Introduction

In its most scientific form, equilibrium is defined as a state in which a particular system at a given temperature has allowed its energy to be distributed in the statistically most probable manner. In easier terms, equilibrium is accomplished when the forces, influences, and reactions involved balance each other out so that there is no net change. For example, your body is said to be in thermal equilibrium when there is no net heat exchange between it and your surroundings. Luckily for you, we have initially made this concept seem more difficult than it really is.

In this experiment you will be introduced to chemical equilibrium. No different from the generalized, and complicated, definitions above, chemical equilibrium is achieved when a reaction and its reverse are proceeding at equal rates. Shown below is the generalized chemical equation for a reversible reaction where the capital letters (A, B, C, and D) represent the species involved in the reaction, and the lower case letter (a, b, c, and d) correlate to the numerical coefficients indicating how many moles of each reactant and product are involved.

The Equilibrium Constant

Perhaps the most important component of equilibrium is the equilibrium constant. Specifically, this value, expressed as K_c when concentrations are used and K_p when partial pressures are used, reveals the position of equilibrium. Further, this constant is generally written as a ratio of the concentration of the products, with their appropriate powers, to the concentration of the reactants, with their respective powers. For our example, the equilibrium constant is:

In the previous paragraph it was mentioned that the equilibrium constant reveals the position of equilibrium. Recalling that the constant is a ratio of products to reactants, it is rather intuitive that a low value of K indicates that the amount of product (C and D) is relatively small compared to the amount of reactant (A and B). In other words, when K is small, the reverse reaction is dominant. The equilibrium expression, as you will see in a later lab, can also indicate just how the equilibrium will shift when reactants or products are added to or taken away from the system.

Example Problem

In order to further clarify the significance of the equilibrium constant and to familiarize you with some of the calculations it involves we will now look at the simple example shown below:

Let’s assume that at a particular temperature, 6.00-mol of NH₃ is introduced into a 3.00-L flask and allowed to reach equilibrium. At equilibrium, it was revealed that only 3.00-mol of NH₃ remained, and that we need to determine the equilibrium constant.

In order to begin this problem, we must first write an expression for the equilibrium constant. Remembering that this is generally given to be a ratio of the concentration of the products to the concentration of the reactants, we arrive at the following:

Since we are given the number of moles of NH₃ remaining at equilibrium and a volume, it is easiest to deal with concentrations here as opposed to partial pressures. With that being said, we then take the 3.00-mol of NH₃ remaining at equilibrium and divide by the volume of our vessel to obtain the equilibrium concentration of ammonia.

Our equilibrium expression now looks like this:

However, we are still left with two unknowns [N₂] and [H₂], values that must be obtained to ascertain our equilibrium constant. Looking back at the equation for this reaction and observing the coefficients for each species, we see that the concentration of NH₃ is twice that of N₂ and two-thirds of H₂. Mathematically, we can arrive at the respective concentrations as shown below:

Finally, we have the correct concentrations of all the species at equilibrium and we can calculate the value of our equilibrium constant

With an equilibrium constant of ~1.7, we can now rationalize that at equilibrium the reaction is dominated by the forward reaction. In other words, we know that for this particular reaction at the given temperature that the concentration of the products will be greater than that of the reactants.

This Week's Experiment

As for this experiment, the calculations will not be as straight forward, but with a good understanding of the example provided above, they are bearable. If you use your resources well, the hardest part of this experiment will not be in the equilibrium constant calculations, but rather in obtaining the equilibrium concentrations of involved species.

The Extinction Coefficient or Molar Absorptivity

Beer’s Law (A = ebc) states that molar absorptivity is constant (and the absorbance is proportional to concentration) for a given substance dissolved in a given solute and measured at a given wavelength. For this reason, molar absorptivities are called molar absorption coefficients or molar extinction coefficients. Because transmittance and absorbance are unitless, the units for molar absorptivity must cancel with units of measure in concentration and light path. Therefore, molar absorptivities have units of M^-1 cm^-1. Standard laboratory spectrophotometers are fitted for use with 1 cm-width sample cuvettes; hence, the path length is generally assumed to be equal to one and the term is dropped altogether in most calculations (A = ec).

In this experiment the extinction coefficient/molar absorptivity is an unknown and will need to be determined experimentally. As we have done previously, we can use the relationship between concentration and absorption to create a calibration graph. As long as one concentration in the reaction remains constant, the graph will then be (hopefully) linear with a slope equal to e.

By creating the solutions in the table below, measuring their absorbance at 447 nm and plotting the concentration of KSCN (x-axis) versus Absorbance (y-axis) you should get a linear graph corresponding to the equation, A = ec, where the extinction coefficient/molar absorptivity is equal to the slope.

Solution	mL 0.200M Fe(NO3)3	mL 0.001M KSCN	mL HNO3	Absorbance
A	5	1	4
B	5	2	3
C	5	3	2
D	5	4	1
E	5	5	0

 

Determining the Equilibrium Constant, K

As with the example problem above, the first step in determining the equilibrium constant is the writing of a balanced chemical equation.

The Reaction

The reaction you will be investigating,

poses some interesting challenges because the exact chemical formula of the product is unknown. The subscripts and coefficients shown in the reaction above could be 1, 2 or 3, yielding three different balanced reactions, respectively:

or or

In order to determine the equilibrium constant, you will need to determine the equilibrium concentrations of all the products and reactants several times starting with different initial concentrations of reactants. The table below contains the initial concentrations of reactants you will use:

Solution	mL 0.002M Fe(NO3)3	mL 0.002M KSCN	Absorbance
F	3	7
G	4	6
H	5	5
I	6	4
J	7	3

Although this seems complicated, all it really means is a little more work. Once you have experimentally determined the concentration of based on the reactant combinations above, you will need to complete calculations in the exact same manner as shown in the example problem. You will just need to do the calculation 3 times; once for each possible x value.

The only question that will remain at that point is which x value is the correct one. The answer to that will be readily apparent when you calculate the average and standard deviations of the equilibrium constant for each run. The value that produces the most consistent value of K (smallest standard deviation)will be the correct value to assume for the reaction.