0620 · 6.3Reversible reactions and equilibrium

Reversible reactions and hydrated compounds

0:00Blue copper(II) sulfate crystals can turn white when heated. Adding water makes the material blue again. The chemical change can happen in either direction, depending on the conditions. This idea helps explain both reversible reactions in the laboratory and the conditions used to manufacture chemicals on an industrial scale. By the end of this video, you will understand how a reaction can continue without changing the overall mixture. We will connect that idea to temperature, pressure, concentration and catalysts.

0:28Then we will explain the conditions used in the Haber and Contact processes. The aim is to understand the choices, rather than memorise disconnected facts. We need three starting ideas from earlier topics. Reactants are the starting substances, while products are the substances formed. Reaction rate describes how quickly a chemical reaction happens. An exothermic reaction transfers energy to the surroundings, whereas an endothermic reaction takes energy in. These ideas will help us separate reaction speed from equilibrium position.

0:58Some chemical reactions are reversible. This means the products can react to form the original reactants again. We show a reversible reaction using two half arrows pointing in opposite directions. The forward reaction goes from left to right in the equation, and the reverse reaction goes from right to left. The symbol tells us that both directions are possible, but it does not mean they happen equally quickly under every condition.

1:20Heating might favour one direction, while adding a substance favours the other. A reversible reaction also does not have to return completely to its starting state. Its behaviour depends on the conditions surrounding the substances. A hydrated compound contains water of crystallisation as part of its crystal structure, whereas an anhydrous compound contains none. Hydrated copper(II) sulfate is blue, while anhydrous copper(II) sulfate is white. Heating the hydrated compound removes its water of crystallisation and produces the anhydrous compound.

1:53This explains the change from blue to white. The word equation is hydrated copper(II) sulfate reacts reversibly to give anhydrous copper(II) sulfate plus water. Heating drives the change towards the anhydrous material. When water is added to the white solid, hydrated copper(II) sulfate forms again. Its blue colour returns because the water becomes associated with the crystal structure. Cobalt(II) chloride gives another example with a different pair of colours. The hydrated compound is pink, and the anhydrous compound is blue.

2:26Heating turns the pink solid blue as its water of crystallisation is removed. When water is added to the blue solid, the pink hydrated form returns. The same chemical idea explains both examples, despite their different colours. For practical questions, describe the observation separately from its explanation. With hydrated copper(II) sulfate, the observation on heating is a change from blue to white. The explanation is that water of crystallisation is removed.

2:51Practical papers may ask you to record these changes or interpret them. Neither compound should be handled as a home experiment. Keep the two colour pairs attached to their correct substances. For copper(II) sulfate, blue means hydrated. For cobalt(II) chloride, blue means anhydrous. Colour alone therefore does not tell you whether a compound contains water of crystallisation. You must identify the substance before interpreting what the observation means.

Dynamic equilibrium in closed systems

3:18This next part is for the Extended paper. A closed system is one in which substances cannot enter or leave. Energy can still pass between the system and its surroundings. Keeping the reacting substances inside allows products to remain available for the reverse reaction. A reversible reaction can then establish equilibrium. Imagine starting with reactants inside a sealed vessel. Initially, the forward reaction forms products, while there may be very little reverse reaction.

3:45As products accumulate, more of them become available to react back. Meanwhile, the reactants are being used up. The two reaction rates therefore change as the composition of the mixture develops. Eventually, the rate of the forward reaction becomes equal to the rate of the reverse reaction. The concentrations of reactants and products are then no longer changing. These are the two essential features of equilibrium in a closed system.

4:08Concentration describes how much of a substance is present in a particular volume. Equilibrium is dynamic, meaning that reactions continue even though the overall composition remains constant. During a given interval, the forward reaction forms as much product as the reverse reaction uses. There is no overall accumulation of either side. Individual particles still undergo chemical changes, but the amounts present remain steady. Think of people moving between two rooms.

4:34If the same number passes through the doorway in each direction every minute, neither room gains people overall. That does not require equal numbers to occupy the rooms. Likewise, chemical equilibrium means equal reaction rates, not equal concentrations. One side may contain much more substance than the other. The position of equilibrium describes whether the equilibrium mixture favours reactants or products. If the position lies towards the right, products are favoured, while a position towards the left favours reactants.

5:04This describes the mixture’s composition, not its reaction speed. A rapidly reacting mixture can still contain mostly reactants at equilibrium.

Predicting equilibrium shifts

5:11This next part is for the Extended paper. Changing temperature, pressure or concentration can disturb an equilibrium mixture. The reaction then shifts in the direction that opposes the effect of the change. This relationship is often called Le Chatelier’s principle. It gives us a practical way to predict the new equilibrium position. The mixture does not deliberately resist a change. The altered conditions affect the forward and reverse reactions differently.

5:37For a time, one direction happens faster, so the amounts of substances change. Eventually, equal rates are established again. The new equilibrium can have a different composition from the original one. For temperature questions, first identify which direction is exothermic. If the forward reaction releases energy, the reverse reaction must take energy in. Increasing temperature favours the endothermic direction because it absorbs energy, whereas cooling favours the exothermic one. The information about energy transfer determines the answer, not simply the direction of the arrow.

6:11Suppose the forward reaction is exothermic. Raising the temperature shifts equilibrium towards the reactants because the reverse reaction is endothermic. With cooling, the position moves towards the products, since the forward reaction releases energy. If the forward reaction were endothermic, these predictions would be reversed. Always use the energy information given for that particular reaction. Pressure changes matter when gases are involved. Increasing pressure by compressing the mixture favours the side with fewer gas molecules.

6:39This direction reduces the total number of gas particles in the vessel. Decreasing pressure favours the side with more gas molecules. Compare the numbers represented by the balanced equation, not the sizes of the molecules. Use the equation coefficients to count the relative numbers of gas molecules. Do not count atoms inside each formula, and do not include solids or liquids in this comparison. If both sides contain the same total number of gas molecules, changing pressure does not shift equilibrium.

7:07Pressure has no preferred direction in that case. Concentration changes are handled by identifying which substance has been added or removed. Adding a reactant shifts equilibrium towards the products, using some of the added material. If a product is added instead, the reverse reaction is favoured. Removing a product shifts equilibrium towards the products because further reaction replaces some of what was removed. Opposing a change does not mean completely cancelling it.

7:33Adding reactant can leave its final concentration higher than it was originally, even after some has reacted. The important prediction is the direction of the shift. Keep the temperature unchanged when considering a concentration change on its own, so the effects do not become confused. A catalyst increases the rates of both the forward and reverse reactions. It provides an alternative reaction pathway with a lower activation energy. Equilibrium is reached more quickly, but its position does not change.

8:00The equilibrium mixture therefore has the same composition at the same temperature, whether a catalyst is present or absent. This distinction separates rate from yield. Rate describes how quickly product forms, whereas equilibrium yield describes how much can be present once equilibrium is established. A catalyst can help a factory produce material more quickly. It cannot increase the equilibrium proportion of product by itself. Faster does not automatically mean a more favourable equilibrium.

Haber process conditions

8:28Let us work through an example using ammonia production. Nitrogen plus hydrogen reacts reversibly to give ammonia. The balanced symbol equation is N₂ plus 3H₂ reacts reversibly to give 2NH₃. All three substances are gases. We are told that the forward reaction is exothermic. First, consider a decrease in temperature while the other conditions remain unchanged. The forward reaction releases energy, so it is the exothermic direction. Cooling favours that direction and shifts equilibrium to the right.

9:01The equilibrium yield of ammonia increases. This prediction concerns composition, rather than how quickly the new equilibrium is reached. Next, consider increasing pressure at constant temperature. The left side represents one nitrogen molecule and three hydrogen molecules, giving four gas molecules altogether. On the right, there are two ammonia molecules. Higher pressure favours the side with fewer gas molecules, so equilibrium shifts right. This also increases the equilibrium yield of ammonia.

9:30Finally, suppose some ammonia is removed while reactants remain available. Removing a product lowers its concentration and disturbs equilibrium. The forward reaction then replaces some of the ammonia that was removed. Equilibrium shifts towards the right until equal reaction rates are established again. More ammonia forms, so the equilibrium yield of ammonia increases. This next part is for the Extended paper. The Haber process manufactures ammonia from nitrogen and hydrogen.

9:58The nitrogen comes from air, while the hydrogen comes from methane. You need to know these sources and the reversible symbol equation. The detailed chemistry used to obtain the starting gases is outside this subtopic. The typical Haber conditions are four hundred and fifty degrees Celsius and twenty thousand kilopascals, or two hundred atmospheres. An iron catalyst is used. These conditions combine a useful reaction rate with a reasonable equilibrium yield.

10:24They also reflect the cost and safety demands of running an industrial process. Because ammonia formation is exothermic, a lower temperature would give a higher equilibrium yield. However, the reaction would then be too slow for efficient production. A higher temperature increases reaction rate but shifts equilibrium towards nitrogen and hydrogen. Four hundred and fifty degrees Celsius is therefore a compromise between speed and equilibrium yield. High pressure favours ammonia because the equation changes four gas molecules into two.

10:53Compressing the gases also increases collision frequency and can increase reaction rate. However, generating and maintaining high pressure requires energy. The vessels and pipework must be strong enough to withstand the forces, which adds to construction and maintenance costs. Pressure also creates safety concerns because equipment failure can release gases violently. Even higher pressure could improve equilibrium yield, but it would demand stronger equipment and greater operating costs. The typical pressure balances these disadvantages against increased ammonia production.

11:25Industrial conditions are chosen for the whole process, not simply for the highest theoretical yield. The iron catalyst makes the reaction faster without moving equilibrium towards ammonia. It helps achieve an economically useful rate at the chosen temperature. After the reacting mixture leaves the vessel, cooling allows ammonia to be separated. Unreacted nitrogen and hydrogen are recycled, so the factory does not simply discard them after one passage. Recycling improves the overall use of the starting materials.

11:54It does not mean that every molecule becomes ammonia during a single passage through the reactor. Equilibrium limits the conversion under those conditions. Separating the product and returning unreacted gases allows repeated opportunities for further reaction, making the overall process more economical.

Contact process conditions

12:10This next part is for the Extended paper. The Contact process includes the conversion of sulfur dioxide into sulfur trioxide. Sulfur dioxide is obtained by burning sulfur or roasting sulfide ores. Roasting means heating those ores in air. The oxygen needed for the conversion also comes from air. Sulfur dioxide plus oxygen reacts reversibly to give sulfur trioxide. The balanced equation is 2SO₂ plus O₂ reacts reversibly to give 2SO₃.

12:41Each substance is a gas. Two sulfur dioxide molecules react with one oxygen molecule to form two sulfur trioxide molecules. The forward reaction is exothermic. The typical conditions are four hundred and fifty degrees Celsius and two hundred kilopascals, or two atmospheres. The catalyst is vanadium(V) oxide. Notice that the temperature matches the typical Haber temperature, but the pressure is much lower. Similar equilibrium principles do not require identical operating conditions for different chemical reactions.

13:12A lower temperature would favour sulfur trioxide because its formation is exothermic. However, lowering the temperature also slows the reaction. Raising it improves reaction rate but reduces the equilibrium yield of sulfur trioxide. Four hundred and fifty degrees Celsius gives a useful compromise. The catalyst helps the reaction proceed quickly without changing the equilibrium position. For pressure, count three gas molecules on the left and two on the right. Increasing pressure would therefore favour sulfur trioxide.

13:41However, a high yield is already achievable at the typical low pressure. A much higher pressure would bring comparatively little benefit while increasing compression costs and the strength required for the equipment. Operating at about two atmospheres avoids the expense and greater safety demands associated with very high pressure. The Contact process therefore does not use the Haber pressure simply because both equations favour fewer gas molecules. The size of the benefit matters alongside its cost.

14:08Economical production depends on the characteristics of the particular reaction. The vanadium(V) oxide catalyst increases both reaction rates and helps equilibrium become established quickly. It does not increase the equilibrium yield of sulfur trioxide. Its value is achieving a useful production rate at the chosen temperature. Further stages of sulfuric acid manufacture are outside this subtopic; here, we focus on this reversible conversion.

Exam recap and answers

14:34Both processes show why choosing conditions requires more than one argument. Because both forward reactions are exothermic, a lower temperature favours the products, but a useful production rate requires sufficient heating. Increased pressure favours the side with fewer gas molecules, although its economic benefit differs between the processes. Catalysts improve rate without shifting equilibrium. Safety and equipment costs constrain the final choices. Common exam mistakes often come from mixing together these different ideas.

15:04Equilibrium does not mean that the reaction has stopped, because both directions continue. Equal rates do not require equal concentrations. A temperature increase does not always favour products, since the endothermic direction depends on the reaction. Each claim needs the correct chemical reason. Another mistake is counting all particles when explaining pressure. Use only the gaseous substances and their coefficients in the balanced equation. Avoid saying that a catalyst shifts equilibrium or increases equilibrium yield.

15:33It speeds up the approach to equilibrium instead. These distinctions let you make predictions even when the equation is unfamiliar. An exam question might ask you to state the features of equilibrium in a closed system. A precise answer gives two separate points. The rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of reactants and products are no longer changing. Do not replace these statements with a vague claim that everything balances.

15:59For the temperature question, a model answer should connect yield and rate. A lower temperature favours the exothermic forward reaction and increases the equilibrium yield of ammonia. However, the reaction rate would be too slow. Four hundred and fifty degrees Celsius is a compromise between a useful rate and a reasonable equilibrium yield. If asked to explain the effect of increased pressure, identify the unequal numbers of gas molecules. There are four on the reactant side and two on the product side.

16:27Higher pressure favours the side with fewer gas molecules. Equilibrium shifts to the right, increasing the equilibrium yield of ammonia. Each sentence contributes a clear, creditworthy point. Use the command words carefully. State asks for a fact, such as a condition or definition. Describe asks what happens, such as a colour change or equilibrium shift. Explain requires the chemical reason for that change. For industrial conditions, a complete explanation often needs both equilibrium position and reaction rate, alongside cost or safety.

16:59The central idea is that reversible reactions can proceed in both directions. Heating hydrated compounds removes water of crystallisation, while adding water restores their hydrated forms. In a closed system at equilibrium, the opposing reactions continue at equal rates. Concentrations stay constant, even though particles are still reacting. That is why equilibrium is dynamic rather than motionless. Changes in temperature, pressure and concentration can alter the equilibrium composition. A catalyst changes reaction speed without moving the equilibrium position.

17:32The Haber and Contact conditions balance rate, yield, economics and safety. Once you keep these ideas separate, the conditions become understandable choices rather than isolated facts to memorise.