Advance Study Assignments Determination of the Equilibrium Constant Reaction 1.

Question

Advance Study Assignments Determination of the Equilibrium Constant Reaction 1. A student mixes 200 M FedNO,, with M KSCN. She finds that in the equilibrium concentration of FescN" 40 x 10 M. Find K., for the reaction Fes (aq) scN-Cad) FesCN" (aq) Step 1. Find the number of moles Fes. and soN- initially present. (Use Eq. modes Fer moles soN- Step 2. How many moles FesCN are in the mixture at equilibrium? What is the volume of the equilibrium mixture? (Use Eq. 3) How many moles of Fe" and scN- are used up in makingthe FesCN hbrium? (Use Step 3. How many moles of Res and SCN-remain in the solution at equil Bq. 4 and the results of Steps 1 and 2) moles Fe' moles SCNT Step 4. What are the concentrations of Fe +.SCN- and FescN2 at equilibrium? What is the volume of the equilibrium mixture? (Usc Eq. 3 and the results of Step 3) M: ISCN-1 Step 5. What is the value of Ke for the reaction? (Use Eq. 2 and the results of Suep 4) K." (continued on following page)

Explanation / Answer

Q1.

Fe+3 = M1*v1 = (2*10^-3)(5*10^-3) = 10^-5 moles of Fe+3

SCN- = M"V2 = (2*10^-3)(5*10^-3) = 10^-5 moles of SCN-

Q2.

in equilbiium:

[FeSCN]eq = 1.4*10^-4 M

Vtotal = 5 mL + 5 mL = 10 mL = 10*10^-3

mol FeSCN = MV = (1.4*10^-4)(10*10^-3) = 0.0000014 moles

moles used:

ratio is 1:1

so

0.0000014 mol of Fe+3 and 0.0000014 mol of SCN-

Q3

Remains

Fe+3 = 10^-5 - 0.0000014 = 8.6*10^-6 moles left

SCN- =10^-5-0.0000014    = 8.6*10^-6 moles left

Q4

Simply, find concentrations

[Fe+3] = (8.6*10^-6) / (10*10^-3) = 0.00086

[SCN-] = (8.6*10^-6) / (10*10^-3) = 0.00086

[FeSCN+2] = 1.4*10^-4

V in equilibirum = 5+5 = 10 mL

Q5.

Kc = [FeSCN+2]/[Fe+3][SCN-]

Kc = (0.00086*0.00086)/(1.4*10^-4) = 0.005282

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.