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Mini-Unit 12: Reaction Rates and Equilibrium

Learn about equilibrium concentrations and the equilibrium constant (Kc) in chemical reactions. Understand how to calculate concentrations using ICE tables. Explore examples and apply the concepts to solve equilibrium concentration problems.

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Mini-Unit 12: Reaction Rates and Equilibrium

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  1. Mini-Unit 12: Reaction Rates and Equilibrium Equilibrium Concentrations and Kc

  2. Recall, equilibrium is reached whenthe rates of the forward and reverse reactions are equal. There is no change in the concentrations of reactants and products.

  3. Equilibrium Constant (Kc) For the general reaction, aA + bB ⇌ cC + dD once the reaction has reached equilibrium, the concentrations of reactants and products are related in the following manner: [C]c[D]d = Kc [A]a[B]b

  4. K>1, products are favored at equilibrium K<1, reactants are favored at equilibrium

  5. Important! • As previously discussed with LeChâtelier’s principle, pure solids and pure liquids have no effect on equilibrium. Do not put them into the equilibrium expression! • Only equilibrium concentrations can be used in the K expression! • Concentrations are always in molarity.

  6. Example For the following reaction, 2CO(g) + O2(g)⇌ 2CO2(g) the equilibrium concentrations for CO is 1.5M and for CO2 is 0.81M. What is the concentration of the O2? (K=15.3) [0.81]2 [CO2]2 Kc= = 15.3 [1.5]2 [x] [CO]2[O2] [O2] = 0.019M

  7. But… what if the concentrations given in the problem are not the concentrations at equilibrium?

  8. Then… “ICE, ICE, baby.”

  9. ICE Tables ICE stands for: Initial concentration, Change in concentration, and Equilibrium concentration. They are setup in the following manner:

  10. Important! • ICE tables are only used if you need to calculate a new concentration. • Allow “x” to represent the change in concentration as the reaction reaches equilibrium. • Always plug concentrations in molarity into the ICE table.

  11. Example The equilibrium constant for the conversion of cis-stilbene to trans-stilbene is 24.0 at 200°C. Suppose that initially only cis-stilbene is present at a concentration of 0.850M. Calculate the equilibrium concentrations of cis- and trans-stilbene. 0.850M 0 -x +x x 0.850-x [trans-stilbene] [x] x = 0.816 = 24.0 Kc= [cis-stilbene] [0.850-x] [cis-stilbene]= 0.850-0.816 [cis-stilbene]= 0.034M [trans-stilbene]= 0.816 [trans-stilbene] = 0.816M

  12. Example For the reaction, H2(g) + I2(g) ⇌ 2HI(g) at a certain temperature, the value of K is 56. If the initial concentration of H2 is 0.0175M and of I2 is 0.0175M, what will be the equilibrium concentrations of each reactant and product? 0.0175M 0.0175M 0 -x -x +2x 2x 0.0175-x 0.0175-x [2x]2 [HI]2 = 56 Kc= x = 0.0138 [I2] [H2] [0.0175-x] [0.0175-x] [H2] and [I2] = 0.0037M [H2] and [I2] = 0.0175-0.0138 [HI] = 2(0.0138) [HI] = 0.0276M

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