Concentration Flammable materials burn faster in pure oxygen than in air because the of O 2 is greater. Hospitals must make sure that no flames are allowed near patients receiving oxygen. Surface Area (not in book) Reality Check: What burns faster, large or small pieces of wood? Reactions occur at the of a solid or a liquid. surface area leads to Rxn Rates Higher surface area leads to more collisions. More collisions lead to higher reaction rates. Simplified reaction for the burning of wood: C(s) + O2(g) à CO2(g) + heat For a reaction to occur, there must be a collision between C & O. Oxygen can t collide with C when it s in the middle of the large piece. For Liquids Spray nozzles create small droplets with greater surface area. (aerosols vs bulk liquids) Example: burning bulk vs sprayed hair spray. Ch 07 Page 15
Catalysts Catalyst: A substance that speeds up a chemical reaction but itself is not used up in the process. They: Do get used up. Provide an alternate pathway for the reaction Lower the Still make the same products Speed up the reaction Do not change the value of. are biological catalysts. With a catalyst, steps with lower Ea values are followed. : A substance that decreases the rate of a reaction. Problem: Ammonia is synthesized by reaction of nitrogen and hydrogen in the presence of an iron catalyst according to the equation: 3 H2(g) + N2(g) à 2 NH3(g) What effect will the following changes have on the reaction rate? a) The temperature is raised from 600 K to 700 K? b) The iron catalyst is removed? c) The concentration of H 2 gas is halved? Ch 07 Page 16
7.7 Reversible Reactions and Chemical Equilibrium Most reactions we have written so far go to completion. e.g. 2 Na (s) + Cl 2(g) " 2 NaCl (s) (single right arrow) What happens when reactants and products are of similar stabilities? Example: Which direction? 2SO 2 + O 2 D 2 SO 3 A reaction which can easily go back and forward in both directions is known as a Reversible Reaction and is shown with arrows. As soon as you start the reaction, a little bit of the product goes back to being reactants. Over time: Reactant concentrations go and eventually become stable. Product concentrations come and eventually become stable. This is a rather than a static process. The concentrations of products and reactants become constant, but the individual molecules making up the reactants and products are constantly changing. Reactant concentrations start high and go down. Products start low and go up. Concentrations reach a value when equilibrium is reached. Ch 07 Page 17
It is not necessary for the concentrations of reactants and products at equilibrium to be equal. The extent to which the forward or reverse reaction is favored over the other is a characteristic property of a reaction under given conditions. Equilibrium and Rates You start with a high rate of forward reaction and a small rate of reverse reaction. The rate of the forward reaction eventually slows down as reactants are used up The rate of the reverse reaction speeds up as more product molecules are formed. Eventually the two rates become equal. (Chemical) Equilibrium: A state in which the rate of forward reaction equals the rate of the reverse reaction. Ch 07 Page 18
7.8 Equilibrium Equations and Equilibrium Constants When substances are in a dynamic equilibrium, a certain ratio of products and reactants remains a constant. This ratio is expressed as follows: For the general reversible reaction: a A + b B c C + d D Equilibrium Equation: K c = C A [ ] c [ D] d [ ] a [ B] b The equation always puts raised to the power of their coefficient. If there is no coefficient for a reactant or product in the reaction equation it is assumed to be. The value of K varies with 25 C is assumed unless specified. Expressed. For reactions that involve pure solids or liquids, these pure substances are omitted when writing the equilibrium constant expression. Only include substances with or state symbols. The value of the equilibrium constant tells us the position of the reaction at equilibrium. If K >> 1 (>10 3 ) mostly products at equilibrium If K is <<1 (<10-3 ) mostly reactants than products If K is @ 1 There are measurable amounts of products and reactants present. Ch 07 Page 19
Example: For the rxn of sulfur dioxide with oxygen to form sulfur trioxide and water, K = 429 2 SO 2 + O 2 D 2 SO 3 K = [SO 3 ] 2 = 429 [SO 2 ] 2 [ O 2 ] Problem: Write equilibrium constant equations for the following reactions: a) N 2 O 4(g) D 2 NO 2(g) b) CH 4(g) + Cl 2(g) D CH 3 Cl (g) + HCl (g) c) 2 BrF 5(g) D Br 2(g) + 5 F 2(g) d) CaCO 3 (s) D CaO(s) + CO 2 (g) Problem: Do the following reactions favor reactants or products at equilibrium? a) Sucrose (aq) + H 2 O (l) D glucose (aq) + fructose (aq) K=1.4 x 10 5 b) NH 3(aq) + H 2 O (l) D NH 4 + (aq) + OH - (aq) K=1.6 x 10-5 c) Fe 2 O 3(s) + 3 CO (g) D 2 Fe (s) + 3 CO 2(g) K=24.2 Ch 07 Page 20
Problem: For the reaction of H 2(g) + I 2(g) D 2 HI (g), equilibrium concentrations at 25 C are [H 2 ] = 0.0250 mol/l, [I 2 ] = 0.0869 mol/l, and [HI] = 0.251 mol/l. What is the value of K at 25 C? Problem: For the reaction N 2 O 4(g) D 2 NO 2(g), the value of K at 25 C is 7.19x10-3. a) Calculate the [N 2 O 4 ] at equilibrium when the [NO 2 ] is 3.50x10-2 mol/l. b) Calculate the [NO 2 ] at equilibrium when the [N 2 O 4 ] is 4.30x10-2 mol/l. Ch 07 Page 21
7.9 LeChatelier s Principle: The Effect of Changing Conditions on Equilibria LeChatelier s Principle: When a stress is applied to a system at equilibrium, the equilibrium to relieve the. The word stress means anything that the original equilibrium. Changes in concentration, temperature, and pressure affect the of the equilibrium, Addition of a catalyst affect the position of the equilibrium, it just lets the system reach equilibrium more quickly. Effects of Concentration Changes Concentration: the number of particles per given volume. Typically measured in moles/l aka M. 1) What would happen if we increased the concentration of a reactant such as CO in the following reaction. When CO was added we had too many reactants (overcrowding) compared to products to keep the ratio of K constant. Some of the reactants had to to reduce the amount of reactants and increase the amount of products to re-establish our constant ratio. When the CO concentration came down, the H2 concentration had to fall as well. The final concentration on CO is still larger than it started out. Ch 07 Page 22
2) What would happen if we increased the concentration of a product such as CH 3 OH instead? Reaction had to. 3) What happens if we decrease the concentration of a reactant? Reaction had to. Better question: HOW can you decrease a concentration? Pluck ions out with tweezers? 4) What happens if we decrease the concentration of a product? Reaction had to. Effects of Temperature Changes Changing the temperature cannot change an endothermic reaction into an exothermic reaction. It is the intrinsic nature of the reaction. You should think of heat as (for endothermic reactions) and as for exothermic reactions. Think of changes in temperature as a change in available heat. (The disturbance) If temperature goes, available heat goes. If temperature goes, available heat goes. Ch 07 Page 23
Example: Which way will the equilibrium shift if the temperature is increased for the following reaction? N 2(g) + O 2(g) + heat D 2 NO (g) If a reaction is exothermic, it will NOT be favored by an increase in temperature. N 2(g) + 3 H 2(g) D 2 NH 3(g) + heat Effects of Pressure Changes Chemical equilibrium is affected by pressure changes only if one of the reactants or products is a. Since gas particles are farther apart than those of liquids or solids, when pressure is increased, the equilibrium tends to shift in the direction of the. Example: Oxygen transport in blood by hemoglobin. Equilibrium equation: Hb + 4 O2 D Hb(O2)4 In the lungs there is an excess on O 2 so the reaction shifts to the to form more Hb(O 2 ) 4 When the blood reaches the tissues where there is a lack of oxygen, the reaction shifts to the and the hemoglobin gives up the oxygen to the tissue. Ch 07 Page 24
Problem: What effect will each of the listed changes have on the position of the equilibrium in the reaction of carbon with hydrogen? C (s) + 2 H 2(g) D CH 4(g) DH = -18 kcal/mol a) increasing temperature C (s) + 2 H 2(g) D CH 4(g) + heat b) Increasing pressure Since there are 2 moles of gas on the left and only one mole of gas on the right c) Allowing CH 4 to escape continuously from the reaction vessel? C (s) + 2 H 2(g) D CH 4(g) + heat Effects of a Catalyst The addition of a catalyst has on the position of the equilibrium. It simply allows the reaction to reach equilibrium faster. Ch 07 Page 25