The purpose of this week\'s lab is to determine the equilibrium constant, K , fo

Question

The purpose of this week's lab is to determine the equilibrium constant, K, for the reaction:

When Fe3+ and SCN- are combined, some FeSCN2+ is formed, and an equilibrium is established among the three ions.  

In order to understand the calculations that you will be making during this experiment, let's look at a simpler, hypothetical reaction:

R = the gas constant (8.314 J/mol K)
T = the absolute temperature (K)
K = the equilibrium constant (described below)
Q = the reaction quotient (described below)

K , the equilibrium constant, is made up of the ratio of the equilibrium concentrations of the reactants and products, raised to their stoichiometric coefficients.  

Q is the reaction quotient, made up of the ratio of the initial concentratons of reactants and products, raised to their stoichiometric coefficients:

Let's assume that we have the following stock solutions to work with:

Part 1

In the first experiment, 10 mL of A are mixed with 10 ml of B and 30 mL of C. To determine the initial concentration of reach reagent, first determine the number of molesof each of the species.
mol A =  
mol B =  
mol C =  

Part 2

Now, divide the moles by the total volume of the solution to get the initial molarity of A, B and C.

[A]i=  M
[B]i =  M
[C]i =  M

Now we can set up a table showing the Initial concentrations, the Changes in concentration and the final Equilibrium concentration of each of the three species.

In order to find the change in the concentrations of A, B and C, when they are allowed to react and come to equilibrium, we need to be able to determine the equilibrium concentration of at least one of these species. We will assume that C is a highly colored compound. If we have a calibration curve relating the absorbance of C to its concentration, we can determine the [C]e spectrophotometrically.

The three reagents are mixed together (all initally at a concentration of 1.00 M) and allowed to come to equilibrium. A sample of the equilibrium mixture is placed in a cuvette and inserted into a Spectronic 20 spectrophotometer that has been properly calibrated. The absorbance of the equilibrium mixture is read as 0.7.

A calibration curve for C is shown below. It is a plot of the [C] v.s. its Absorbance at different concentrations. This will allow us to convert absorbance, A, to [C].  

Explanation / Answer

From the given data

Part 1 :

Initial moles of A = 5 M x 10 ml = 50 mmol

Initial moles of B = 5 m x 10 ml = 50 mmol

Initial moles of C = 1.67 M x 30 ml = 50.1 mmol

Part 2 : Initial molar concentrations

[A]i = 50 mmol/50 ml = 1 M

[B]i = 50 mmol/50 ml = 1 M

[C]i = 50.1 mmol/50 ml = 1.002 M

Plot standard diluted solutions of known concentration of [C] against absorbance values

The stragiht line plot is obtained.

Calculate slope of the line

slope = molar absorptivity of [C]

Now, from the mixture of A, B and C we get absorbance values.

Find concentration of [C] in mixture = absorbance of mixture/molar absorpticvity (slope)

This is equilibrium concentration of [C]

Substract this concentration of [C] from initial concentrations of [A]i, [B]i and [C]i to get equilibrium concentrations of all three in the mixture.

Equilibrium constant = K = [C]eq/[A]eq.[B]eq

ICE chart

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