Chem 112 Exam AID Course Pack

Example carbon 3 has increased in

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Unformatted text preview: rium this is the effective concentration which are called activities. §༊ In sufficiently dilute solutions stoichiometric concentrations and effective concentrations are the same §༊ Pure liquids and pure solids (H2O(l) and NaCl(s)) are assigned relative activities equal to 1 Ex. To find the activity of hydrogen if it’s partial pressure is 2.9atm a[O2 ] = 2.9 atm = 2.9 1.0 atm Types of Equilibria Homogeneous equilibriumà༎ when all reactants and products are all of the same phase Heterogeneous equilibriumà༎ when not all reactants and products are all of the same phase € • In a heterogeneous system involving a gas the final equilibrium is independent of the amount of pure solid or liquid there is because they have relative activities (and concentrations) of 1 Reaction Quotient • Calculated in the same way as K except it can be calculated at any point during the reaction (not just at equilibrium like K) which allows us to determine whether the reaction is at equilibrium and if it is not we can determine which way the reaction will shift to establish equilibrium: o If Q=K the system is at equilibrium Ethan Newton & Barry Zhang for SOS Winter 2012 43 o o If Q>K there is greater product than reactant and reaction will shift towards the reactants (Left) If Q<K there is greater reactant than product and the reaction will shift towards the products (right) Example Q calculation: Q = [products] [reactants] Factors Effecting Equilibrium Le Chatelier’s Principleà༎ the reaction will shift in response to a change to re- establish equilibrium • Pressure will never change the value of K but changing pressure (by removing or adding a € substance in the system) the reaction will shift to restore equilibrium • Temperature will always change the value of K because adding heat to a system will change the rate of forward reaction differently than the reverse rate (example: adding heat to a highly exothermic reaction generally will slow the forward reaction and decrease value of K because there will be more reactants than products relatively at equilibrium) • When the equation is reversed K becomes 1/K • When a reaction equation is multiplied by a constant (n) the equilibrium constant becomes Kn • When you add two equilibrium equations together the new K value is equal...
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This note was uploaded on 10/02/2012 for the course CHEM 112 taught by Professor Carran during the Winter '08 term at Queens University.

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