162 Determination of the Equilibrium Constant for a Chemical Reaction 2. Assume

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162 Determination of the Equilibrium Constant for a Chemical Reaction 2. Assume that the reaction studied in Problem lis Fe aq 2 SCN (aq Fe(SCN) (aq). Find K for this reaction, given the data in Problem l. except that the equilibrium concentration of Fe(SCN, is equal to 0.6 x 10 M. a. Formulate the expression for K for the alternate reaction just cited. b. Find K, as you did in Problem I: ake due account of the fact that two moles SCN are used up per mole Fe(SCN)2 formed. Step 1 Results are as in Problem 1. Step 2 How many moles of Fe(SCN, are in the mixture at equilibrium (Use Eq. 3.? moles Fe(SCN) How many moles of Fe and SCN- are used up in making the Fe(SCN), moles Fe moles SCN Step 3 How many moles of Fe and SCN remain in solution at equilibrium? Use the results of Steps 1 and 2, noting that no. moles SCN at equilibrium original no. moles SCN (2 x no moles Fe(SCN)2'). moles Fe moles SCN Step 4 what are the concentrations of Fest, SCN-, and Fe(SCN)' at equilibrium? (Use Eq. 3 and the results of Step 3.) [Fe3+] M; SCN M: [Fe (SCN)2 Step 5 Calculate K. on the basis that the alternate reaction occurs. (Use the answer to Part 2a.)

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