In the reaction: Fe^3+ + SCN^− ⇌ [Fe(NCS)]2+ The initial concentration of Fe^3+ was 0.109 and...

In the reaction:

Fe^3+ + SCN^− ⇌ [Fe(NCS)]2+

The initial concentration of Fe^3+ was 0.109 and the initial concentration of SCN^− was 0.105. After equilibrium was established, the concentration of the complex was 0.09. What is the equilibrium constant?

Solutions

Expert Solution

Steps for calculating equilibrium constant are

Write the equilibrium expression for the reaction.
Determine the molar concentrations or partial pressures of each species involved.
Determine all equilibrium concentrations or partial pressures using an ICE chart.
Substitute into the equilibrium expression and solve for K

Fe ³⁺ + SCN⁻ ⇌ [Fe(NCS)]²⁺

Initial Concentration 0.109 0.105 0

Change in Concentration -x - x =x

Equilibrium Concentration 0.109- x 0.105- x 0.09

therefore x=0.09

Equilibrium Concentrations 0.109-.09 0.105-0.09 0.09

0.019 0.015 0.09

Hence, Equilibrium constant K_c = 0.09/0.019x0.015 = 315 or 3.15 x 10²

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

Using the initial concentrations of [Fe(H2O)6]^3+, SCN^-, and the equilibrium concentration of the [Fe(H2O)5SCN]^2+ complex, calculate...

Using the initial concentrations of [Fe(H2O)6]^3+, SCN^-, and the equilibrium concentration of the [Fe(H2O)5SCN]^2+ complex, calculate the equilibrium concentrations of both [Fe(H2O)6]^3+ and SCN^-. All mol/L. Ice table is needed. Initial concentrations of [Fe(H2O)6^3+: Initial concentrations of SCN: [Fe(H2O)5SCN]^2+ equilibrium: Fe(H2O)6 - SCN - EQ Fe(H2O)5SCN 0.390 1.54 23.42 0.390 3.12 35.99 0.390 4.84 58.09 0.390 6.25 74.81 0.390 7.96 98.97

Consider the following reaction : Fe3+ (aq) + SCN- (aq) ---> Fe(SCN)2+ (aq) Starting with 4.00...

Consider the following reaction : Fe3+ (aq) + SCN- (aq) ---> Fe(SCN)2+ (aq) Starting with 4.00 mL of .200 M Fe3+ (aq) in a cuvette, 0.10 mL increments of 0.00100 M SCN- (aq) will be added. Assume that because [Fe3+ (aq)] >> [SCN- (aq)], the [Fe(SCN)2+] concentration can be calculated from the limiting reagent, SCN-. Calculate [Fe(SCN)2+]. Volume .00100 M KSCN mL [Fe(SCN)2+] (M) .10 .20 .30 .40 .50 .60 .70 .80 .90 1.00

If acidic H2O2 is mixed with Fe^2+, which reaction occurs: the oxidation of Fe^2+ to Fe^3+...

If acidic H2O2 is mixed with Fe^2+, which reaction occurs: the oxidation of Fe^2+ to Fe^3+ or the reduction of Fe^2+ to Fe? Give the balanced redox reaction for each possibility and explain your reason for selecting the reaction which will occur. Assume standard solution concentrations. For a) I have: red: H2O2 + 2H+ + 2e- ---> 2 H2O E=+1.78 V oxi Fe2+ ----> Fe3+ + e- E=-0.77V Ecell = 1.01V I'm stuck on b) but for the explanation I...

An experiment with Le Chatelier's Principle resulted in this equation: Fe3+ + SCN- <=> Fe(SCN)2+ First...

An experiment with Le Chatelier's Principle resulted in this equation: Fe3+ + SCN- <=> Fe(SCN)2+ First off, should the first reactant be Fe2+ or Fe3+? Our TA said Fe2+, but I'm not sure why it's not Fe3+ since we were adding Fe3+ to observe changes in equilibrium. (The lab manual specifies that Fe(SCN)2+ in a product.) Second, it was obvious that adding Fe3+ and SCN- shifted the equilibrium to the right toward products. However, addition of NaH2PO4 shifted the equilibrium...

You mix Fe^3+ with SCN^- and form the product [FeSCN]^2+. After mixing the reactants and allowing...

You mix Fe^3+ with SCN^- and form the product [FeSCN]^2+. After mixing the reactants and allowing the system to reach equilibrium, which statements are true? More than one statement is true. a) the forward and reverse reactions continue to occur b) the reaction stops (no molecules convert between reactant/product) c) the forward rate constant is equal to the reverse rate constant d) the amount of product present is constant e) the rate of the forward reaction is equal to the...

thanks Fe+3 + SCN- ----> <---- FeSCN+2 You mix the following: Volume of KSCN solution 4.27...

thanks Fe+3 + SCN- ----> <---- FeSCN+2 You mix the following: Volume of KSCN solution 4.27 mL Original concentration of KSCN solution 0.00495 M Volume of ferric ion solution 4.74 mL Original concentration of ferric ion solution 0.02488 M Voluume of distilled water 5.25 mL You measure the %T of the blank and sample at 447 nm using the same cuvet. Only FeSCN2+ absorbs at 446 nm! KSCN and Fe+3 are transparent. %T of blank 100.6 %T of sample 42.7...

Given: Molar Concentration of Fe(NO3)3 is 0.2 Molar concentration of NaSCN is 0.001 Volume of Fe(NO3)3...

Given: Molar Concentration of Fe(NO3)3 is 0.2 Molar concentration of NaSCN is 0.001 Volume of Fe(NO3)3 is 10.00 ml (in 0.1 M HNO3) Volume of NaSCN is 2.00 ml (in 0.1 M HNO3) 10.00 ml Fe(NO3)3 + 2.00ml NaSCN + 13.00 ml HNO3= 25 mL Calculate: a) Moles of SCN- b) [SCN-] (25.0 ml) c) [FeNCS2+]

Fe(III) forms a complex with thiocyanate, Fe(SCN)2+, that has a molar absorptivity of 7.00 x 103...

Fe(III) forms a complex with thiocyanate, Fe(SCN)2+, that has a molar absorptivity of 7.00 x 103 at 580 nm. A 5.00 ml aliquot of a solution that contains 47.4 ppm Fe3+ is reacted with excess thiocyanate and diluted to 25 ml. a) What fraction of 580 nm light is absorbed in a 1.00 cm pathlength cell by this solution? Consider 5.00 ml of this solution diluted to a volume of 50ml b) What fraction of 580 nm light is absorbed...

in part 1 of this experiment how are solution of unknown Fe(SCN)2+ made?

Fe(NO3)3+KSCN ---- FeSCN^2+KNO3 the assumption was made that all of the SCN^- which had been added...

Fe(NO3)3+KSCN ---- FeSCN^2+KNO3 the assumption was made that all of the SCN^- which had been added to the standard solutions had been converted to FeSCN^2+. Is this assumption reasonable. Explain considering the initial concentration of SCN- relative to that of Fe3+ in each sample soln 0.200M Fe(NO3)3 (ml) 0.00200M KSCN (ml) water (ml) FeSCN2+ (ml) 1 5.00 0.20 4.80 4x10^-5 2 5.00 0.40 4.60 8x10^-5 3 5.00 0.60 4.40 1.2x10^-4 4 5.00 0.80 4.20 1.6x10^-4 5 5.00 1.00 4.00 2.00x10-4

Subjects