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Ch. 19 Reaction Rates and Equilibrium

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1 Ch. 19 Reaction Rates and Equilibrium

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3 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 which takes millions of years. This chapter deals with reaction rates of chemical reactions.

4 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.

5 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

6 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.

7 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.

8 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”.

9 Activation Energy Activated Complex or transition state

10 (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

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

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

13 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

14 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 represents the reaction mechanism, but gives no indication of the steps in the mechanism

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

16 Factors Affecting Rate – Collision Theory Demos
Temperature Increasing temperature usually increases the rate of a reaction, because it raises the kinetic energy of the particles when they collide. 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.

17 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. Remember to list catalysts on top of the arrow! (Pt) 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.

18 Endothermic Reaction w/Catalyst

19 Exothermic Reaction w/Catalyst

20 Reaction Rates Simulation
reactions-and-rates_en.jar file on hard drive Web based address (if you are running this at home)

21 Section 19.2 Reaction Equilibrium

22 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

23 Reversible Reactions 2 SO2 + O SO3

24 Reversible Reactions

25 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.

26 Equilibrium position – indicates favored direction
Imagine this reaction: A B 1% % Here the formation of B is favored (note length of arrows). Now what about this one: A B 99% % 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).

27 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)? No! Catalysts do not affect the amounts of reactants and products present at equilibrium, just the time it takes to establish equilibrium.

28 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. Video showing “see-saw” model Henry Le Chatelier

29 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.

30 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

31 Le Chatelier Example #1b
The reaction below is at equilibrium. Then heat is added. Add heat ← direction of shift 2SO2(g) + O2(g) SO3(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.

32 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

33 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

34 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.

35 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

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

37 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.

38 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.

39 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).

40 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 mol N2O4 and mol NO2. Write the expression for the equilibrium constant and calculate the equilibrium constant Keq for the reaction: N2O NO2 𝑲 𝒆𝒒 = [𝑵𝑶 𝟐 ] 𝟐 [𝑵 𝟐 𝑶 𝟒 ] 𝟏 = ( 𝟎.𝟎𝟑𝟎 𝒎𝒐𝒍/𝑳) 𝟐 (𝟎.𝟎𝟎𝟒𝟓 𝒎𝒐𝒍 𝑳 ) =𝟎.𝟐𝟎 Keq typically is given without units

41 Sample problem 19-3 (more difficult)
One mole of colorless hydrogen gas and one mole of violet iodine vapor are sealed in a 1L flask and allowed to react at 450 oC. At equilibrium, 1.56 mole of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate Keq for the reaction: H2 (g) + I2 (g) HI (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 = 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

42 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

43 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 ]

44 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] Note: if no value on top or bottom, just put a 1 there, as shown above.

45 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]

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

47 Related Videos to help you
Equilibrium – Crash Course Chemistry-10’ video Le Chatelier’s Principle https://www.youtube.com/watch?v=g5wNg_dKsYY Equilibrium Equations – Crash Course Chemistry 9:30 video Keq and how to do RICE tables https://www.youtube.com/watch?v=DP-vWN1yXrY

48 RICE tables RICE stands for R = reaction - write the balanced equation
I = initial concentrations of reactant and product C = change in concentrations of reactants and products E = equilibrium concentrations of reactants These are often called “ICE” tables as well because they assume you have the good sense to write down the reaction first anyway.

49 RICE example 1 from video
Let’s use the reaction H2(g) + F2(g) ↔ 2HF(g) Note that we will have 2x the amount of HF than either H2 or F2 R H2(g) + F2(g) ↔ 2HF(g) I C E

50 RICE example 1 from video
Let’s use the reaction H2(g) + F2(g) ↔ 2HF(g) Let’s say we start with 3.00 mol H2(g) and 6.00 mol of F2(g) in a 3.0 L container. The concentration of H2(g) =3.00 mol/3L = 1.00M. The concentration of F2(g) =6.00 mol/3L = 2.00M. The concentration of HF is zero because reaction hasn’t started yet. R H2(g) + F2(g) ↔ 2HF(g) I 1.00 M 2.00 M 0.00 M C E

51 RICE example 1 from video
We don’t know the change in concentration of the reactants, but we are being asked to find the final concentration. So for now, let’s say the concentration of H2 and F2 both decrease by x, because the coefficients in the balanced equation are both 1. Because the coefficient for HF is 2, the amount of product made is 2x R H2(g) + F2(g) ↔ 2HF(g) I 1.00 M 2.00 M 0.00 M C -x +2x E

52 RICE example from video
Add the I and C rows to get the expression for the equilibrium concentration of each reactant and product. Then we will be substituting these into the Keq expression: 𝐾 𝑒𝑞 = [𝐻𝐹] 2 [ 𝐻 2 ] [ 𝐹 2 ] = [2𝑥] −𝑥 [2.00−𝑥] R H2(g) + F2(g) ↔ 2HF(g) I 1.00 M 2.00 M 0.00 M C -x +2x E 1.00-x 2.00-x 2x

53 RICE example 1 from video
Add the I and C rows to get the expression for the equilibrium concentration of each reactant and product. Then we will be substituting these into the Keq expression: 𝐾 𝑒𝑞 = [𝐻𝐹] 2 [𝐻 2 ] [ 𝐹 2 ] = [2𝑥] −𝑥 [2.00−𝑥] solve for x Keq for this reaction can be looked up in tables to be 115 115= [2𝑥] −𝑥 [2.00−𝑥] 115 1−𝑥 2−𝑥 =4 𝑥 2 115[2−3𝑥+ 𝑥 2 ]=4 𝑥 2

54 RICE example 1 from video
115[2−3𝑥+ 𝑥 2 ]=4 𝑥 2 230 −345𝑥+115 𝑥 2 =4 𝑥 2 111 𝑥 2 −345𝑥+230=0 Use quadratic equation 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 𝑥= +345 ± −4(111)(230) 2∗111 𝑥= 345 ±

55 RICE example from video
𝑥= 345 ± Each time you do this, only one of the answers will make sense and be chemically feasible. If +, then x = 2.13 M If -, then x = M Since you only started out with x = 1.00 M or 2.00 M, it can’t be that it changed by 2.13M because you didn’t have that much to begin with. So that leaves the correct answer: x = M

56 RICE example 1 from video
So x = M Therefore [H2]eq = = 0.032M And [F2] = = 1.03 M And [HF] = 2 x = 1.94 M R H2(g) + F2(g) ↔ 2HF(g) I 1.00 M 2.00 M 0.00 M C -x +2x E 1.00-x 2.00-x 2x

57 RICE/ICE Ex. 2 R 2SO2(g) + O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C ??
Here’s a more complicated example of how to set up an ICE table when the values of how much it changed (the C) are given to you. Let’s say moles of SO2 and moles of O2 are placed in a 1 L container (so that’s 0.6M of each). At equilibrium, [SO3] = 0.250M R 2SO2(g) O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C ?? M E 0.250 M

58 RICE/ICE Ex. 2 R 2SO2(g) + O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C - 0.250M
How do you know what to put for the change in the reactants? Use stoichiometry! If 0.250M of SO3 was created, which is 2SO3 then if 2SO2 was consumed, because the mole ratio of 2SO2 / 2SO3 is a ratio of 1, it must be that 0.250M goes in the box for SO2 R 2SO2(g) O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C M ?? M E 0.250 M

59 RICE/ICE Ex. 2 R 2SO2(g) + O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C - 0.250M
If 0.250M of SO2 was consumed, then it must be that ½ that amount of O2 was consumed because the mole ratio of O2 to SO2 is ½. ½ of 0.250M is M, as shown in the table below. Now it is a simple matter to subtract to get the equilibrium concentrations of the reactants, in green. R 2SO2(g) O2(g) ↔ 2SO3(g) I 0.600 M 0.00 M C M M M E 0.350 M 0.475 M 0.250 M

60 RICE/ICE Tables RICE/ICE tables are a major part of this unit because you need the ability to solve for equilibrium concentrations of reactants and products, given initial concentrations and an equilibrium constant. Pay attention to the POGIL called Equilibrium – it helps teach you the concept. Watch the video that is embedded in this power point (go to my website, open the power point file, go to those slides on slideshow, click on the links and it will go to the correct youtube file). You may need to watch it more than once. Mostly, you will learn ICE tables by really working through Worksheet B Eq. Calcs using Ice, and Worksheet C Equilibrium Calculations. YOU NEED TO BE ABLE TO DO THESE KINDS OF PROBLEMS FOR THE TEST.

61 Section 19.3 Spontaneity Free energy is energy that is available to do work. Energy that is used to do work is typically not used very efficiently. For example, an internal combustion engine in a car maybe only about 30% efficient. That means that of the energy produced by burning gasoline, only 30% goes into making the car go forward. The rest is lost as heat or friction. Also, energy can only be obtained from a reaction if the reaction actually occurs. Reactions may not always be spontaneous (they may not occur on their own). A really great efficiency for energy use would be 70%.

62 19.3 Spontaneity This reaction normally would occur (it is spontaneous, given a spark): C(s) + O2(g) → CO2 (the burning of something organic) Spontaneous means the reaction favors the formation of products at equilibrium. Spontaneous reactions release free energy. This reaction normally would not occur: CO2 → C(s) + O2(g) We do not normally see carbon dioxide falling apart to become solid carbon plus oxygen gas. Since the second reaction is non-spontaneous, this reaction does not give substantial amounts of product at equilibrium.

63 19.3 Spontaneity In nearly all reversible reactions, one direction is favored over the other. It is important to note that the terms spontaneous and non-spontaneous do not refer to HOW FAST the reactants go to products. Some spontaneous reactions go so slowly that they appear to be nonspontaneous. The reaction of C6H12O6 + O2 ↔ CO2 + H2O, however a bowl of sugar in air does nothing, so you might suspect that the reaction favors the reactants. The reaction actually does favor the products, but the reaction rate is really slow. If you supply energy in the form of heat, then the reaction goes quite quickly towards the products (burning sugar).

64 19.3 Spontaneity Some reactions that are nonspontaneous at one set of conditions are spontaneous at other conditions. For example, photosynthesis: 6CO2 + 6H2O → C6H12O6 + 6O2 This nonspontaneous reaction can be driven to completion by the addition of sunlight as energy. Sometimes if a nonspontaneous reaction is coupled with a spontaneous reaction (that releases free energy), the free energy released by the spontaneous reaction can enable the non-spontaneous reaction to go forward.

65 Spontaneity and Entropy
You might expect that based on heat being released for spontaneous reactions, only exothermic reactions would be spontaneous. But some other reactions are spontaneous even though they absorb heat! Consider: ice + heat → water Ice melts spontaneously. That seems to violate the rule that in spontaneous processes, the direction of the change in energy is from higher to lower energy (free energy is released). Some factor other than heat must help determine whether a physical or chemical process is spontaneous.

66 Entropy The other factor is related to order.
The disorder of a system is measured as entropy Increasing entropy

67 Entropy junk

68 A more advanced definition of entropy
Entropy is a measure of the “disorder” of a system. What disorder refers to is really the number of different microscopic states a system can be in, given that the system has a particular fixed composition, volume, energy, pressure and temperature. By “microscopic states” we mean the exact states of all the molecules making up the system. Suppose you put a marble in a large box, and shook it up, and didn’t look inside afterwards. Then the marble could be anywhere inside the box. Because the box is large, there are many places inside the box that the marble could be, so the marble in the box has a high entropy. Now suppose you put the marble into a tiny box, and shook it. You have better knowledge of where the marble is because the box is small, so the box has low entropy.

69 Entropy If entropy is considered, then it makes sense why the melting of ice to form liquid water is spontaneous, because it raises the entropy of the system.

70 4 Entropy Guidelines (p. 552/553)
For a given substance, the entropy of the gas is greater than the entropy of the liquid or solid. So entropy increases in reactions in which solid reactants form liquid or gaseous products. Entropy increases when liquid reactants form gaseous products too. Entropy increases when a substance is divided into parts, for example when a crystalline solid like NaCl dissolves in water, because solute particles are more separated than in the crystalline solid form. Entropy tends to increase when the total number of product molecules is > total number reactant molecules. Entropy increases when temperature increases because disorder increases as molecules move faster.

71 More entropy guidelines (advanced)
Entropies of large, complicated molecules are greater than those of smaller, simpler molecules. Examples for So - CH4 : C2H6: C3H8: C4 H10 : 310.0 Entropies of ionic solids are larger when the bonds within them are weaker: Examples for So – NaF: NaCl: 72.13 NaBr: 86.82 NaI:

72 Example: ice going to water

73 Example: water going to ice (doesn’t happen!)

74 Section 19.4 - Calculating Entropy and Free Energy
The symbol for the entropy of a system is S, with units of J/K or J/K mol when it is given per mole of substance. The standard entropy of a liquid or solid substance at 25oC is designated So. The pressure at So for gases is kPa. The theoretical entropy for a perfect crystal at 0 K is zero. Other substances have positive entropies, even at absolute zero (0 K). Entropy change is calculated as follows: ∆ 𝑆 𝑜 = 𝑆 0 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)− 𝑆 0 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠)

75 Table 19.2 gives standard entropies for a variety of compounds.

76 Example 19-5 Calculate the standard entropy change when 1 mol H2O(g) at 25oC and kPa condenses to 1 mol H2O(l) at the same temperature. Look up values in table on previous page: H2O(g) : J/K mol H2O(l) : J/K mol ∆ 𝑆 𝑜 (𝑐𝑜𝑛𝑑𝑒𝑛𝑠𝑎𝑡𝑖𝑜𝑛)= 𝑆 0 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)− 𝑆 0 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) ∆ 𝑆 𝑜 = 𝐽 𝐾 𝑚𝑜𝑙 − 𝐽 𝐾 𝑚𝑜𝑙 =− 𝐽 𝐾 𝑚𝑜𝑙 The negative sign indicates that entropy decreases (it becomes more orderly)

77 Example 19-6 What is the standard change in entropy for the following reaction at 25o C and kPa: NO(g) + O2(g) → 2NO2 (g) From Table 19.2 NO(g) So = J/K mol NO2 (g) So = J/K mol O2(g) So = J/K mol First balance the reaction (I’ve done that above) Then multiply the per-mole values by the number of moles and subtract [products – reactants] ΔSo = 2(240.5) – [1(205.0) + 2(210.6)] = J/K

78 Free-Energy Calculations
In every spontaneous process, some energy becomes available to do work. This energy, the Gibbs free-energy change (ΔG) is the maximum amount of energy that can be coupled to another process to do useful work. The change in Gibbs free energy is related to the change in entropy (ΔS) and the change in enthalpy (ΔH) of the system by the following equation: ΔG = ΔH – TΔS The temperature in this equation must be in Kelvin. ΔGo = ΔHo – TΔSo You can calculate the standard free energy change using this equation

79 Free energy calculations
If ΔH and ΔS are unknown, you can use ΔGfo , the standard free energy change for the formation of substances from their elements, to calculate ΔGo for a given reaction. ΔGo = ΔGfo(products) - ΔGfo(reactants) This is similar to how we calculate standard heats of formation ΔHfo .

80 Note that ΔGfo = 0 for elemental substances in their standard state.
Note that this is at 25oC and kPa only. Why do you think Br2 gas has more free energy than Br2 liquid? (Liquid is the std. state for bromine)

81

82 Free Energy Consider the reaction CaCO3 (solid, formula unit) →
CaO (solid, formula unit) + CO2 (gas, molecule) In this reaction, entropy increases because one of the products formed from the solid reactant is a gas. This entropy increase is not sufficient for the reaction to be spontaneous at ordinary temperatures because the reaction is endothermic. The enthalpy of the reactants is lower than that of the products. The effect of an entropy increase is magnified as the temperature increases. At temperatures above 850oC, the TΔS term outweighs the unfavorable enthalpy term ΔHo, and the reaction becomes spontaneous.

83 Example 19-7 Using values for enthalpy and entropy,
determine if this reaction is spontaneous: C(s, graphite) + O2 (g) → CO2(g) Note that when I have looked up the values for So in Table 19.2, they are in J, and I have divided by 1000 to convert them to kJ to match ΔHfo values Substance ΔHfo (kJ/mol) So (kJ/K mol) C(s, graphite) O2 (g) CO2 (g) ∆ 𝑆 𝑜 = 𝑆 0 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)− 𝑆 0 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) = ( ) = kJ/K mol ∆ 𝐻 𝑜 = 𝐻 𝑓 0 (𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠)− 𝐻 𝑓 0 (𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠) = – ( ) = kJ/mol ΔGo = ΔHo – TΔSo = kJ/mol – (298.15K x kJ/Kmol) = kJ/mol The reaction is spontaneous due to the decrease in enthalpy and increase in entropy (answer is negative). Example 19-7

84 Sample problem 19-8 Using the ΔGfo for the reactants and products, determine whether the reaction from the last example is spontaneous: C(s, graphite) + O2 (g) → CO2(g) C(s, graphite) ΔGfo = 0 kJ/mol O2 (g) ΔGfo = 0 kJ/mol CO2 (g) ΔGfo = kJ/mol ΔGo = ΔGfo(products) - ΔGfo(reactants) ΔGo = kJ/mol – ( ) ΔGo = kJ/mol Of course these results are identical to the last problem, nd the reaction is spontaneous, with a large release of free energy, the ΔGo is negative.

85 19.5 Rate Laws and Reaction Mechanisms
The rate of a reaction depends in part on the concentrations of the reactants. Take this one step reaction (with just one activated complex between reactants and products): A → B The rate at which A transforms to B is given by the change in A with respect to the change in time: 𝑅𝑎𝑡𝑒= ∆𝐴 ∆𝑡 α [A] The last part of that equation says that the rate of disappearance of A is proportional to the concentration of A (recall square brackets for concentrations). The constant of proportionality can be called k, and yields 𝑅𝑎𝑡𝑒= ∆𝐴 ∆𝑡 =k[A]

86 19.5 Rate Laws and Reaction Mechanisms
A rate law is an expression relating the rate of a reaction to the concentration of reactants. The specific rate constant (k) for a reaction is a proportionality constant relating the concentrations of reactants to the rate of the reaction. 𝑅𝑎𝑡𝑒= ∆𝐴 ∆𝑡 =k[A] The order of a reaction is the power to which the concentration of a reactant must be raised to give the experimentally observed relationship between concentration and rate. In a first order reaction, the reaction rate is directly proportional to the concentration of only one reactant (even if there are other reactants): Rate = k[A]1

87 19.5 Rate Laws and Reaction Mechanisms
How about a more general reaction, like a double replacement? aA + bB → cC + dD Ro a one-step reaction of A with B, the rate of reaction is dependent on the concentrations of both A and B: Rate = k[A]a [B]b When each of the exponents in the reaction above are 1, then reaction is said to be first order in A, first order in B, and second order overall. The overall order of a reaction is the sum of the exponents for the individual reactants. For “ideal” one step reactions, the coefficents a and b are also the exponents, but in “real” reactions, these are experimentally determined because it may not be a simple one-step reaction.

88 19.5 Rate Laws and Reaction Mechanisms
Initial concentration of A (mol/L) Initial rate (mol/L sec) 0.050 3.0 x 10-4 0.10 12 x 10-4 0.20 48 x 10-4 What can we observe about what order this rate equation should be? When we 2x [A] the rate is multiplied by 4 When we 4x [A] the rate is multiplied by 16 What does that suggest about the order? The rate must be 2nd order in [A] given the factors above ( 22 = 4, 42 = 16)

89 19.5 Rate Laws and Reaction Mechanisms
If you were to graph all the energy changes that occur as reactants are converted to products in a chemical reaction, that would be a reaction progress curve. An elementary reaction is one where reactants are converted to products in a one step reaction. This reaction would have one activated complex between reactants and products, and therefore only one activation energy peak. Most chemical reactions, however, consist of a number of elementary reactions. For a complex reaction, the reaction progress curve resembles a series of hills and valleys. The hills correspond to the energy of the activated complexes, and the valleys represent intermediate products (the reactants of the next part of the reaction).

90 Example: The three peaks in this energy diagram correspond to activation energies for the intermediate steps of the reaction. The middle hump represents the highest energy barrier to overcome; therefore, the reaction involving N2O2 + 2H2 is the rate determining step.


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