1. Review 2 Chapter 16 Chemical Equilibrium

2. Equilibrium Condition: Study equilibrium tells us more about whether a reaction will occur or not. Closed system Example: At 500K, the reaction PCl 3(g) + Cl 2(g)  PCl 5(g) in a closed system If start with 1M of reactants, at equilibrium, [PCl 3(g) ] = [Cl 2(g) ] = 0.52M, [PCl 5(g) ]= 0.48M this gives Kc = 1.8 If start with 1M of product, at equilibrium, [PCl 3(g) ] = [Cl 2(g) ] = 0.52M, [PCl 5(g) ]= 0.48M this gives Kc = 1.8 If start with 2M of reactants, at equilibrium, [PCl 3(g) ] = [Cl 2(g) ] = 0.82M, [PCl 5(g) ]= 1.18M this gives Kc = 1.8 If add 1M of Cl 2, [PCl 3(g) ] = 0.30, [Cl 2(g) ] = 1.3M, [PCl 5(g) ]= 0.70M this gives Kc = 1.8 Depend on concentration, energy and organization

3. K: Law of mass action Expression (Heterogeneous vs. Homogeneous reactions): K p, K c Calculate K of reversed reaction, ½ of a reaction or doubled Example:2SO 2(g) + O 2(g)  2SO 3(g) 4SO 2(g) + 2O 2(g)  4SO 3(g) SO 2(g) + ½ O 2(g)  SO 3(g) 2SO 3(g)  O 2(g) + 2SO 2(g) Example: 2CO (g) + O 2(g)  2CO 2(g) K = 2.75 x 10 20 @1000K CO 2(g)  CO (g) + ½ O 2(g) K = 6.03 x 10 -11 @1000K If the coef in the reaction is:Then K is: DoubledSquared halvedSquare root Reversed in signInverted Multiplied by a constant nRaised to the n th power Summarize:

4. 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: SO 2(g) + ½ O 2(g)  SO 3(g) K = 1.84 at 1000K CO 2(g)  CO (g) + ½ O 2(g) K = 6.03 x 10 -11 Find K for the reaction: SO 2(g) + CO 2(g)  CO (g) + SO 3(g) K = 1.11 x 10 -10

5. 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: SO 2(g) + NO 2(g)  NO (g) + SO 3(g) K = 1.96 @ 1200K Compare Q and K for shift of reaction. Pressures of inert materials or non-reactive materials do not influence K eq. 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, K eq 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

6. 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!! Ice chart has to be in MOLARITY or PARTIAL pressure, no other units such as moles etc allowed