AN INTRODUCTION TO CHEMICALEQUILIBRIUM KNOCKHARDY PUBLISHING

CONTENTS Concentration change during a chemical reaction Dynamic equilibrium Equilibrium constants Le Chatelier’s Principle The equilibrium constant K c Calculations involving K c CHEMICAL EQUILIBRIUM

In an ordinary reaction; all reactants end up as products; there is 100% conversion 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 SLOWS DOWN AS REACTANTS ARE USED UP TOTAL CONVERSION TO PRODUCTS THE STEEPER THE GRADIENT, THE FASTER THE REACTION

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. EQUILIBRIUM REACTIONS 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 FASTEST AT THE START NO BACKWARD REACTION FORWARD REACTION SLOWS DOWN AS REACTANTS ARE USED UP BACKWARD REACTION STARTS TO INCREASE AT EQUILIBRIUM THE BACKWARD AND FORWARD REACTIONS ARE EQUAL AND OPPOSITE

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. 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 DYNAMIC EQUILIBRIUM

Simply states “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” There are several forms of the constant; all vary with temperature. K c the equilibrium values are expressed as concentrations of mol dm -3 K p the equilibrium values are expressed as partial pressures The partial pressure expression can be used for reactions involving gases THE EQUILIBRIUM LAW

for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c

for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c Example Fe 3+ (aq) + NCS¯(aq) FeNCS 2+ (aq) K c = [ FeNCS 2+ ]with units of dm 3 mol -1 [ Fe 3+ ] [ NCS¯ ]

for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c VALUE OF K c AFFECTED bya change of temperature NOT AFFECTED bya change in concentration of reactants or products a change of pressure adding a catalyst

”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.” Everyday example A rose bush grows with increased vigour after it has been pruned. Chemistry example If you do something to a reaction that is in a state of equilibrium, the equilibrium position will change to oppose what you have just done LE CHATELIER’S PRINCIPLE

CONCENTRATION FACTORS AFFECTING THE POSITION OF EQUILIBRIUM 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 K c 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.

CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c 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 FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c 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 decreasing [H 2 O]- 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

CONCENTRATION example CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) the equilibrium constant K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 4 (at 298K) [CH 3 CH 2 OH] [CH 3 COOH] increasing [CH 3 CH 2 OH]- will make the bottom line larger so K c 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 decreasing [H 2 O]- 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 INCREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE RIGHT THE EFFECT OF CHANGING THE CONCENTRATION ON THE POSITION OF EQUILIBRIUM DECREASE CONCENTRATION OF A REACTANTEQUILIBRIUM MOVES TO THE LEFT INCREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE LEFT DECREASE CONCENTRATION OF A PRODUCTEQUILIBRIUM MOVES TO THE RIGHT FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

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

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

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. 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. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM 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. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM Predict the effect of an increase of pressure on the equilibrium position of.. 2SO 2 (g) + O 2 (g) 2SO 3 (g) H 2 (g) + CO 2 (g) CO(g) + H 2 O(g) 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. INCREASE PRESSUREMOVES TO THE SIDE WITH FEWER GASEOUS MOLECULES DECREASE PRESSUREMOVES TO THE SIDE WITH MORE GASEOUS MOLECULES THE EFFECT OF PRESSURE ON THE POSITION OF EQUILIBRIUM Predict the effect of an increase of pressure on the equilibrium position of.. 2SO 2 (g) + O 2 (g) 2SO 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. 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  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT 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  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT Predict the effect of a temperature increase on the equilibrium position of... H 2 (g) + CO 2 (g) CO(g) + H 2 O(g)  = + 40 kJ mol -1 2SO 2 (g) + O 2 (g) 2SO 3 (g)  = - ive 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  INCREASE TEMP DECREASE TEMP EXOTHERMIC - TO THE LEFTTO THE RIGHT ENDOTHERMIC+TO THE RIGHTTO THE LEFT Predict the effect of a temperature increase on the equilibrium position of... H 2 (g) + CO 2 (g) CO(g) + H 2 O(g)  = + 40 kJ mol -1 moves to the RHS 2SO 2 (g) + O 2 (g) 2SO 3 (g)  = - ive moves to the LHS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

CATALYSTS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY Catalysts work by providing an alternative reaction pathway involving a lower activation energy.

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. 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. FACTORS AFFECTING THE POSITION OF EQUILIBRIUM

for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c

VALUE OF K c AFFECTED bya change of temperature NOT AFFECTED bya change in concentration of reactants or products a change of pressure adding a catalyst for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant THE EQUILIBRIUM CONSTANT K c

construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm 3 ) to get the equilibrium concentrations in mol dm -3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for K c. substitute values from third step and calculate the value of K c with any units CALCULATIONS INVOLVING K c

construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm 3 ) to get the equilibrium concentrations in mol dm -3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for K c. substitute values from third step and calculate the value of K c with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CALCULATIONS INVOLVING K c

construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm 3 ) to get the equilibrium concentrations in mol dm -3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for K c. substitute values from third step and calculate the value of K c with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH 3 COOH(l) + C 2 H 5 OH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) CALCULATIONS INVOLVING K c

construct the balanced equation, including state symbols (aq), (g) etc. determine the number of moles of each species at equilibrium divide moles by volume (in dm 3 ) to get the equilibrium concentrations in mol dm -3 (If no volume is quoted, use a V; it will probably cancel out) from the equation constructed in the first step, write out an expression for K c. substitute values from third step and calculate the value of K c with any units Example 1 One mole of ethanoic acid reacts with one mole of ethanol at 298K. When equilibrium is reached it is found that two thirds of the acid has reacted. Calculate the value of Kc. CH 3 COOH(l) + C 2 H 5 OH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) moles (initially) moles (at equilibrium) 1 - 2/ /3 2/3 2/3 Initial moles of CH 3 COOH = 1 moles reacted = 2/3 equilibrium moles of CH 3 COOH = 1/3 For every CH 3 COOH that reacts; a similar number of C 2 H 5 OH’s react (equil moles = 1 - 2/3) a similar number of CH 3 COOC 2 H 5 ’s are produced a similar number of H 2 O’s are produced CALCULATIONS INVOLVING K c