# Identify data, plan and perform a first-hand investigation to model an equilib- rium reaction

This experiment really just aims at showing you what equilibrium ‘looks’ like by using a model. Provided you understand Le Chatelier’s Principle thoroughly, I would not overly concerned about this experiment.

Materials:

• Two 50mL measuring cylinders
• Two pipettes of differing diameters (e.g. one 10ml, one 5ml)

Procedure:

1. Fill one 50mL measuring cylinder with water, leaving the other measuring cylinder
1. Place one pipette into first measuring cylinder, letting the pipette lightly touch the bottom. Place your finger over the top of the pipette and move its contents to the second measuring cylinder, letting your finger off the top and releasing the Do so carefully to ensure minimum spillage.
2. Place the other pipette into the second measuring cylinder, repeating the second step by empting the contents of the second pipette into the first measuring
3. Note the amount of water in both measuring cylinders and tabulate the These results represent the results after ’Cycle 1’. Repeat steps 2 & 3 until Cycle 30 is recorded. Graph the results against each other, with volume on the y-axis and cycle-count on the x-axis.
4. Repeat steps 1 to 4, but transfer 10mL of water from the second measuring cylinder to the first after the 15th cycle and continue as normal. Graph the results using the same axis as before.

Expected results:

You will find that equilibrium will be seen as the water levels on both cylinders tend towards, but do not reach, the halfway mark- 25ml of water in each measuring cylinder. The change will be relatively quick at first, but will eventually slow down to the point where the water level does not change considerably.

Step number 5 will lead to a temporary spike in the volume of water in the first measuring cylinder, and a decrease in the second measuring cylinder. However this will also be rapidly corrected at first, and then more gradually after a while.

# Choose equipment and perform a first-hand investigation to gather information and qualitatively analyse an equilibrium reaction

Given that the nature of this experiment is qualitative, this experiment is quite easily accomplished, as all you need to do is disturb a system in equilibrium, and note the changes. Clearly you must pick a disturbance which has a visible effect, and the simplest disturbance is simple a change in temperature.

Select two ampoules of nitrogen dioxide (NO2) gas. Place one ampoule in a beaker of warm-hot water, and one in a beaker of cold water. You will soon notice that one ampoule- the one which was heated- has turned a reddish brown, whereas the other- the one that was placed in cold water- has become almost completely colourless.

In effect, you are simply observing the following equilibrium:

$N_2O_{4 (g)}$ ↽⇀ $2 NO_{2 (g)}$

As $NO_2$ is a reddish brown, and $N_2O_4$ is colourless, we can effectively deduce that this equation is endothermic (Note that if the reaction was reversed such that nitrogen dioxide gas was a reactant it would be exothermic).

Confirm this by stating the effects as per Le Chatelier’s Principle. If the equation is endothermic, then an increase in heat should see a shift in the equilibrium towards the right, making the gas appear more reddish as more NO2 is produced. If there is a decrease in heat, then the gas would become colourless as more N2O4 is produced. The results are consistent with this hypothesis.

# Explain the effect of changing the following factors on identified equilibrium reactions: Pressure, volume, concentration and temperature

This dotpoint is simply a review of Le Chatelier’s Principle.

Using the following equation as an example, and assuming it is an exothermic reaction:

$2 A_{(g)} + B_{(g)}$ ↽⇀ $C_{(g)} + D_{(g)}$

An increase in pressure would shift the equilibrium to the right. An increase in the volume of the substances is essentially an increase in pressure (Same space, more atoms). As such the equilibrium would shift once more to the right to account for this. An increase in the concentration of one substance will result in the system working to minimise this change. For example, an increase in B would result in a decrease in A and B and an increase in C and D. Therefore the equilibrium has shifted to the right. An increase in temperature in an exothermic reaction will result in a shift in the equilibrium to the left as the system works to reduce the amount of heat produced.

Remember- A closed system will always work to minimise the impact of a disturbance upon the system.

# Interpret the equilibrium constant expression (no units required) from the chemical equation of equilibrium reactions

Let the following equation be any given chemical equation, where A, B, C, and D are any given substances, and a, b, c, and d are their respective molar ratios:

$aA + bB$ ↽⇀ $cC + dD$

When this equation is in equilibrium, a constant, K also known as the equilibrium constant, can be obtained using the following expression:

It helps to keep in your mind that the products are always on top of the reactants when calculating

1. K. One way to remember this is that P comes before R alphabetically.

K provides much information if it can be read successfully, as it tells us how far to completion an equilibrium is currently at. Think of it as at a scale. At one end no reactions have occurred and

there are only reactants and no products. At the other end, everything has been reacted and there

are no reactants, only products. A point of equilibrium always lies between these two points. The value of K tells us exactly where on this scale the equilibrium is currently at.

The smaller the value of K, the lower on the scale you’ll find the equilibrium- Mostly reactants, very little product. (Generally where \$latex K < 10^{−4}\$). The larger the value of K, the higher up on the scale you’ll find the equilibrium- Mostly products, very little reactant. (Generally where [latex]K < 10^4[/latex]). In between these two values, there is a moderate mixture of products and reactants, with the ratio depending upon the exact value of K, and therefore where on the scale the equilibrium can be found. When a reaction is not at equilibrium, Q replaces K in the same expression. If Q is less than K, then the reaction is lower on the scale than the point equilibrium, and thus the concentration of products must be increased in order to achieve equilibrium. Conversely, if Q is larger than K, then the reaction is higher on the scale than the point of equilibrium, and thus the concentration of reactants must be increased in order to achieve equilibrium. Remember- K is the equilibrium constant, calculated by multiplying the concentrations of the products to the power of their respective number of moles. A lower value of K indicates an equilibrium where very few products are produced, whereas a high value of K indicates that there is an equilibrium which is close to completion.

Example:

$N_{2(g)} + 3 H_{2(g)}$ ↽⇀ $2 NH_{3(g)}$

If the above equation is at equilibrium when there are 3mol of nitrogen gas, 2mol of hydrogen gas, and 2mol of ammonia, then:

# Process and present information from secondary sources to calculate K from equilibrium conditions

The easiest way to grasp how to calculate K is really to go through a worked example.

Continuing from the Haber process example, let us begin with the equation:

$N_{2 (g)} + 3 H_{2 (g)}2 NH_{3 (g)}$

At the beginning of an experiment, there were 2.1mol of nitrogen gas, and 6.9mol of hydrogen gas. The reaction was allowed to proceed to equilibrium in a 10L container, at which point 1.2mol of N2 was remaining. What is the value of K, assuming a fixed temperature?

The first step is to note the concentration of each item individually at equilibrium. If 1.2mol of N2 is remaining, then 0.9mol must have been converted. Therefore 3 x 0.9mol of H2 must also have been converted, leaving 0.3mol of H2 (For every one mole of nitrogen gas, 3 moles of hydrogen gas are converted, as per the chemical equation). In addition, there must now be 2 x 0.9mol of NH3.

We now know the concentrations of each substance. The next trick is to note that, when dealing with gases, you must account for the size of the container, as it is very rarely a simple 1L in exams. You must do this because concentration is proportional to pressure, and the size of a container influences the pressure.

Once the 10L has been taken into account, K can be calculated:

Note that the indices have been adjusted (divided by 10) to account for the 10L container.

# Identify that temperature is the only factor that changes the value of the equilibrium constant (K) for a given equation

If volume, concentrations or pressure change, then the numerator and denominator used in the calculation of the equilibrium constant K shift correspondingly, cancelling out the effect of one another. As such, these disturbances do not impact upon K.

However, when the temperature is changed, then K does in fact change.

For endothermic reactions, if the temperature increases then K increases (Remember how equilibrium ‘shifts to the left’ according to Le Chatelier’s Principle. This means it moves up the scale.) Thus for exothermic reactions, if the temperature increases then K decreases.

Remember- A change in temperature is the only factor which changes the value of K. Changes in volume, concentration, and pressure all have no effect on the value of the equilibrium constant.