FURTHER TOPICS ON CHEMICALEQUILIBRIUM KNOCKHARDY PUBLISHING 2008 SPECIFICATIONS

INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING CHEMICAL EQUILIBRIUM (2)

CONTENTS The Equilibrium Law The equilibrium constant K c Calculations involving K c Calculations involving gases Mole fraction and partial pressure calculations Calculations involving K p CHEMICAL EQUILIBRIUM (2)

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 in 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

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

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 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm 3 ) of the equilibrium mixture 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 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm 3 ) of the equilibrium mixture K c = [CH 3 COOC 2 H 5 ] [H 2 O] [CH 3 COOH] [C 2 H 5 OH] 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 equilibrium concs. 1/3 / V 1/3 / V 2/3 / V 2/3 / V V = volume (dm 3 ) of the equilibrium mixture K c = [CH 3 COOC 2 H 5 ] [H 2 O] = 2/3 / V. 2/3 / V = 4 [CH 3 COOH] [C 2 H 5 OH] 1/3 / V. 1/3 / V CALCULATIONS INVOLVING K c

Example 2 Consider the equilibrium P + 2Q R + S (all species are aqueous) One mole of P and one mole of Q are mixed. Once equilibrium has been achieved 0.6 moles of P are present. How many moles of Q, R and S are present at equilibrium ? P + 2Q R + S Initial moles At equilibrium 0·6 0·2 0·4 0·4 (0·4 reacted) (2 x 0·4 reacted) (get 1 R and 1 S for every P that reacts) 1- 0·6 remain 1- 0·8 remain Explanation if 0.6 mol of P remain of the original 1 mol, 0.4 mol have reacted the equation states that 2 moles of Q react with every 1 mol of P this means that 0.8 (2 x 0.4) mol of Q have reacted, leaving 0.2 mol one mol of R and S are produced from every mol of P that reacts this means 0.4 mol of R and 0.4 mol of S are present at equilibrium CALCULATIONS INVOLVING K c

CALCULATIONS INVOLVING GASES Method carried out in a similar way to those involving concentrations one has the choice of using K c or K p for the equilibrium constant when using K p only take into account gaseous species for the expression use the value of the partial pressure of any gas in the equilibrium mixture pressure is usually quoted in Nm -2 or Pa - atmospheres are sometimes used as with K c, the units of the constant K p depend on the stoichiometry of the reaction

CALCULATIONS INVOLVING GASES Method carried out in a similar way to those involving concentrations one has the choice of using K c or K p for the equilibrium constant when using K p only take into account gaseous species for the expression use the value of the partial pressure of any gas in the equilibrium mixture pressure is usually quoted in Nm -2 or Pa - atmospheres are sometimes used as with K c, the units of the constant K p depend on the stoichiometry of the reaction USEFUL RELATIONSHIPS total pressure =sum of the partial pressures partial pressure =total pressure x mole fraction mole fraction =number of moles of a substance number of moles of all substances present

MOLE FRACTION AND PARTIAL PRESSURE Example A mixture of 16g of O 2 and 42g of N 2, exerts a total pressure of Nm -2. What is the partial pressure of each gas ?

MOLE FRACTION AND PARTIAL PRESSURE Example A mixture of 16g of O 2 and 42g of N 2, exerts a total pressure of Nm -2. What is the partial pressure of each gas ? moles of O 2 = mass / molar mass = 16g / 32g = 0.5 mol moles of N 2 = mass / molar mass = 42g / 28g= 1.5 mol total moles = = 2.0 mol

MOLE FRACTION AND PARTIAL PRESSURE Example A mixture of 16g of O 2 and 42g of N 2, exerts a total pressure of Nm -2. What is the partial pressure of each gas ? moles of O 2 = mass / molar mass = 16g / 32g = 0.5 mol moles of N 2 = mass / molar mass = 42g / 28g= 1.5 mol total moles = = 2.0 mol mole fraction of O 2 = 0.5 / 2 = 0.25 mole fraction of N 2 = 1.5 / 2 = 0.75 sum of mole fractions = = 1 (sum should always be 1)

MOLE FRACTION AND PARTIAL PRESSURE Example A mixture of 16g of O 2 and 42g of N 2, exerts a total pressure of Nm -2. What is the partial pressure of each gas ? moles of O 2 = mass / molar mass = 16g / 32g = 0.5 mol moles of N 2 = mass / molar mass = 42g / 28g= 1.5 mol total moles = = 2.0 mol mole fraction of O 2 = 0.5 / 2 = 0.25 mole fraction of N 2 = 1.5 / 2 = 0.75 sum of mole fractions = = 1 partial pressure of O 2 = mole fraction x total pressure = 0.25 x Nm -2 = 5000 Nm -2 partial pressure of N 2 = mole fraction x total pressure = 0.75 x Nm -2 = Nm -2

MOLE FRACTION AND PARTIAL PRESSURE Example A mixture of 16g of O 2 and 42g of N 2, exerts a total pressure of Nm -2. What is the partial pressure of each gas ? moles of O 2 = mass / molar mass = 16g / 32g = 0.5 mol moles of N 2 = mass / molar mass = 42g / 28g= 1.5 mol total moles = = 2.0 mol mole fraction of O 2 = 0.5 / 2 = 0.25 mole fraction of N 2 = 1.5 / 2 = 0.75 sum of mole fractions = = 1 partial pressure of O 2 = mole fraction x total pressure = 0.25 x Nm -2 = 5000 Nm -2 partial pressure of N 2 = mole fraction x total pressure = 0.75 x Nm -2 = Nm -2

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p.

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p. N 2 (g) + 3H 2 (g) 2NH 3 (g) moles (initially) moles (equilibrium)1 - x 3 - 3x 2x (x moles of N 2 reacted)

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p. N 2 (g) + 3H 2 (g) 2NH 3 (g) moles (initially) moles (equilibrium)1 - x 3 - 3x 2x (x moles of N 2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P )

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p. N 2 (g) + 3H 2 (g) 2NH 3 (g) moles (initially) moles (equilibrium)1 - x 3 - 3x 2x (x moles of N 2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 10 6 Pa.

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p. N 2 (g) + 3H 2 (g) 2NH 3 (g) moles (initially) moles (equilibrium)1 - x 3 - 3x 2x (x moles of N 2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 10 6 Pa. applying the equilibrium law K p = (PNH 3 ) 2 with units of Pa -2 (PN 2 ). (PH 2 ) 3

CALCULATIONS INVOLVING K p ExampleWhen nitrogen (1 mole) and hydrogen (3 moles) react at constant temperature at a pressure of 8 x 10 6 Pa, the equilibrium mixture was found to contain 0.7 moles of ammonia. Calculate K p. N 2 (g) + 3H 2 (g) 2NH 3 (g) moles (initially) moles (equilibrium)1 - x 3 - 3x 2x (x moles of N 2 reacted) mole fractions (1-x)/(4-2x) (3-3x)/(4-2x) 2x/(4-2x) (total moles = 4 - 2x) partial pressures P.(1-x)/(4-2x) P.(3-3x)/(4-2x) P.2x/(4-2x) (total pressure = P ) at equilibrium there are 0.7 moles of ammonia, so 2x = 0.7 and x = 0.35 the total pressure (P) = 8 x 10 6 Pa. applying the equilibrium law K p = (PNH 3 ) 2 with units of Pa -2 (PN 2 ). (PH 2 ) 3 Substituting in the expression gives K p = 1.73 x Pa -2

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