Ch. 19 Reaction Rates and Equilibrium

Reaction Rates Objective: Describe what is meant by the rate of a chemical reaction. Some chemical reactions occur very rapidly, such as an explosion. Other chemical reactions take years to come to completion, such as decomposition of organic material. Coal is produced by the decomposition of plants under pressure and it takes millions of years. This chapter deals with reaction rates of chemical reactions.

Rates A rate is some measurement per unit time. For example, take two runners competing in a 100 m race. If the first runner takes 12 seconds to complete the race, their rate is 100 m in 12 seconds or 𝟏𝟎𝟎 𝒎 𝟏𝟐 𝒔𝒆𝒄 =𝟖.𝟑 𝒎 𝒔 If the second runner takes 15 seconds to complete the race, their rate is 100 m in 15 seconds or 𝟏𝟎𝟎 𝒎 𝟏𝟓 𝒔𝒆𝒄 =𝟔.𝟕 𝒎 𝒔 The second runner has a lower rate, they are slower.

Reaction Rates Fast reaction rates: Burning a candle Explosions Medium reaction rates: Rusting Aging (human) Decomposition of organic material (rotten food) Slow reaction rates: Formation of coal, diamond

Collision Theory According to collision theory, atoms, ions and molecules can react to form products when they collide, provided that the particles have enough kinetic energy. If they don’t have enough energy, they may just bounce off each other.

Collision Model Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). Reactants must have proper orientation to allow the formation of new bonds.

Activation Energy How do you know if the colliding particles have enough energy? The minimum amount of energy that they must have in order to react is the activation energy. Activation energy is like a barrier that they have to cross to for reactants to be converted to products. Otherwise the reaction won’t happen. During a reaction there may be some intermediate products, called the “activated complex” that form momentarily (10-13 seconds) and then turn into the products. Another name for activated complex is “transition state”.

Activation Energy Activated Complex or transition state

(The minimum energy required to produce an effective collision) Activation Energy The minimum energy required to transform reactants into the activated complex (also known as the transition state) (The minimum energy required to produce an effective collision) Flame, spark, high temperature, radiation are all sources of activation energy

Exothermic Processes Processes in which energy is released as it proceeds, and surroundings become warmer Reactants  Products + energy

Endothermic Processes Processes in which energy is absorbed as it proceeds, and surroundings become colder Reactants + energy  Products

Reaction Rates: 1. Can measure disappearance of reactants 2NO2(g)  2NO(g) + O2(g) Reaction Rates: 1. Can measure disappearance of reactants 2. Can measure appearance of products 3. Are proportional stoichiometrically

The Reaction Mechanism The reaction mechanism is the series of steps by which a chemical reaction occurs. A chemical equation does not tell us how reactants become products; it is a summary of the overall process. Reactants  Products The  sign has represents the reaction mechanism, but gives no indication of the steps in the mechanism

The Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of reaction.

Factors Affecting Rate Temperature Increasing temperature usually increases the rate of a reaction, because it raises the kinetic energy of the particles. Surface Area Increasing surface area increases the rate of a reaction because it increases the surface area that is exposed to participate in the reaction. Concentration Increasing concentration USUALLY increases the rate of a reaction, because it increases the frequency of collisions. Presence of Catalysts and/or Inhibitors. A catalyst serves to lower the activation energy and allow the reaction to proceed more easily. An inhibitor interferes with the action of the catalyst.

Catalysis Catalyst: A substance that speeds up a reaction by lowering activation energy. A catalyst is not actually consumed in the reaction, it just serves as an intermediate, so it is neither a reactant nor a product. Recall you list catalysts on top of the arrow (Pt here) Pt 2H2 + O2 → 2H2O Enzyme: A large molecule (usually a protein) that catalyzes biological reactions (makes them occur at lower temperatures like your body temperature!) Homogeneous catalyst: Present in the same phase as the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

Endothermic Reaction w/Catalyst

Exothermic Reaction w/Catalyst

Reaction Rates Simulation reactions-and-rates_en.jar file on hard drive Web based address (if you are running this at home) http://phet.colorado.edu/en/simulation/reactions-and-rates

Section 19.2 Reaction Equilibrium

Chemical Equilibrium Reversible Reactions: A chemical reaction in which the products can react to re-form the reactants. In a reversible reaction, the reactions occur simultaneously in both directions. Chemical Equilibrium: When the rate of the forward reaction equals the rate of the reverse reaction and the concentration of products and reactants remains unchanged 2HgO(s)  2Hg(l) + O2(g) Arrows going both directions (  ) indicates equilibrium in a chemical equation

Reversible Reactions 2 SO2 + O2 2SO3

Reversible Reactions

Which substance(s) have the highest concentration at equilibrium? If you start out with an excess of SO3, which way does the reaction proceed? Chemical equilibrium is shown at the right hand side of both graphs – when the rate at which the forward and reverse reactions take place is equal. NOTE: rates are equal, but concentrations are not.

Equilibrium position – indicates favored direction Imagine this reaction: A B 1% 99% Here the formation of B is favored (note length of arrows). Now what about this one: 99% 1% In this reaction, the formation of A is favored. In principle every reaction is reversible, but if it is very unbalanced like this, it can be considered irreversible (reversible part is negligible).

Catalysts – forward and reverse reactions Does a catalyst speed up the reaction in only one direction or both? A catalyst speeds up the forward and reverse reactions exactly the same because the reverse reaction is exactly the opposite of the forward reaction. In other words, the catalyst lowers the activation energy barrier when looked at from either side. Does a catalyst shift the location of the equilibrium position (amount reactants/products)? Nope! Catalysts do not affect the amounts of reactants and products present at equilibrium, just the time it takes to establish equilibrium.

Le Chatelier’s Principle When a system at equilibrium is placed under Stress (something that disrupts the balance at equilibrium), the system will undergo a change in such a way as to relieve or counter that stress. http://www.youtube.com/watch?v=dIDgPFEucFM Video showing “see-saw” model Henry Le Chatelier

Le Chatelier Translated: When you take something away from a system at equilibrium, the system shifts in such a way as to replace what you’ve taken away. When you add something to a system at equilibrium, the system shifts in such a way as to use up what you’ve added. Some things that could change include concentrations of reactant or product, changes in temperature and changes in pressure.

Le Chatelier Example #1a A closed container of ice and water at equilibrium. The temperature is raised. Ice + Heat Energy  Water The equilibrium of the system shifts to the _______ to use up the added energy. *** ENDOTHERMIC EXAMPLE *** right

Le Chatelier Example #1b The reaction below is at equilibrium. Then heat is added. Add heat ←direction of shift 2SO2(g) + O2(g) 2SO3(g) + heat Remove heat (cool) direction of shift → This reaction as shown is EXOTHERMIC, so heat can be considered to be a product here. If you add more heat (products) then the reaction is driven backwards towards reactants to try to restore equilibrium. If you remove heat, then reaction shifts to the right.

Le Chatelier Example #2 A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the container. N2O4 (g) + Energy  2 NO2 (g) The equilibrium of the system shifts to the _______ to use up the added NO2. left

LeChatelier Example #3 A closed container of water and its vapor at equilibrium. Vapor is removed from the system. water + Energy  vapor The equilibrium of the system shifts to the _______ to replace the vapor. right

Pressure and Le Chatelier’s If pressure is increased, it drives the reaction to the side that has fewer moles, to reduce the number of molecules in order to offset the pressure increase.

LeChatelier Example #4 A closed container of N2O4 and NO2 at equilibrium. The pressure is increased. N2O4 (g) + Energy  2 NO2 (g) The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation. left

CoCl2 LeChatelier’s video Reaction is as follows: CoCl4-2(aq) + 6H2O(l) ↔ Co(H20)6+2 + 4Cl-(aq) + heat BLUE PINK So the reaction is exothermic in the forward direction as shown. If we write that reaction backwards: Co(H20)6+2 + 4Cl-(aq) + heat ↔ CoCl2-2(aq) + 6H2O(l) PINK BLUE Now can you see that the reaction is endothermic in that direction? http://www.youtube.com/watch?v=cWr3UDo-WeU Video using CoCl2 to demonstrate LeChatelier’s

Sample problem 19-1 What effect do each of the following changes have on the equilibrium position for this reversible reaction? PCl5 + heat PCl3 + Cl2 addition of Cl2 Shifts the equilibrium to the left, forming more PCl5 increase in pressure Shifts equil. to left (fewer moles) to decrease P. removal of heat Shifts equil. to left to produce more heat. removal of PCl3 as it forms Shifts equil. to right to produce more PCl3.

Equilibrium Constants aA + bB cC + dD For the reaction above, the equilibrium constant, Keq, is the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation. 𝑲 𝒆𝒒 = 𝑪 𝒄 𝒙 𝑫 𝒅 𝑨 𝒂 𝒙 𝑩 𝒃 The exponents are the coefficients of the balanced chemical reaction.

Equilibrium Constants 𝑲 𝒆𝒒 = 𝑪 𝒄 𝒙 𝑫 𝒅 𝑨 𝒂 𝒙 𝑩 𝒃 Equilibrium constants provide valuable chemical information. They show whether reactants (A and B) or products (C and D) are favored at equilibrium. If Keq > 1, then products are favored at equil. If Keq < 1, then reactants are favored at equil. Exclude any pure solids or pure liquids (their concentration is undefined anyway).

Sample Problem 19-2 Dinitrogen tetroxide (N2O4), which is colorless, and nitrogen dioxide (NO2), which is brown, exist in equilibrium with each other. A liter of the gas mixture at 10oC at equilibrium contains 0.0045 mol N2O4 and 0.030 mol NO2. Write the expression for the equilibrium constant and calculate the equilibrium constant Keq for the reaction: N2O4 2NO2 𝑲 𝒆𝒒 = (𝑵𝑶 𝟐 ) 𝟐 ( 𝑵 𝟐 𝑶 𝟒 ) 𝟏 = ( 𝟎.𝟎𝟑𝟎 𝒎𝒐𝒍/𝑳) 𝟐 (𝟎.𝟎𝟎𝟒𝟓 𝒎𝒐𝒍 𝑳 ) =𝟎.𝟐𝟎 𝐦𝐨𝐥/𝐋

Sample problem 19-3 One mol of colorless hydrogen gas and one mol of violet iodine vapor are sealed in a 1L flask and allowed to react at 450 oC. At equilibrium, 1.56 mol of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate Keq for the reaction: H2 (g) + I2 (g) 2HI (g) From the equation, 1 mole each hydrogen and iodine will make two moles of HI, so that means you need 0.78 mol H2 and I2 to make 1.56 mole HI. That means there is 1-0.78 = 0.22 moles left of hydrogen and iodine reactants at equilibrium. 𝑲 𝒆𝒒 = [𝑯𝑰] 𝟐 𝑯 𝟐 [ 𝑰 𝟐 ] = [𝟏.𝟓𝟔 𝒎𝒐𝒍 𝑳 ] 𝟐 𝟎.𝟐𝟐 𝒎𝒐𝒍 𝑳 𝟎.𝟐𝟐 𝒎𝒐𝒍 𝑳 =𝟓𝟎

Sample problem 19-4 Bromine Chloride (BrCl) decomposes to form chlorine and bromine. 2BrCl (g) Cl2(g) + Br2(g) At some temperature, the equilibrium constant is 11.1, and the equilibrium mixture is 4.00 mol Cl2. (That means there must be 4.00 mol Br2 as well, why??) How many moles of Br2 and BrCl are in the mixture? Assume the volume is 1 L. 𝑲 𝒆𝒒 = 𝑪𝒍 𝟐 [ 𝑩𝒓 𝟐 ] [𝑩𝒓𝑪𝒍] 𝟐 𝒔𝒐 𝟏𝟏.𝟏= 𝟒.𝟎𝟎 𝒎𝒐𝒍 𝑳 [𝟒.𝟎𝟎 𝒎𝒐𝒍 𝑳 ] [𝑩𝒓𝑪𝒍] 𝟐 solve for BrCl concentration: [BrCl] = 1.20 mol/L

Equilibrium Constant with Heterogeneous Reactions What about this reaction? CaCO3 (s) ↔ CaO(s) + CO2 (g) note the solids You would think the equilibrium constant would be: 𝑲 𝒆𝒒 = [ 𝑪𝑶 𝟐 ] [𝑪𝒂𝑶] [𝑪𝒂𝑪 𝑶 𝟑 ] But experimental results show that the position of a heterogeneous equilibrium does not depend on the amounts of pure solids or pure liquids present. The reason is because the concentrations of pure solids and liquids cannot change. So the equilibrium expression for this reaction is: 𝑲 𝒆𝒒 =[𝑪 𝑶 𝟐 ]

Heterogeneous Reactions Remember - If more than one phase of matter is present in a reaction be aware that equilibrium does not depend on the amount of solid, or pure liquid present. Pure solids and liquids are excluded from the Keq expression. They have undefined concentrations. NH3 (g) + HCl (g) → NH4Cl (s) 𝐾 𝑒𝑞 = 1 [𝑁𝐻 3 ][𝐻𝐶𝑙]

Heterogeneous reactions Example: PCl5(s) PCl3(l) + Cl2(g) Keq = [products]/[reactants] Keq = [PCl3] [Cl2] / [PCl5] Pure liquids and solids are omitted So Keq = [Cl2]

Heterogeneous Equilibrium Now you try these ones: 2H2O (l) ↔ 2H2 (g) + O2 (g) Keq = 2H2O (g) ↔ 2H2 (g) + O2 (g)