Chapter 14 Chemical Equilibrium Review 2 Chapter 14 Chemical Equilibrium

Equilibrium Condition: Study equilibrium tells us more about whether a reaction will occur or not. Closed system Example: At 500K, the reaction PCl3(g) + Cl2(g)  PCl5(g) in a closed system If start with 1M of reactants, at equilibrium, [PCl3(g)] = [Cl2(g)] = 0.52M, [PCl5(g)]= 0.48M this gives Kc = 1.8 If start with 1M of product, at equilibrium, [PCl3(g)] = [Cl2(g)] = 0.52M, [PCl5(g)]= 0.48M this gives Kc = 1.8 If start with 2M of reactants, at equilibrium, [PCl3(g)] = [Cl2(g)] = 0.82M, [PCl5(g)]= 1.18M this gives Kc = 1.8 If add 1M of Cl2, [PCl3(g)] = 0.30, [Cl2(g)] = 1.3M, [PCl5(g)]= 0.70M this gives Kc = 1.8 Depend on concentration, energy and organization

Expression (Heterogeneous vs. Homogeneous reactions): Kp, Kc K: Law of mass action Expression (Heterogeneous vs. Homogeneous reactions): Kp, Kc Calculate K of reversed reaction, ½ of a reaction or doubled Example: 2SO2(g) + O2(g)  2SO3(g) 4SO2(g) + 2O2(g)  4SO3(g) SO2(g) + ½ O2(g)  SO3(g) 2SO3(g)  O2(g) + 2SO2(g)   Example: 2CO(g) + O2(g)  2CO2(g) K = 2.75 x 1020 @1000K CO2(g)  CO(g) + ½ O2(g) K = 6.03 x 10-11 @1000K Summarize: If the coef in the reaction is: Then K is: Doubled Squared halved Square root Reversed in sign Inverted Multiplied by a constant n Raised to the nth power

Kp = Kc (RT)n n = change in gas moles. It is always product GAS moles – reactant GAS moles T = temp in Kelvin , R is gas conatant Reaction Quotient Q: Use the initial concentrations (or partial pressure pressures) of the reactants and products to predict the direction of the reaction. QK (shift left), QK (shift right), Q = K (at equilibrium). Calculations Example: SO2(g) + ½ O2(g)  SO3(g) K = 1.84 at 1000K CO2(g)  CO(g) + ½ O2(g) K = 6.03 x 10-11 Find K for the reaction: SO2(g) + CO2(g)  CO(g) + SO3(g) K = 1.11 x 10-10

Le Chatelier’s Principle If a system in equilibrium is altered in any way, the system adjust itself by shifting the reaction to minimize the effect of the change (Use Q to predict the shift of the reaction) 1) Change of the concentration or partial pressure of one reactant or product. An example: SO2(g) + NO2(g)  NO(g) + SO3(g) K = 1.96 @ 1200K Compare Q and K for shift of reaction. Pressures of inert materials or non-reactive materials do not influence Keq. 2) Change of the volume of the container. In general:  V shift to the side has more # moles. If equal # moles of product and reactant then V has no influence. 3) Change the pressure of the container. (Change the partial pressure of an inert gas or a non-reactive material has nothing to do with the K)  in Pressure favors side with low gas moles 4) Change of temperature of the environment (depend on exothermic or endothermic reaction, Keq changes with temperature). Inc in temp favors direction of endothermic 5) Addition of a Catalyst, no effect on K, Equilibrium is achieved faster 6) Adding Inert gas : No effect on K

Review Problems with Equilibrium concentrations given and find Kc or Kp Problems with initial concentrations and K, find eqm concentrations Need ICE chart in this one!! For ice chart problems, can you use Approximations?? K*100 << [reactant]0 YES!! [X] *100/ [reactant]0 < 5% YES!! Ice chart has to be in MOLARITY or PARTIAL pressure, no other units such as moles etc allowed