Chapter Seven Chemical Reactions: Energy, Rates, and Equilibrium Fundamentals of General, Organic and Biological Chemistry 6th Edition James E. Mayhugh Copyright © 2010 Pearson Education, Inc.

“Chemical” energy The energy released by substances when bonds are _____________

“Chemical” energy The energy released by substances when bonds are FORMED!

- The “Energy in Chemical Bonds Myth” (Ok… “myth” is a bit overstated… how about “oversimplification”?) Without exception, energy is released when new bonds form Without exception, energy must be absorbed in order for a bond to be broken.

- The “Energy in Chemical Bonds Myth” - So…. Why have you often heard that: - “Energy stored in chemical bonds is released when those bonds are broken.” - ???

- The “Energy in Chemical Bonds Myth” - So…. Why have you often heard that: - “Energy stored in chemical bonds is released when those bonds are broken.” - ??? - Because they don’t go on to tell you that…. after those bonds are broken (which takes IN energy), new bonds are going to form (which RELEASES energy)…

- The “Energy in Chemical Bonds Myth” - So…. Why have you often heard that: - “Energy stored in chemical bonds is released when those bonds are broken.” - ??? - And the NET result is that more energy was released than absorbed.

►Heat of reaction: Represented by  H, is the difference between the energy absorbed in breaking bonds and that released in forming bonds.  H is also known as enthalpy change. (“Heat energy” & “enthalpy” are synonyms)

- Enthalpy Diagrams

►Heat of reaction: Represented by  H ►Endothermic: A process or reaction that absorbs more heat than it releases ►and has a positive  H ►and feels cold to the touch. ►Exothermic: A process or reaction that releases more heat than it absorbs ►and has a negative  H ►and feels hot to the touch.

7.3 Exothermic and Endothermic Reactions

7.3 Energy Stoichiometry ! ►Predicting the amount of heat released or absorbed by a given reaction with given starting amounts of reactants.

►Example: How much heat is released when 6.1 mol of O 2 reacts with excess C 3 H 8 ? C 3 H O 2  3 CO H 2 O ΔH = -531 kcal/mol rxn

7.4 Spontaneous Reactions ►Spontaneous process: A process that, once started, proceeds without any external influence. ►“Spontaneous” does NOT mean instantaneous ►The 2 “driving forces” that push a reaction toward being spontaneous: ►If the reaction is exothermic, that is a favorable condition for spontaneity (ΔH = - ) ►If the products are more disordered than the reactants, that is a favorable condition.

7.4 Disorder in Reactions –Entropy ►Entropy: The amount of disorder of a substance. ►The symbol S is used for entropy ►An increase in entropy (ΔS = +) is favorable ►Gases have high entropy, solids have low entropy, liquid and aqueous substances are in-between ►If the physical states are the same, then the more independent particles a sample has, the greater the entropy. ►If all else is equal, higher temperature would lead to higher entropy

3 H 2 (g) + N 2 (g)  2 NH 3 (g) During this reaction 1.entropy decreases and  S = –. 2.entropy decreases and  S = +. 3.entropy increases and  S = –. 4.entropy increases and  S = +.

3 H 2 (g) + N 2 (g)  2 NH 3 (g) During this reaction 1.entropy decreases and  S = – since all of the physical states are the same, and there are fewer moles of products than reactants, so there is less disorder.

3Pb(NO 3 ) 2 (aq) + 2FeCl 3 (aq)  3PbCl 2 (s) + 2Fe(NO 3 ) 3 (aq) For the above reaction we predict that 1.entropy decreases and  S = –. 2.entropy decreases and  S = +. 3.entropy increases and  S = –. 4.entropy increases and  S = +.

3Pb(NO 3 ) 2 (aq) + 2FeCl 3 (aq)  3PbCl 2 (s) + 2Fe(NO 3 ) 3 (aq) For the above reaction we predict that 1.entropy decreases and  S = – since there is a solid product which is a lot less disordered than the aqueous reactants

7.4 Is the Reaction Spontaneous?? ►If BOTH driving forces are favorable… ►ΔH = - (exothermic) ►ΔS = + ( products are more disordered than reactants ) Then the reaction IS spontaneous… And will continue to completion by itself once we get it started.

7.4 Is the Reaction Spontaneous?? ►If BOTH driving forces are unfavorable… ►ΔH = + (endothermic) ►ΔS = - (products are less disordered than react.) Then the reaction is NOT spontaneous… And will require a constant input of energy to keep the reaction going.

7.4 Is the Reaction Spontaneous?? ►But what if the 2 Driving forces CONFLICT??? ►Then we must rely on….

Chapter Seven DA JUDGE!

7.4 Gibbs Free Energy ►  G =  H - T  S Da Judge!

7.4 Gibbs Free Energy ►  G =  H - T  S Temp MUST be in Kelvin !!!

7.4 Gibbs Free Energy ►  G =  H - T  S Must be the same units as each other

►If  G ends up being negative, the reaction will be spontaneous. ►Also known as an exergonic reaction ►If  G ends up being positive, the reaction will NOT be spontaneous (or nonspontaneous) ►Also known as an endergonic reaction

HH SS GG (-) favorable(+) favorable (+) unfavorable(-) unfavorable (-) favorable(-) unfavorable (+) unfavorable(+) favorable  G =  H - T  S

HH SS GG (-) favorable(+) favorable(-) spontaneous always (+) unfavorable(-) unfavorable (-) favorable(-) unfavorable (+) unfavorable(+) favorable  G =  H - T  S

HH  S (“magnified” by temp!) GG (-) favorable(+) favorable(-) spontaneous always (+) unfavorable(-) unfavorable(+) nonspontaneous always (-) favorable(-) unfavorable (+) unfavorable(+) favorable  G =  H - T  S

HH  S (“magnified” by temp!) GG (-) favorable(+) favorable(-) spontaneous always (+) unfavorable(-) unfavorable(+) nonspontaneous always (-) favorable(-) unfavorable(-) Low T (+) High T (+) unfavorable(+) favorable  G =  H - T  S

HH  S (“magnified” by temp!) GG (-) favorable(+) favorable(-) spontaneous always (+) unfavorable(-) unfavorable(+) nonspontaneous always (-) favorable(-) unfavorable(-) Low T (+) High T (+) unfavorable(+) favorable(+) Low T (-) High T  G =  H - T  S

7.5 Reaction Rates ►The value of  G indicates whether a reaction will occur but… ► it does not say anything about how fast the reaction will occur or … ►about the details of the molecular changes that takes place.

7.5 The “Collision Model” ►Reactant molecules must collide in order to “swap atoms” ►The Rate at which a reaction occurs is determined by: ►1. How often the molecules are colliding ►2. And how many of those collisions actually work! ►We will look at these factors one at a time…

7.6 Factors Affecting How Often Molecules Collide ►1. How “concentrated” the reactants are: ►Large quantities of reactants in a small volume will collide more often than when they are more spread out… ►So: Higher Concentration  Faster Reaction Rate

7.6 Factors Affecting How Often Molecules Collide ►2. How fast the molecules are moving: ►Molecules at higher temperatures move faster… ►So: higher temperatures  Faster Reaction Rate

7.6 Factors Affecting Whether a Collision Will Work ►A collision that leads to the formation of products is called an “effective” collision ►1. For a collision to be “effective”, the molecules must collide with the correct orientation:

7.5 Factors Affecting Whether a Collision Will Work ►2. For a collision to be “effective”, the molecules must collide hard enough to break the necessary bonds ►Activation energy (E act ): The amount of energy the colliding particles must have for productive collisions to occur.

►Activation energy (E act ): ►Some collisions are harder than others… some collide hard enough to react, some don’t. ►The lower the activation energy threshold is, the greater the number of collisions that will react. ►So: Low E act  Faster Reaction Rate

Examples

►Catalyst—a substance that speeds up a chemical reaction by lowering the Activation Energy. ►It does this by “pre-weakening” the bonds of the reactants… ►So they don’t have to collide as hard to break each other’s bonds… ►So more of the collisions are now effective ►A catalyst always gets re-made by the time the reaction completes (or else it would just be another “reactant”) 7.6 Effects of Catalysts on Reaction Rates

►Enzymes are the catalysts in the human body… ►They speed up the thousands of biochemical reactions taking place

Catalyst Example Atmospheric Ozone depletion:

Catalyst Example Atmospheric Ozone depletion: light O 3 O 2 + O ∙

Catalyst Example Atmospheric Ozone depletion: light O 3 O 2 + O ∙ O 3 + Cl ∙  O 2 + ∙ OCl

Catalyst Example Atmospheric Ozone depletion: light O 3 O 2 + O ∙ O 3 + Cl ∙  O 2 + ∙ OCl O ∙ + ∙ OCl  O 2 + Cl ∙

When a catalyst is added to a reaction, the… 1.activation energy is lowered and  G becomes more negative. 2.activation energy is lowered and  G remains unchanged. 3.activation energy is raised and  G becomes more positive. 4.activation energy is raised and  G remains unchanged.

7.7 Reversible Reactions and Chemical Equilibrium ►Some reactions are reversible

7.7 Reversible Reactions and Chemical Equilibrium ►Some reactions are reversible ►The products collide and re-form the reactants

7.7 Reversible Reactions and Chemical Equilibrium ►We will call the “left-to-right” reaction the forward reaction ►Right-to-left: reverse reaction

7.7 Reversible Reactions and Chemical Equilibrium ►As the forward reaction uses up its reactants, its rate will slow down

7.7 Reversible Reactions and Chemical Equilibrium ►As the forward reaction uses up its reactants, its rate will slow down ►As the reverse reaction gains more of ITS reactant (NH 3 ), its rate will Speed up.

►At some point, the two rates will become equal… ►Meaning that the reactants are depleting at the same rate that they are being remade… ►And the same goes for the products. ►We call this chemical equilibrium.

7.7 Reversible Reactions and Chemical Equilibrium ►Chemical Equilibrium: ►When the rates of the forward and reverse reactions are equal ►The point at which the amounts of the reactants and products stop changing

7.8 The Mathematics of Reversible Reactions ►We can’t use normal stoichiometry for reversible reactions, because… ►None of the substances totally reacts away… there’s no limiting reactant ►Instead, we make use of a very handy relationship… the equilibrium constant of a reversible reaction.

7.8 Equilibrium Equations and Equilibrium Constants ►Consider the following general equilibrium reaction: aA + bB + … mM + nN + … - OMIT all ( l ) and (s) substances from the Equilibrium Equation (replace with “1”)

7.8 Equilibrium Equations and Equilibrium Constants Where [M] means “concentration of M” in mol once the reaction reaches equilibrium L

And at any given temperature, the value for K is always the same for a given reaction…. The concentrations will always “land” in such a way that they will divide out to whatever its K value is.

►Let’s consider the qualitative meaning of the value of K: ►K is larger than 1: ►Numerator (the products) must be greater than the denominator (the reactants), so… ►More products than reactants are present at equilibrium. ►We say that the “products are favored”

►K is less than 1: ►Denominator is greater than the numerator, so… ►More reactants than products are present at equilibrium. ►We say that the “reactants are favored”

7.8 Equilibrium Equations and Equilibrium Constants ►Now for quantitative equilibrium math: ►1. Calculate the equilibrium constant, K ►Given: concentrations of all substances at equilibrium… ►Calculate: The value for K ►Solution: Set up the equilibrium equation and substitute in the given values… solve for K

7.8 Equilibrium Equations and Equilibrium Constants ►2. Calculate the equilibrium concentration of a given substance ►Given: K and all equilibrium values except for one ►Calculate: The equilibrium concentration of the remaining substance ►Solution: Set up the equilibrium equation and substitute in the given values… solve for the unknown concentration

Example 1: S 2 (g) + 2 H 2 (g)  2 H 2 S (g) The above reaction is allowed to reach equilibrium, and found to contain the following concentrations: [S 2 ] = 0.21 mol/L [H 2 ] = 0.35 mol/L [H 2 S] = mol/L What is the equilibrium constant, K, for this reaction? K = 0.16

Example 2: 2 NH 3 (g)  3 H 2 (g) + N 2 (g) K = 15 The above reaction is allowed to reach equilibrium, and found to contain the following concentrations: [NH 3 ] = ?? mol/L [N 2 ] = 0.73 mol/L [H 2 ] = 1.3 mol/L What is the equilibrium concentration of NH 3 ? [NH 3 ] = 0.33 mol/L

7.9 Le Châtelier's Principle: The Effect of Changing Conditions to Reactions AT Equilibrium ►Le Châtelier's Principle: When a stress is applied to a reaction at equilibrium, the reaction will “shift” to relieve that stress.

►What happens to the concentrations of H 2 and CH 3 OH if the concentration of CO is increased? ►What’s the stress? ►There’s too much CO now ►Which direction will the reaction “shift” to relieve that stress? ►To the “right” in order to react away some of the extra CO

►What happens to the concentrations of H 2 and CH 3 OH if the concentration of CO is increased? ►What’s the effect of the “shift” on the other substances? ►The H 2 concentration drops as it is reacted away by the extra CO ►The CH 3 OH conc. increases as a result

►In summary: ►If a substance’s concentration is increased… ►The reaction will shift AWAY from that substance ►If a substance’s concentration is decreased… ►The reaction will shift TOWARD that substance

►A stress, though, can be ANY change that disturbs original equilibrium. ►Concentration change ►Temperature change ►Pressure change (by changing the container volume… applies to gases only) ►NOTE: Temperature change is the ONLY stress that changes the value of K.

Predicting Shift Direction with a Temperature change “stress” ►If the reaction is exothermic (ΔH = -), write “heat” or “energy” as a product ►If the reaction is endothermic, (ΔH = +), write “heat” as a reactant. ►Then, treat temperature changes as increasing or decreasing the “concentration” of heat, and apply the same rules as for concentration.

Predicting Shift Direction with a Pressure change “stress” ►If the reaction substances are put into a smaller container (thus increasing the pressures they exert)… ►The reaction will shift to the side with fewer moles of gas ►If the reaction substances are put into a larger container (thus decreasing the pressures they exert)… ►The reaction will shift to the side with more moles of gas

►Addition of a catalyst will NOT cause the reaction to shift… it is NOT a “stressor” ►Catalysts just help slow reactions reach their normal equilibrium point faster.

What happens to the amount of H 2 when… 1. The temperature of the reaction is INCREASED? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 2. The pressure of the system is increased by decreasing the volume of the container? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 3. A catalyst is added to the reaction container? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 4. Some O 2 is removed? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 5. Some H 2 O is removed? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 6. Some O 2 is added? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 7. The pressure of the system is increased by adding an inert gas, like helium? 1.Increase 2.Decrease 3.Stay the same

What happens to the amount of H 2 when… 8. The reaction is cooled? 1.Increase 2.Decrease 3.Stay the same

End of Chapter 7