Chapter 17 Equilibrium

We know reactions take place. We even know how they take place. But, why are some slow, some fast? Why do we have refrigerators? How can we prevent rust?

17.1 How chemical reactions occur Chemicals have to collide with each other to react. They must touch each other They must touch each other hard enough to react. Called collision model

17.1 This spells it all out: if we can keep them from colliding, slower reactions Get them to collide more often, speedier reactions

17.2 Conditions that affect reaction rates What Conditions will affect reaction rates? If we can increase the Concentration of the reactants the reaction should go faster There is a better chance that collision will occur Ex. put more O2 in a fire and whoosh!

17.2 Conditions that affect reaction rates Temperature also increases the chance of collision Moving faster means better chance of smacking into the other reactant opposite is reason for refrigeration!

17.2 Conditions that affect reaction rates But Temperature plays another role Merely colliding does not mean a reaction will occur The reactants must hit with the right amount of Energy or they bounce away The amount of Energy required for a collision to be a good one is called activation energy (Ea)

17.2 Conditions that affect reaction rates Having Ea will make sure the old bonds are broken so that atom exchange can take place

17.2 Conditions that affect reaction rates Particle size will affect reaction rate Particles can only collide at the surface. Smaller particle size, bigger surface area thus faster reaction Smallest possible particle size is ions or molecules That’s why dissolving speeds up reactions. Getting two solids to react is very slow

17.2 Conditions that affect reaction rates Catalyst This is a substance that speeds up a reaction without being consumed In your body the catalysts are called enzymes It speeds up a reaction by giving the reaction a new path The new path has a lower activation energy thus goes faster. An inhibitor is something that blocks the catalyst

17.2 The catalyst lowers the Ea so more molecules can go over the hill This appears to our eyes as a speed up in the reaction

17.3 Heterogeneous Reactions So far we have considered reactions that take place in only one phase 2 aqueous solutions mixed together 2 gases mixed together Those are homogeneous reactions But a heterogeneous reaction would involve two different phases e.g., Zn (solid) tossed into HCl (aqueous)

17.3 These reactions depend on surface area e.g. a block of Zn reacts a lot slower than Zn dust

17.3 The reason some grain silos in the Midwest blow up each year

a summary

17.4 The Equilibrium Condition So far we have assumed that reactions proceed until completion – until one of the reactants “runs out”. Some reactions will stop before completion an equilibrium has been achieved Equilibrium means balance. For chemistry: the exact balancing of two processes, one of which is the opposite of the other. Remember dynamic equilibrium & vapor pressure?

17.4 The Equilibrium Condition Example NO2(g) + NO2(g) N2O4(g) The brown gas NO2 becomes colorless N2O4 in a flask, but the flask doesn’t become colorless. Equilibrium was achieved. The concentration of all reactants & products remain the same. ie the reaction is going forward & backward at the same time. R->P and P->R

17.4 The Equilibrium Condition When reactants are first put together the forward reaction occurs. Since there are no products the reverse reaction does not occur. As the forward reaction proceeds the reactants are used up and the reaction slows. The products build up and the reverse reaction speeds up.

17.5 Chemical Equilibrium Eventually you reach a point where the reverse reaction is going as fast as the forward reaction. This is dynamic equilibrium The rate of the forward reaction is equal to the rate of the reverse reaction. The concentrations of the reactant and products stay the same, but the reactions are still running.

17.5 Chemical Equilibrium Equilibrium position- how much product and reactant there are at equilibrium. Shown with the double arrow Reactants are favored. Products are favored. Catalysts speed up both forward and reverse reactions so don’t effect equilibrium position.

17.6 The Equilibrium Constant Science is based on results of experiments. The equilibrium between reactants and products is described by an equilibrium constant. For the balanced reaction: aA + bB  cC + dD The equilibrium constant, Keq is defined as: [C]c [D]d  products Keq = --------- [A]a [B]b  reactants where the [ ] brackets indicate the concentration of the chemical species in mol/L. K is the equilibrium constant.

17.6 The Equilibrium Constant Rules for Writing Keq Expressions Products are always in the numerator. Reactants are always in the denominator. Express gas concentrations as partial pressure, P, and dissolved species in molar concentration, []. The partial pressures or concentrations are raised to the power of the stoichiometric coefficient for the balanced reaction. Leave out pure solids or liquids and any solvent.

17.6 The Equilibrium Constant Example: Zn (s) + 2 H+(aq)  Zn2+(aq) + H2 (g) PH2 [Zn2+] Keq = ----------- [H+]2 where PH2 is the partial pressure of H2, [Zn2+] and [H+] are the molar concentrations of Zn2+ and H+, respectively, and Zn (s) is left out of the Keq expression because it is a pure solid.

17.6 The Equilibrium Constant Example What is the equilibrium expression for the following reaction? H2O(l)  H+(aq) + OH-(aq) [H+] [OH-]  products Keq = ------------- [H2O]  reactants

17.6 The Equilibrium Constant Example What is the equilibrium expression for the following reaction? 4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g) [NO2]4 [H2O]6 Keq = ------------------ [NH3]4 [O2]7

17.6 The Equilibrium Constant Example Calculate the value of Keq for the following reaction N2(g) + 3H2(g)  2NH3(g) if [NH3 (g)] = 0.34M, [H2(g) ] = 2.1 x 10-3M and [N2(g)] = 4.9 x 10-4M [NH3]2 [0.34]2 Keq = ------------------  Keq = ------------------------ [N2] [H2]3 [4.9 x 10-4] [2.1 x 10-3]3 ANSWER: Keq = 2.5 x 1010

17.8 Le Chatelier’s Principle Le Chatelier's Principle: When a system at equilibrium is disturbed, the equilibrium conditions shift to counteract the disturbance. Example: N2O4 (g)  2 NO2 (g)    K = 11 (at 25oC) Sample equilibrium conditions: PN2O4 = 0.027 atm, PNO2 = 0.546 atm in a 4.0 L container at 25oC. What would disturb this equilibrium? Changing pressure. Changing temperature. Changing the amounts of the reactants or products in the container.

17.8 Le Chatelier’s Principle What can the Keq tell us? When a Keq is calculated A Keq value greater than 1 means that products are favored over reactants; Keq > 1 A Keq value less than 1 means that reactants are favored over products; Keq < 1

17.8 Le Chatelier’s Principle Changing Pressure Example: Decrease the volume of the container in the above example from 4.0 L to 1.0 L (Remember PV = nRT. P and V are inversely proportional!) New partial pressures: PNO2 = 0.546 atm x 4 = 2.18 atm PN2O4 = 0.027 atm x 4 = 0.108 atm In which direction will the reaction proceed to reestablish equilibrium? Calculate the reaction the Keq for the new conditions Keq= (2.18 atm)2 ÷ 0.108 atm = 44 atm Keq > 1 so the reaction will proceed in the reverse direction favoring the products.

How do we find the new equilibrium partial pressures? Start with the new initial conditions: PNO2 = 2.18 atm and PN2O4 = 0.108 atm Q tells us the reaction will shift in the reverse direction, N2O4 (g) 2 NO2 (g). The problem is set up in the following: N2O4NO2Po0.108 atm2.18 atmP+0.5x atm-x atmPeq(0.108+0.5x) atm(2.18-x) atmK = 11 = (2.18-x)2 / (0.108+0.5x) Find x, then find PNO2 and PN2O4