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Experiment 23 Advance Study Assignment: Determination of the Equilibrium Constant for a Chemical Reaction A student mixes 5.00 mL 2.00 × 10-"M FeNO), in l M HNO, with 3.00 mL 2.00 x 10-1 M KSCN and 2.00 mL of water. She finds that in the equilibrium mixture the concentration of FeSCN2+ is 7.0 × 10 Find K. for the reaction Fee (aq) + SCN-(aq)·n FSC02+ (aq). 1. 5 M. Step 1 Find the number of moles Fe and SCN" initially present. (Use Eq. 3.) IL 1moles Femoles SCN Step 2 How many moles of FeSCN are in the mixture at equilibrium? What is the volume of the equi- librium mixture? (Use Eq. 3.1 IL How many moles of Fe and SCN" are used up in making the FeSCN+ Step 3 How many moles of Fe" and SCN remain in the solution at equilibrium? (Use Eq. 4 and the results of Steps I and 2.) Step 4 What are the concentrations of Fe", SCN , and FeSCN at equilibrium? (Use Eq. 3 and the results of Step 2 and Step 3.) OIL 0.01L 001L Step 5 What is the value of K, for the reaction? (Use Eq. 2 and the results of Step 4) x 10 5 (9.3 x 0.9(53 x K,-142 142.6 ) (continued on following page) 94 Experiment 23 Detemination 2. Opsional Assume that the reaction studied in Problem 1 isFe (aq)+2 SCN (aq) of the Equilibrium Constant for a Chemical Reaction e(SCN), "(aq) Find K, for this reaction, given the data in Problem I. Formulate the expression for K, for the alternate reaction just cited. b Find K, as you did in Problem I: take due account of the fact that two moles SCN are used up per mole Fe(SCN) formed as you carry out the following calculations: Step 1 Results are as in Problem 1 Step 2 How many moles of Fe(SCN) are in the mixture at equilibrium? (This will be the same as in Problem 1.) moles Fe(SCN)2 How many moles of Fel" and SCN are used up in making the FerSCN),"7 moles Fe"; moles SCN Step 3 How many moles of Fe3 and SCN remain in solution at equilibrium? Use the results of Steps 1 and 2, noting that moles SCN at equilibrium original moles SCN -(2 X moles Fe(SCN)2) moles Fe";moles SCN Step 4 What are the concentrations of Fe, SCN. and Fe(SCN)2 at equilibrium? (Use Eq. 3 and the results of Step 3.) M: [Fe(SCN)2+1- M Step 5 Calculate K, based on the assumption that the alternate reaction occurs.(Use the answer to Part 2a.)

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Step 1/5

Moles of `Fe3+` initially: `2.00×10−3mol/L×0.005L=1.00×10−5mol`Moles of `SCN` − initially: `2.00×10−3mol/L×0.003L=6.00×10−6mol`

Explanation:

Here is the Calculate Initial Moles

Step 2/5

The number of Moles of `FeSCN2+` at equilibrium: `7.0×10−3mol/L×0.01L=7.00×10−5mol`

Moles of `Fe3+` and `SCN− ` used up: `7.00×10−7mol`

Explanation:

Here we Determine Equilibrium Moles

Step 3/5

Remaining moles of `Fe3+: ` `1.00×10−5−7.00×10−7=9.3×10−6mol`Remaining moles of `SCN−` `6.00×10−6−7.00×10−7=5.3×10−6mol`

Explanation:

Here we Calculate Remaining Moles

Step 4/5

Concentration of `Fe3+: 0.0129.3×10−6​=7.75×10−4M` Concentration of `SCN−: 0.0125.3×10−6​=4.42×10−4M` Concentration of `FeSCN2+: 7.00×10−5/0.012=5.83×10−3M`

Explanation:

Here we Determine Equilibrium Concentrations

Step 5/5

`Kc​=[Fe3+]⋅[SCN−][FeSCN2+]​=` `(7.e75×10−4)⋅(4.42×10−4)5.83×10−3​` `≈142.01`Therefore, the value of `Kc​≈142.01.`

Explanation:

Here we Calculate Equilibrium Constant `(Kc​)`

Final Answer

Here the Determination of the Equilibrium Constant for a Chemical Reaction

Therefore, `K c ​ ≈142.01.`

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