3. Assume that the reaction studied in Question 2 is Fe\"(aq) + 2 SCN(aq) Fe(SCN

Question

3. Assume that the reaction studied in Question 2 is Fe"(aq) + 2 SCN(aq) Fe(SCN)'(aq) Find Kc for this reaction 10 M KSCN), concentrations of the reactants are still the same as in Question 2 , given the data in Question 2 (5.0 mL 2.00 x 10' M Fe(NO,)h with 5.0 mL 2.00x ,except the equilibrium concentration of Fe(SCN)2 is equal to 0.6x10 M. Note: The initial a. Formulate the expression for Kc for the alternate reaction just cited. Find Kc as you did in Question 2; take due account of the fact that two moles SCN are used up per mole Fe(SCN)2 formed. Calculate the equilibrium molarity of Fe and SCN b. Fe "(aq) + 2 SCN-(aq) Fe(SCN)'(aq) Fe3+ SCN Fe(SCN)2 0.6 × 10" c. Calculate Kc 63

Explanation / Answer

Kc =     [Fe(SCN)2]+ / [Fe+3][SCN-]2

Moles of Fe+3 = 2 X 10-3 X 5 = 10 X 10-6 moles

Moles of [SCN-]= 2 X 10-3 X 5 = 10 X 10-6 moles

                          Fe+3 + 2SCN- ===> [Fe(SCN)2]+

Iniital                 10-5    10-5    0

Change              -x             -2x              +x

Equili               10-5 - x        10-5 - 2x        x

x = moles of Fe(SCN)+ at equilibrium = 0.6 X 10-4 X 10 = 6 X 10-7 moles

[Fe(SCN)]+ = moles / total volume = 6 X 10-7 moles / 10mL = 6 X 10^-5 M

equilibrium moles [Fe+3] = 10-5 - 6 X 10-7 moles = 9.4 X 0-6 moles

[Fe+3] = moles / total volume = 9.4 X 0-6 moles / 10mL = 9.4X 10-4 M

Equilibrium [SCN-]= 10-5 - 2(6 X 10-7 )moles = 8.8 X 0-6 moles

[SCN-]= moles / total volume = 8.8 X 0-6 moles / 10mL = 8.8 X 10-4 M

Kc = [Fe(SCN)2]+ / [Fe+3][SCN-]2

Kc = 6 X 10-5 M / 9.4X 10-4 X [8.8 X 10-4]2

Kc = 0.00824 X 107 = 8.24 X 104

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