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Equilibrium problems

Question 1

The easiest of these types of problems are those where you are given the concentration of each species at equilibrium and then you have to substitute those into the expression for the equilibrium constant. There are samples of these in the text so I won't do them here.
What about the problems where you are given the quantities of the starting material?

3 moles of ethanol were mixed with 3 moles of ethanoic acid in a 1 L vessel and the reaction was allowed to reach equilibrium at 80°C

At equilibrium, it was found that 2 moles of each product had been formed. Write down the expression for the equilibrium constant and calculate it’s value at 80°C.

Worked Answer

Notice the temperature has been included. This is because equilibria (and their constants) change at different temperatures, so a temperature must always be specified. However, you do NOT need to use the temperature in your calculation.
The general equation for equilibrium constants is

This is a simple one because the coefficients in the balanced equation are all 1 - so no higher powers in our K expression.
Remember, too, products go on top!

Now, don’t be in a rush to use the 2 mol for each of the products and 3 mol for each of the reactants. 2 mol for the products is OK but those 3 mol quantities were the starting amounts and, by now, some of this has been used up. So…the next step is to calculate the amounts present AT equilibrium.

To do this, use the amounts of product formed to calculate how much reactant was used up…I’ll use water as the product because it’s easier to type. It works with either one!
Mole ratio of H₂O to ethanol is 1:1. therefore 2 mol H₂O produced means 2 mol of ethanol used up. There were 3 mol of ethanol to start with so, at equilibrium, there is 3-2=1 mol left. Use the same reasoning to get 1 mol of ethanoic acid as well.
The equilibrium constant requires concentrations not moles, so using the container size, convert no. of mol to concentration. This is particularly easy when the vessel size is 1 L!

Finally, we can substitute these values into the expression for K to get

Here is another problem in the same style taken from a past exam paper.

Question 2

Carbonyl bromide, COBr₂, dissociates at 75°C according to the equation

Carbonyl bromide (2.00mol)was placed in a 2.00L container at 75°C and, after a period of time at this temperature, the amount of COBr₂ in the container was found to be constant at 1.28 mol. Determine the value of the equilibrium constant for the reaction at 75°C.

OK. Quite often exam questions will start with a basic calculation like the one above, then extend it by changing conditions. You have to draw on your knowledge of Le Chatelier's Principle to predict what happens to the equilibrium. Here's an example adapted from a 1997 paper

Question 3

Hydrogen, used for the synthesis of ammonia, can be made according to

Equal number of mole of methane and water were added to an empty 1000cm³ reaction vessel at 1030K in the presence of a nickel catalyst. After the system reached equilibrium, the concentrations of methane and water were each 0.012M and the concentration of carbon monoxide was 0.0083M.
(a) What was the hydrogen concentration?
(b) Write the expression for the equilibrium constant and calculate its value at 1030K.

Here is the working out. View only as much as you need! * * We don’t know how much methane or water there was to start with, but we know that, at equilib, the concentration of CO is 0.0083M. Because the volume is 1L, the no. of mol of CO formed = 0.0083mol. By the mole ratio, 3 moles of hydrogen are formed for every mol of CO so the amount of H2 formed at equilib. = 3 x 0.0083 = 0.0249mol. Because the volume is 1 L, the concentration of H2 at equilib = 0.0249M and that’s the answer to part (a) To do (b), we need the expression for K which is K=[CO][H2]³ / [CH4][H2O] =0.0083 x (0.0249)³ / 0.012 x 0.012 M² = 0.00089 M²

There are further parts to the question:-
If more CO gas was added at 1030K, what would happen to the temperature of the vessel? Why?

Check your answer here. * * First look at the delta H value. It's positive, so that means the reaction is endothermic, ie.takes in heat. So the more heat put in, the more the reaction moves to the right. It also means that if the reaction moves to the left, heat would be given off. In this case, more product has been added. By Le Chatelier's principle, the applied change will be oppposed. So the reaction will move to the left to lessen the amount of product. As mentioned above, the back reaction is exothermic, so heat will be released and the temp of the contents will rise.

If the total pressure of the system was increased by decreasing the volume of the vessel, how would this affect the yield of hydrogen? Why?

Check your answer here. * * Compare the amount of moles of gas on each side of the equation by looking at the coefficients and you will see that the left hand side has 2 moles while the right has 4 moles in total. This means the products is the high pressure side and the more product formed, the higher the pressure. If the pressure is increased externally, then, by Le Chatelier's principle (don't forget to name drop by the way!!), the change will be opposed. That means the equilib. will move to lower pressure, ie. to the left. A movement to the left would mean less hydrogen gas formed!

If the total pressure of the system was increased by adding argon(an inert gas) to the vessel without changing its volume, how would this affect the yield of hydrogen? Why?

Check your answer here. * This is the same as the question before. This is just another way of increasing the pressure in the system. Sneaky wasn't it?

One last one, in the form of three multiple choice questions. Scroll down for the answers.

Question 4

An equilibrium mixture of NO₂ and N₂O₄ is cooled in the container at a fixed volume. The colours of each gas are shown below

As the mixture cools, the brown colour becomes fainter. The equilibrium constant has

1. increased indicating the reaction is exothermic
2. decreased indicating the reaction is exothermic
3. increased indicating the reaction is endothermic
4. decreased indicating the reaction is endothermic

Answer. * Answer is A. Fainter colour means equilib moves to right. If cooling of the system causes a move to the right the actual reaction must give off heat, ie. be exothermic. (Putting heat in would drive it backwards.) More product formed means that the value of K will be higher since products are on the top of the expression not the denominator.

At a particular temperature, the equilibrium concentration of nitrogen dioxide is 0.60M and that of dinitrogen tetroxide is 0.486M. The equilibrium constant at this temperature is

1. 0.39 /M
2. 0.44 /M
3. 0.81 /M
4. 1.4 /M

Answer. * Answer = D K=[N2O4]/[NO2]² = 0.486/(0.60)² = 1.35 =1.4 (given to 2 sig. figs)

At a particular temperature, the pressure is increased by pushing in the piston. In order to maintain equilibrium, the ratio [NO₂]/[N₂O₄]

1. becomes smaller
2. remains the same
3. becomes larger

Answer. * Answer is A. The right side is the low pressure side so increasing the pressure externally will cause a shift to the right to oppose the change. (Whose principle was that again?) More N2O4 formed will mean the ratio of [NO2]/[N2O4] is smaller.

Really there are only so many ways to ask these questions and they are always there. Make sure you know what happens when a product is removed. Or more reactant is added. And DON'T forget to drop Le Chatelier's name whenever it suits. You are expected to!!