CHEMICAL EQUILIBRIUM CALCULATIONS Please spend extra time understanding and learning the material on the slides with Learning Objectives for the topic Chemical Equilibrium Calculations After having studied the class notes watching the Lesson videos doing the Homework accessing the additional resource materials and attending this Lesson Review students should be able to Correctly construct an ICE table Determine when an ICE Table s construction is required to solve a problem Solve a quadratic equation if needed on your calculator or using a formula Decide when the 5 rule should be used Correctly label the variable used in an ICE table Problem Reviewing Stoichiometric Change in a 1 L Say you started the reaction below with only 10 moles of SO3 container Say x moles of O2 were made when the reaction reaches equilibrium Complete the Table below based on this information 2SO3 g 2SO2 g O2 g Mole ratio of SO3 SO2 O2 2 2 1 Initial of moles in 1L conc Changes in moles for reaction to reach equilibrium Moles in 1L at equilibrium conc ICE Initial Change Equilibrium Box or Table These Tables are constructed in order to ascertain or use equilibrium concentrations or pressures for a reversible reaction E g For the reaction 2A g B g C g where you begin with 2 moles of A and 1 mole of B in a 1L container and x is defined as the change in the moles of C per liter that happened in order for equilibrium to be reached the ICE Table would look like this A B C Initial concentration M 2 1 0 Change in concentration 2x x x Equilibrium concentration M 2 2x 1 x x Problem 3 00 moles of pure SO3 g are introduced into an 8 00 L container at 1105 K At equilibrium 0 580 mol of O2 g have been formed Calculate K Define x as the change in moles per liter O2 needed for equilibrium to be reached The reaction is 2SO3 g 2SO2 g O2 g Initial concentration M Change in concentration Equilibrium concentration M K SO2 2 O2 SO3 2 Problem The initial concentration of I2 in the reaction shown is 1 00 M Calculate the equilibrium concentration of I if at 600 K K 3 30 x 10 5 The reaction is I2 g 2I g Let x be the change in the moles of I2 per liter in order for equilibrium to be reached Initial concentration M Change in concentration Equilibrium concentration M K I 2 I2 Making Approximations If K is relatively small the reaction will not proceed very far to the right If the initial reactant concentration is relatively large we can make the assumption that x is small relative to the initial concentration of reactant So for the former problem involving I2 g 2I g K 3 30 x 10 5 The equilibrium concentration of I2 was 1 00 x K products reactants Since K is small the reaction will not proceed very far to the right and therefore x will also be small So 1 00 x can be approximated to be 1 00 Making Approximations the 5 Rule When K is small it can be approximated that the added or subtracted x 2x 3x etc stated in a species equilibrium concentration can be ignored ONCE THE 5 RULE IS OBEYED 5 RULE When x or 2x 3x etc is less than 5 of the number the approx was made against the assumption to ignore x or 2x 3x etc is valid and the problem can be solved by avoiding the use of a quadratic equation If this rule fails you must use a quadratic equation to solve the problem Let s Approach the Former Problem Again The initial concentration of I2 in the reaction I2 g 2I g is 1 00 M Calculate the equilibrium concentration of I if at 600 K K 3 30 x 10 5 I2 g 2I g Equilibrium concentration M 1 00 x 2x K I 2 I2 4x2 1 00 x 3 30 x 10 5 Small value of K Problem If the initial concentration of I2 in the reaction shown is 0 200 M Calculate the equilibrium concentration of I if at 2 00 x 103 Kelvin K 0 209 I2 g 2I g Let x be the change in the moles of I2 per liter in order for equilibrium to be reached Relevant Equation Initial concentration M Change in concentration Equilibrium concentration M K I 2 I2 The 5 Rule Failed so Now What Formula for solving quadratic equations x b b2 4ac 2a From our quadratic equation 4x2 0 209 x 0 0418 0 Solve this a 4 b 0 209 c 0 0418 Answer I 0 1588M Note An amount of a substance with physical reality cannot have a negative value Problem Part 1 If for the following reaction 1 0 mol NOCl and 1 0 mol Cl2 are initially placed in a 1 0 L flask what are the expressions for the equilibrium concentrations for all species Let x be the change in the moles of Cl2 per liter in order for equilibrium to be reached in your ICE Table 2NOCl g 2NO g Cl2 g K 1 6 x 10 5 at 35oC Relevant rxn Initial concentration M Change in concentration Equilibrium concentration M Problem Part 2 Using the equilibrium expressions from Part 1 is the 5 rule obeyed when solving for x using approximations Relevant equation K 1 6 x 10 5 at 35oC 2NOCl g 2NO g Cl2 g 1 0 2x 2x 1 0 x from last problem Equilibrium concentration M