EQUILIBRIUM When non reversible chemical reactions proceed to completion, the concentration of the reactants gradually decrease, until there is NO limiting reactant remaining. Most chemical reactions, however, do not proceed to completion, even if the mixture is allowed to react over a long period of time. If there is sufficient energy available, products will collide to form activated complexes, which will then recombine to form reactants. Therefore, a reverse reaction occurs. During the course of a reversible chemical reaction, the concentration of products and reactants will change, and ultimately reach a value that remains constant. At the instant that the individual concentrations are no longer changing, the system is in uilibrium. Consider the reversible system initially consisting of reactants only. If reactants have sufficient energy when they collide, products will be formed. The initial rate of conversion of reactants to products (the forward reaction) is high, due to the presence of large numbers of reactant molecules which fruently collide. As the reaction proceeds, the concentration of reactant(s) decreases. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 16
As the reactant concentration decreases, the rate of the forward reaction will decrease (fewer reactant numbers result in less fruent collisions and hence fewer successful collisions). At the same time, the concentration of products increases. As the product concentration increases, the rate of the reverse reaction will increase (greater product numbers will result in more fruent collisions and hence more successful collisions). The rate of the forward reaction continues to decrease and the rate of the reverse reaction continues to increase until they become ual i.e. there is no net reaction occurring. At this point, the concentrations of the reactants and products are constant, and the system is in uilibrium. Forward Reaction Reactants Products Reverse/Back Reaction Equilibrium is a dynamic state in which the rates of the forward and reverse reactions are ual i.e. The rate at which the product is being produced is balanced by the rate at which the reactant is being consumed. Note: If the rate of the forward reaction (R f ) is not ual to the rate of the reverse/back reaction (R b ), the system is NOT in uilibrium. Net forward reactions: If the rate of the forward reaction (R f ) is greater than the rate of the reverse/back reaction (R b ), an overall or net forward reaction is occurring. Products are being formed at the expense of reactants. If the rate of the forward reaction (R f ) is less than the rate of the reverse/back reaction (R b ), an overall or net back reaction is occurring. Reactants are being formed at the expense of products. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 17
IMPORTANT POINTS REGARDING SYSTEMS AT EQUILIBRIUM If a reversible reacting system is not at uilibrium, it must eventually attain uilibrium. When uilibrium has been reached, the forward and reverse (back) reactions continue to occur and at ual rates i.e. the system does NOT stop. At uilibrium, individual reactant and product concentrations remain constant i.e. individual concentrations do not change (unless the system is removed from uilibrium). This does NOT imply that the reactant and product concentrations are EQUAL in value. The uilibrium mixture contains every reactant and product. The speed at which a system reaches uilibrium varies from reaction to reaction. Some reactions will reach uilibrium in a matter of seconds. For some reactions, it takes many years to reach a steady state. As the products are able to recombine to produce reactants, the amount of product produced in uilibrium systems will always be less than the amount calculated using standard stoichiometric manipulations. Not all chemical reactions will reach uilibrium. For example, uilibrium is unlikely to be attained in open systems. Equilibrium can be approached from either direction via the forward or reverse (back) reaction. When a system is at uilibrium: The temperature of the system will remain constant. In gaseous systems, the temperature, volume and pressure will be constant. There will be no observable colour changes. The reaction is incomplete. The amounts and concentrations of reactants and products do not change. It appears to the observer that the reaction has stopped, but this is an illusion. There is no net reaction occurring, but both reactions continue to take place at exactly the same rate. Equilibrium can be recognised by measuring: Conductivity ph Pressure Temperature When a system is in uilibrium, it exists in a state of maximum stability. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 18
THE EQUILIBRIUM CONSTANT The extent to which a chemical reaction proceeds is described by the Equilibrium Constant K or K or K ). ( c The uilibrium constant also indirectly describes: The ratio of products to reactants at uilibrium. The proportion (or percentage) of reactants that will be converted to products. The uilibrium constant is an experimentally deduced relationship which describes the relative concentrations of products to reactants when the system has reached uilibrium. The general rule for calculating the uilibrium constant is called the uilibrium law and is described mathematically as follows: c d [ C] [ D] Given aa + bb cc + dd, then the uilibrium constant is K = a b [ A] [ B] i.e. The Equilibrium Constant is calculated by substituting the concentrations of each chemical species when the system is in uilibrium into the above expression. Note: The uilibrium constant is reported for the forward reaction at a particular temperature. The K value for a particular reaction is temperature dependant. If the temperature of the system is changed, the K will also change. Changes to other parameters like concentration or dilution do not change K. Different chemical reactions have different uilibrium constants. The uilibrium law expression refers to the uation as written ( K is uation related). For example: Given that A+ B C, K = 10 i.e. K [ C] A B = 10 If the mole ratios are doubled, the K value will also change. A+ B C, [ C] [ A] [ B] [ C] [ A][ B] 10 100 K = = = = [ ][ ] Whatever the change in the mole ratios, raise K to that value. If the reaction is written in reverse, the value of K for this reaction will be the reciprocal of the forward reaction. K reverse = K 1 forward The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 19
c d [ C] [ D] When the system is not in uilibrium the value of a b [ A] [ B] concentration fraction or reaction quotient (Q). is referred to as the Regarding uilibrium constants for reactions involving gases: We can use the concentrations of gaseous substances to calculate uilibrium constants, however, it is often more convenient to use partial pressures. The partial pressure of a gas is directly proportional to the amount present in mole, which is a measure of concentration. i.e. Using PV = nrt, n P= RT = crt P c V Therefore, given aa( + bb( g) cc( g) + dd( g), the expression for the uilibrium constant may be written as K = pc pd. c d a b pa pb To determine the units for the uilibrium constant: Substitute the unit M (molar) into each concentration in the fraction. Then simplify units by applying the appropriate index laws. Important Notes: As the [ H O( l) ] in aqueous solutions is constant ( 55.5 M ), we omit [ H O( l) ] when calculating the units of the corresponding uilibrium constant. + For example: CH 3COOH ( aq) + H O( l ) CH 3COO ( aq) + H 3O ( aq) + K [ CH3COO ] [ H3O ] = M [ CH COOH ][ H O ] Units = 3 M M M As the [solid] in gaseous and aqueous systems is constant, we omit [solid] when calculating the units of the corresponding uilibrium constant. For example: C ( s) + H O( CO ( + H ( K [ CO][ H ] = M Units = [ H O][ C] M M M Include H O( l) and solids in uilibrium law expressions, but ignore these species when determining the corresponding uilibrium constants and their units. If asked for Ka or K, DO NOT include H O ( l ). w The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 0
QUESTION 13 Write balanced uations for the reactions with the following uilibrium laws: (a) K = H ( g) CO ( g) CH OH 3 ( (b) K HS ( = S( g) H( g) QUESTION 14 Consider the reaction: A( s) + 3B( l) C( aq) + 4D( aq) Which of the following is the correct expression for the uilibrium constant? A 3 [ A] [ B] 4 [ C][ D] B [ C][ D ] 4 C D [ C][ 4D] [ A][ 3B] 4 [ C][ D] [ A][ B] The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 1
THE SIGNIFICANCE OF K The uilibrium constant provides us with valuable information regarding the composition of a reaction mixture, and whether it consists mainly of reactants or products when the system is at uilibrium. Informally, the uilibrium constant represents the proportion of the system that is in the product phase i.e. the percentage conversion of reactants to products. A large K ( K uilibrium. 4 > 10 ) indicates that there are more products than reactants at A small K ( K uilibrium. < 10 4 ) indicates that there are less products than reactants at 4 A medium K ( 10 4 to 10 ) indicates that there are significant amounts of reactants and products at uilibrium. For values of K between Note: the value of K increases. 4 10 and 4 10, the concentration of products increases as The uilibrium constant can never be ual to zero as all reactions in uilibrium systems proceed to some extent, even if negligibly. The uilibrium constant does NOT provide an indication of the rate of a reaction. A high K value DOES NOT imply a fast reaction rate. Catalysts do not change the position of the uilibrium but will allow the system to reach uilibrium more quickly. The uilibrium constant has NO bearing on the spontenaity of a reaction. A high K value does not indicate that a reaction is spontaneous. The extent of a chemical reaction may also be described in terms of the position of uilibrium. If a reaction favours the formation of products, we say that the position of uilibrium lies to the right. If a reaction favours the formation of reactants, we say that the position of uilibrium lies to the left. Reactants Products Left Right The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page
QUESTION 15 Consider the following reactions: I + I I A ( aq) ( l ) 3( aq) Cu + 4 NH Cu ( NH ) + B [ ] ( aq) 3( aq) 3 4 ( aq) + K = 7.1 10 13 K = 1.4 10 Answer the following questions by choosing one of the options below. A B C (a) Reaction A Reaction B Answer cannot be determined from the given information Identify the reaction which is most complete. (b) Which reaction will display the fastest reaction rate? (c) Which reaction has the smallest activation energy barrier? (d) Which reaction is spontaneous? QUESTION 16 The uilibrium constant for the reaction below at 5 o C is AgCl(s) Ag + ( aq) + Cl( aq) 1.7 10 M 10. Which of the following answers represents the uilibrium constant at 50 o C? A 1.7 10 M 10 B 1 1.7 10 C ( ) M 10 1.7 10 M 10 D K cannot be determined The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 3
QUESTION 17 The graphs below illustrate the changes in the reactant and product concentrations in a catalysed and uncatalysed reaction system, If the other reaction conditions are identical,which set of curves (solid lines or broken lines), best represents the uncatalysed system? THE CONCENTRATION FRACTION/REACTION QUOTIENT When the system is not in uilibrium the expression concentration fraction or reaction quotient (Q). c [ C] [ D] a [ A] [ B] d b is referred to as the The concentration fraction provides a wealth of information, and is fruently used to determine: If a system is in uilibrium. When the Concentration Fraction is ual to the Equilibrium Constant, the system is in uilibrium. How far a system is from attaining uilibrium. The smaller the difference between the concentration fraction and the uilibrium constant, the closer the system is to a state of uilibrium. The direction in which uilibrium is being approached. When Q < K the system is said to be approaching uilibrium via a net forward reaction. The concentration of the products will increase and the reactant concentrations will decrease, until the values of Q and K become ual. When Q > K, uilibrium will be approached via a net back reaction. The concentration of the products will decrease and the reactant concentrations will increase, until the values of Q and K become ual. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 4
CALCULATIONS INVOLVING SYSTEMS AT EQUILIBRIUM METHOD: Step 1: Write an expression for the uilibrium constant. Step : Substitute the uilibrium concentrations and/or uilibrium constant into the uilibrium expression. Step 3: Solve for the ruired value. WATCH OUTS: (a) Have I been given data in terms of mole or concentration? If you are given information in terms of mole, do not forget to convert data to concentrations before you substitute values into the uilibrium law. (b) Is the given data at uilibrium, or have initial concentrations been supplied? You must only substitute concentrations at uilibrium into the uilibrium expression (otherwise, a concentration fraction will be obtained). Never substitute initial concentrations. If initial concentrations or number of mole are given, use the mole ratios in the balanced chemical uation to determine how much of each substance is reacting. Use this information to then determine how much of each species is present at uilibrium. (c) Have there been any changes in temperature? If a temperature change occurs - you cannot use the previous uilibrium constant. If new concentrations are not given or a new uilibrium constant is not supplied, you will be unable to calculate an answer. ADDITIONAL POINTS: Do not forget to include the appropriate units. Note that at times, there are no units!!! If asked to determine whether a system is in uilibrium, calculate the concentration fraction for the system. If this value is ual to the uilibrium constant, then the system is in uilibrium. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 5
QUESTION 17 Calculate the uilibrium constant for the reaction H( + O( g) H O( g) if the uilibrium concentrations are as follows: [ H ] = 0. 30 M, [ O ] = 0. 30 M, [ H O] = 1.8 10 M K c [ HO ] (1.8 10 ) = = = [ H ] [ ] 0.30 0.30 ( 1 0.01 M ( g) O( g) QUESTION 18 The formation of nitrosyl chloride from nitric oxide was investigated at a particular temperature. NO + O NOCl ( ( g) ( g) The pressures if gases at uilibrium were p = 0.atm, p pnocl K p NO O = 0.11atm and = 0.3atm. Calculate the uilibrium constant, K p, for the reaction. ( p ) NOCl (0.3) = = = 19 atm (0.) (0.11) ( pno ) ( pcl ) 1 QUESTION 19 Consider the reaction: A+ 3B C+ D. Analysis of an uilibrium mixture in a.0 L container shows that.0 mol of A, 0.50 mol of 1 B and 3.0 mol of D are present. If the uilibrium constant of the reaction is 0.04 M, calculate the amount of C, in mol, in the uilibrium mixture. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 6
QUESTION 0 The following reaction was allowed to reach uilibrium: U + V xw. When the system is at uilibrium, the concentration of each species present is [ U ] = 0. 3M [ V ] = 0. 4M [ W ] = 0. 5M If the value of K is 1.04 for the given temperature of this experiment, then the value of x is A 1 B C 3 D 4 QUESTION 1 Given the uilibrium: A ( + 4C( AC ( K1 = 4. 8 1 It follows that, for the reaction AC( A ( + C( K = X. X would be: A 1 4.8 B. 4 C D 1.4 1 4.8 The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 7
QUESTION 3 Some hydrogen and nitrogen are placed in a 1.0 dm flask at 500 o C. A reaction occurs and ammonia is formed. At uilibrium, p( N) = 50atm, p( H) = 150 atm and 6 K =.65 10 atm. (a) Write an expression for the uilibrium constant for this reaction. (b) Calculate the partial pressure of ammonia. (c) The experiment was repeated at the same temperature and the partial pressures of the gases were found to be p( N) = 1.85 atm, p( H) = 5.35 atm and p( NH3) = 1.75 atm (i) Show that the system is not in uilibrium. (ii) In which direction is uilibrium approaching? Via a net back or forward reaction? (d) State the uilibrium constant for the reaction NH3( 3H( g) + N( g) at 500 o C 3 in a.0 dm flask. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 8
EQUILIBRIUM CALCULATIONS BASED ON INITIAL MOLE/CONCENTRATIONS If initial concentrations or number of mole are given, we must first determine the amounts present at uilibrium before calculating uilibrium constants. METHOD: Step 1: Determine the initial number of mole of each species. If the initial number of mole of product is not given, assume that its amount is ual to zero. Step : Calculate the amount in mole of each species at uilibrium. Use the mole ratios in the balanced chemical uation, together with the given data to determine the number of mole being transferred during the reaction. Then use common sense to determine whether to add or subtract the number of mole being transferred to/from the initial number of mole. Step 3: Calculate the concentrations at uilibrium. Step 4: Calculate the uilibrium constant. Problems of this nature are best resolved using the table below: Reactants Products Mole Ratio Initial Number of Mole Change in Mole Final Mole at Eq Concentration at Eq The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 9
QUESTION 3 One of the steps in the production of sulfuric acid is the oxidation of sulfur dioxide ( SO ). The reactants form an uilibrium with the product, sulfur trioxide ( SO 3 ), according to the uation: SO ( + O( SO 3(..000 mole of SO,.000 mole of O were placed in 10.00 L vessel and allowed to reach uilibrium. When uilibrium was established, 0.150 mole of SO remained. Calculate the value of the uilibrium constant at this temperature. SO O SO 3 Mole Ratio 1 Initial Number of Mole Change in Mole.000.000.000-0.150 1.850 = 1.850.000-0.95 Final Mole at Eq 0.150 = 1.075 0.150 Concentration C = = 0.0150M 1.075 at Eq 10 C = = 0.1075M 10 0 1.850 1.850 1.850 C = = 0.1850M 10 K [ SO ] (0.1850) = = = 1.41 10 [ SO ] [ ] (0.0150) (0.1075) 3 O M 3 1 The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 30
QUESTION 4 3+ Fe ions react with SCN ions according to the uation: Fe 3+ ( aq) + SCN( aq) + ( SCN) ( aq) Fe. 3+ If 0.060 mole of Fe and 0.13 mole of added to 500 ml of water and at uilibrium, 0.16 mole of value for the uilibrium constant. SCN and 0.10 mole of Fe are initially + ( SCN ) ( aq) SCN are present, calculate a Mole Ratio Initial Number of Mole Change in Mole Final Mole at Eq Concentration at Eq The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 31
QUESTION 5 For the reaction HI( H ( + I(, 4.00 mole of HI ( was allowed to react in a.0 L vessel at a constant temperature. When uilibrium had been attained, it was found that 5% of the HI molecules had dissociated. Calculate the uilibrium constant for the reaction. HI ( g) H ( + I( The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 3
FORCED CHANGES TO EQUILIBRIUM SYSTEMS If a system is removed from uilibrium by an external change, it has been experimentally determined that the concentration of reactants and products will change until a new uilibrium position is established. In effect, the system will oppose the change which was made to it. This idea is referred to as Le Chatelier s Principle. If an uilibrium system is subjected to an external change the system will adjust itself to partially oppose the effect of the change The new uilibrium position may be attained by favouring a net forward or reverse reaction. The direction of the net reaction will simply depend upon which reaction (the forward or back reaction) will oppose the change which was introduced. Factors that will cause changes to uilibrium systems include: Concentration changes at constant temperature and volume. Volume changes at constant pressure and temperature. Temperature changes. The effects of an external change on a system may be determined using a number of different approaches, including: Le Chatelier s Principle The Concentration Fraction Rate/Time and Concentration/Time graphs. The safest (and most fool proof) method is by analysing the corresponding Rate/Time and Concentration/Time graphs. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 33
IMPORTANT NOTES Unless otherwise specified or implied, always assume that when a change is introduced, that the temperature of the system remains constant. Read questions carefully to determine whether changes in amounts or concentrations or positions of uilibrium are ruired. When uilibrium is re-established: The value of K will remain unchanged unless a temperature change has occurred. The proportions of reactants and products alter in such a way as to keep the ratio in the K expression unchanged. The concentrations of the reactants and products will be different from that in the initial uilibrium system. The rate of the forward reaction will be ual to the rate of the reverse reaction, but both reaction rates will be different from the previous uilibrium system. If the imposed change increases the concentration of particles, final reaction rates of both the forward and back reaction will increase. If the imposed change decreases the concentration of particles, final reaction rates of both the forward and back reaction will be lower than in the previous uilibrium system. QUESTION 6 According to Le Chatelier s principle, when disrupted by a change at constant temperature, an uilibrium system A Will restore original conditions. B Will not reach uilibrium again. C Will reach a new uilibrium position but with a different value for K. D Will reach a new uilibrium position with the same value of K. The School For Excellence 011 The Essentials Unit 4 Chemistry Book 1 Page 34