Chemical Equilibrium A reversible reaction can go in

Chemical Equilibrium A reversible reaction can go in either the forward or reverse directions A + B C + D Equilibrium is dynamic as both the forward and reverse reactions continuously occur At Chemical equilibrium the rate of forward reaction and rates of reverse reaction are equal

CONCENTRATION CHANGE IN A REACTION As the rate of reaction is dependant on the concentration of reactants. . . the forward reaction starts off fast but slows as the reactants get less concentrated FASTEST AT THE START THE STEEPER THE GRADIENT, THE FASTER THE REACTION In an ordinary reaction; all reactants end up as products; there is 100% conversion SLOWS DOWN AS REACTANTS ARE USED UP TOTAL CONVERSION TO PRODUCTS

EQUILIBRIUM REACTIONS Initially, there is no backward reaction but, as products form, it speeds up and provided the temperature remains constant there will come a time when the backward and forward reactions are equal and opposite; the reaction has reached equilibrium. FASTEST AT THE START NO BACKWARD REACTION FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP BACKWARD REACTION STARTS TO INCREASE In an equilibrium reaction, not all the reactants end up as products; there is not a 100% conversion. BUT IT DOESN’T MEAN THE REACTION IS STUCK IN THE MIDDLE AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

The conditions and properties of a system at equilibrium are summarized below: 1. The system must be closed, meaning no substances can enter or leave the system. 2. Equilibrium is a dynamic process. Even though we don’t observe any changes, both the forward and reverse reactions are still taking place. 3. The rates of the forward and reverse reactions must be equal. 4. The amounts of reactants and products do not have to be equal. However, after equilibrium is attained, the amounts of reactants and products (their concentration) will remain constant.

DYNAMIC EQUILIBRIUM IMPORTANT REMINDERS • a reversible chemical reaction is a dynamic process • everything may appear stationary but the reactions are moving both ways • the position of equilibrium can be varied by changing certain conditions Trying to get up a “down” escalator gives an excellent idea of a non-chemical situation involving dynamic equilibrium. Summary When a chemical equilibrium is established. . . • both the reactants and the products are present at all times • the equilibrium can be approached from either side • the reaction is dynamic - it is moving forwards and backwards • the concentrations of reactants and products remain constant

THE EQUILIBRIUM LAW “If the concentrations of all the substances present at equilibrium are raised to the power of the number of moles they appear in the equation, the product of the concentrations of the products divided by the product of the concentrations of the reactants is a constant, provided the temperature remains constant” for an equilibrium reaction of the form. . . a. A + b. B then (at constant temperature) where Kc c. C + d. D [C]c. [D]d = a constant, (Kc) [A]a. [B]b [ ] denotes the equilibrium concentration in mol/L is known as the Equilibrium Constant VALUE OF Kc AFFECTED by a change of temperature NOT AFFECTED by a change in concentration of reactants or products a change of pressure adding a catalyst

The value of the equilibrium constant for any reaction does not depend on the starting concentrations, so the equilibrium constant has the same value regardless of the initial amounts of each reaction component. It does, however, depend on the temperature of the reaction.

Large Kc > 1 products are "favoured" they dominate the mixture Kc = 1 neither reactants nor products are favoured Small Kc < 1 reactants are "favoured" they dominate the mixture Kc = [Products] [Reactants] Kc give a ratio of the product to reactants Kc at a particular temperature is constant

Equilibrium Constant • The value of Kc is only applies at equilibrium • The value of Kc is temperature dependent i. e. constant at a particular temperature • It is not changed by pressure or concentration of reactant or products because it will shift to relieve the stress • Units of Kc depends on relative no. of moles. If no. of moles equal on both sides then Kc has no units.

• If there is the same number of molecules in the forward and reverse reactions, it is possible to calculate the equilibrium constant even if the volume of the equilibrium mixture is unknown • No need to know the volume when the number of molecules is equal on both sides!! Some cases X= -b +/- √b 2 -4 ac 2 a

The position (i. e. rate of forward vs. reverse) of an equilibrium is temporary affected by many things. This means that the relative concentrations of reactant and products can differ. The equilibrium position will shift to try and make things equal again. This means the reverse reaction will speed up a bit, until the concentrations are where they want to be, and then it will return to a constant dynamic equilibrium - forward and reverse reactions being the same.

LE CHATELIER’S PRINCIPLE ”When a change is applied to a system in dynamic equilibrium, the system reacts in such a way as to oppose the effect of the change. ” If you do something to a reaction that is in a state of equilibrium, the equilibrium position (not Kc) will change to oppose what you have just done (in order to maintain same Kc value)

FACTORS AFFECTING THE POSITION (not value) OF EQUILIBRIUM CONCENTRATION The equilibrium constant is not affected by a change in concentration at constant temperature. To maintain the constant, the composition of the equilibrium mixture changes. If you increase the concentration of a substance, the value of Kc will theoretically be affected. As it must remain constant at a particular temperature, the concentrations of the other species change to keep the constant the same.

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) the equilibrium constant Kc = CH 3 COOC 2 H 5(l) + H 2 O(l) [CH 3 COOC 2 H 5] [H 2 O] = 4 (at 298 K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH] decreasing [H 2 O] - will make the bottom line larger so Kc will be smaller - to keep it constant, some CH 3 CH 2 OH reacts with CH 3 COOH - this reduces the value of the bottom line and increases the top - eventually the value of the constant will be restored - will make the top line smaller - some CH 3 CH 2 OH reacts with CH 3 COOH to replace the H 2 O - more CH 3 COOC 2 H 5 is also produced - this reduces the value of the bottom line and increases the top

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM SUMMARY REACTANTS PRODUCTS THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM INCREASE CONCENTRATION OF A REACTANT EQUILIBRIUM MOVES TO THE RIGHT DECREASE CONCENTRATION OF A REACTANT EQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCT EQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCT EQUILIBRIUM MOVES TO THE RIGHT Predict the effect of increasing the concentration of O 2 on the equilibrium position 2 SO 2(g) + O 2(g) 2 SO 3(g) Predict the effect of decreasing the concentration of SO 3 on the equilibrium position EQUILIBRIUM MOVES TO RHS

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM PRESSURE When studying the effect of a change in pressure, we consider the number of gaseous molecules only. The more particles you have in a given volume, the greater the pressure they exert. If you apply a greater pressure they will become more crowded (i. e. they are under a greater stress). However, if the system can change it will move to the side with fewer gaseous molecules - it is less crowded. No change occurs when equal numbers of gaseous molecules appear on both sides. THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM INCREASE PRESSURE MOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSURE MOVES TO THE SIDE WITH MORE GASEOUS MOLECULES Predict the effect of an increase of pressure on the equilibrium position of. . 2 SO 2(g) + O 2(g) 2 SO 3(g) MOVES TO RHS : - fewer gaseous molecules H 2(g) + CO 2(g) CO(g) + H 2 O(g) NO CHANGE: - equal numbers on both sides

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM TEMPERATURE • • • temperature is the only thing that can change the value of the equilibrium constant. altering the temperature affects the rate of both backward and forward reactions it alters the rates to different extents the equilibrium thus moves producing a new equilibrium constant. the direction of movement depends on the sign of the enthalpy change. REACTION TYPE DH INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFT TO THE RIGHT ENDOTHERMIC + TO THE RIGHT TO THE LEFT Predict the effect of a temperature increase on the equilibrium position of. . . H 2(g) + CO 2(g) 2 SO 2(g) + O 2(g) CO(g) + H 2 O(g) 2 SO 3(g) DH = + 40 k. J mol-1 DH = - ive moves to the RHS moves to the LHS

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM CATALYSTS Catalysts work by providing an alternative reaction pathway involving a lower activation energy.

FACTORS AFFECTING THE POSITION OF EQUILIBRIUM CATALYSTS An increase in temperature is used to speed up chemical reactions but it can have an undesired effect when the reaction is reversible and exothermic. In this case you get to the equilibrium position quicker but with a reduced yield because the increased temperature moves the equilibrium to the left. In many industrial processes a compromise temperature is used (see Haber and Contact Processes). To reduce the problem one must look for a way of increasing the rate of a reaction without decreasing the yield i. e. with a catalyst. Adding a catalyst DOES NOT AFFECT THE POSITION OF EQUILIBRIUM. However, it does increase the rate of attainment of equilibrium. This is especially important in reversible, exothermic industrial reactions such as the Haber or Contact Processes where economic factors are paramount.

In Summary Stress Change Increase in concentration of X Reaction that removes X is favoured Decrease in concentration of X Reaction that forms X is favoured Increase in temperature Endothermic reaction is favoured Decrease in temperature Exothermic reaction is favoured Increase in Pressure Reaction that produces fewer molecules is favoured Decrease in Pressure Reaction that produces more molecules is favoured Adding a Catalyst No Change

Le Chateliers Principle - Experiments If a stress is applied to a system at equilibrium, the system adjusts to relieve the stress Co. Cl 42 - + Blue H 2 O Co(H 2 O)62+ + 4 Cl- Pink 1. Add the cobalt chloride crystals to water, state which side the equilibrium favours and state how you came to this conclusion? 2. Add HCL, State and Explain what you observed 3. Add Water, State and explain what you observed 4. Heat in Beaker of hot water, State and explain what you observed = blue

Le Chatelier’s Principle - Experiments If a stress is applied to a system at equilibrium, the system adjusts to relieve the stress Cr 2 O 72 - + orange H 2 O Cr. O 42 - + 2 H+ yellow 1. Add sodium dichromate to deionised water, note colour observed and explain 2. Add Na. OH and note observation and explain 3. Add HCL, Note observation and explain

Le Chateliers Principle -Experiments If a stress is applied to a system at equilibrium, the system adjusts to relieve the stress Fe. Cl 3 yellow + CNS- Fe(CNS)2 - + 3 Cl- red 1. Add iron (III) chloride and potassium thiocyanate, note colour and explain why 2. Add HCL, note colour change and explain why 3. Add Iron (III) chloride, note colour change and explain why

Haber process Manufacture of ammonia NH 3 for the fertilizer industry N 2 + 2 NH 3 3 H 2 2 NH 3 Kc = N 2 x H 2 3 H= - 92. 4

HABER PROCESS N 2(g) + 3 H 2(g) Conditions Pressure Temperature Catalyst 2 NH 3(g) : DH = - 92 k. J mol-1 20000 k. Pa (200 atmospheres) 380 -450°C iron Equilibrium theory favours low temperature exothermic reaction - higher yield at lower temperature high pressure decrease in number of gaseous molecules Kinetic theory favours high temperature greater average energy + more frequent collisions high pressure more frequent collisions for gaseous molecules catalyst lower activation energy Compromise conditions Which is better? A low yield in a shorter time or a high yield over a longer period. The conditions used are a compromise with the catalyst enabling the rate to be kept up, even at a lower temperature.

HABER PROCESS IMPORTANT USES OF AMMONIA AND ITS COMPOUNDS MAKING FERTILISERS 80% of the ammonia produced goes to make fertilisers such as ammonium nitrate (NITRAM) and ammonium sulphate NH 3 + HNO 3 ——> 2 NH 3 + H 2 SO 4 ——> MAKING NITRIC ACID NH 4 NO 3 (NH 4)2 SO 4 ammonia can be oxidised to nitric acid is used to manufacture. . . fertilisers (ammonium nitrate) explosives (TNT) polyamide polymers (NYLON)

Haber process Manufacture of ammonia NH 3 for the fertilizer industry N 2 + 3 H 2 2 NH 3 H= - 92. 4 Le Chatelier’s principle predicts the yield of NH 3 is maximised by 1. Low temperature 2. High pressure Actual temp used is 500 o. C (as lower temp reduces the rate ) Pressure used is 200 ATM

Contact Process • Sulfuric acid is used to manufacture a variety of substances (paints, detergents, fertilisers, plastics, fibres, car batteries) • The name Contact Process comes from the fact that one of the stages in the manufacture involves passing sulfur dioxide gas (obtained by burning sulfur in air) and oxygen gas over a catalyst. • Need very close contact between the two gases

Vanadium Pentoxide Sulfur Dioxide+ oxygen Sulfur trioxide V 2 O 5 2 SO 2 + O 2 2 SO 3 H = -196 Kj/mol Sulfer trioxide is then reacted with water to form sulfuric acid. Platinum can be used as catalyst but it is easily poisened by impurities What conditions do you think are ideal to improve percentage yield?

Contact process Manufacture of Sulfuric acid H 2 SO 4 2 SO 2 + O 2 2 SO 3 H= - 196 Le Chatelier’s principle predicts the yield of SO 3 is maximised by 1. Low temperature 2. High pressure Actual temp used is 450 o. C (as low temp reduces the rate ) Pressure used is 1 ATM

8 Chemical Equilibrium explain what is meant by a reversible reaction explain what is meant by dynamic equilibrium explain what is meant by chemical equilibrium state the equilibrium law (Kconly) write expressions for Kc perform calculations involving equilibrium constants (Kc) state Le Chatelier’s principle use Le Chatelier’s principle to predict the effect (if any ) on equilibrium position of concentration, pressure, temperature and catalyst perform simple experiments to demonstrate Le Chatelier’s principle using the following equilibrium mixtures Co. Cl 42– + 6 H 20 Co(H 20)62+ + 4 Cl– (to demonstrate the effects of both temperature changes and concentration changes on an equilibrium mixture) Cr 2 O 72– + H 2 O 2 Cr. O 42– + 2 H+ Fe 3+ + CNS– Fe(CNS)2+ (to demonstrate the effects of concentration changes on an equilibrium mixture) discuss the Industrial application of Le Chatelier’s principle in the catalytic oxidation of sulfur dioxide to sulfur trioxide and in the Haber process