 M x 10 ml = 1.4 x 10^-3 mmol

Step 3. moles [Fe3+] remained = 0.01 - 1.4 x 10^-3 = 8.6 x 10^-3 mmol

moles [SCN-] remained = 0.01 - 1.4 x 10^-3 = 8.6 x 10^-3 mmol

Step 4. Equilbrium concentration of,

[Fe3+] = 8.6 x 10^-3 mmol/10 ml = 8.6 x 10^-4 M

[SCN-] = 8.6 x 10^-3 mmol/10 ml = 8.6 x 10^-4 M

[FeSCN^2+] = 1.4 x 10^-4 M

Step 5. equilibrium constant Kc

Kc = [FeSCN^2+]/[Fe3+][SCN-]

     = 1.4 x 10^-4/(8.6 x 10^-4)(8.6 x 10^-4)

    = 189.3

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