1. Energy Changes, Reaction Rates, and Equilibrium

2.  The capacity to do work ◦ The ability to move or change something  Change in position  Change in speed  Change in state (form of matter)  Stored energy = potential energy  Moving energy = kinetic energy potentialkinetic

3.  The total energy in a system does not change. Energy cannot be created or destroyed. It can only be changed from one form into another or transferred from one object to another.

4.  Stored energy ◦ Exists in natural attractions and repulsions  Chemical energy ◦ PE possessed by chemicals ◦ Stored in chemical bonds ◦ Breaking bonds requires energy ◦ Forming bonds releases energy ◦ PE may be released and converted to heat  Heat is a form of kinetic energy due to motion of particles  A compound with lower potential energy is more stable than a compound with higher energy

5.  A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 o C.  A joule (J) is a SI unit of energy. Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal) 1 cal = 4.184 J 1,000 J = 1 kJ 1,000 cal = 1 kcal 1 kcal = 4.184 kJ

6.  A gummy bear is 9.000 Calories (nutritional calories). How much energy is stored in a gummy bear in units of Joules? 9.000 Cal = 9.000 kcal x 1000 cal = 9000. cal 1 kcal 9000. cal x 4.184 J = 37660. J = 37.66 kJ 1 cal

7.  Breaking bonds requires energy  Forming bonds releases energy Cl To cleave this bond, 58 kcal/mol must be added.  H = +58 kcal/mol Endothermic To form this bond, 58 kcal/mol is released.  H = −58 kcal/mol Exothermic  H is the energy absorbed or released in a reaction; it is called the heat of reaction or the enthalpy change.

8.  The bond dissociation energy is the  H for breaking a covalent bond by equally dividing the e − between the two atoms.  Bond dissociation energies are positive values, because bond breaking is endothermic (  H > 0).  Bond formation always has negative values, because bond formation is exothermic (  H < 0). HH H + H  H = +104 kcal/mol H + HHH  H = −104 kcal/mol

9.  The stronger the bond, the higher its bond dissociation E.

10.   H indicates the relative strength of the bonds broken and formed in a reaction:  H negative: Exothermic reaction: more energy is released in forming bond than is need to break bonds: products have stronger bonds.  H positive: Endothermic reaction: more energy is needed to break bonds than is released in forming bonds: products have weaker bonds.

12.  For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds.

13.  The orientation of the two molecules must be correct as well.

14.  E a, the energy of activation, is the difference in energy between the reactants and the transition state. It can be thought of as the energy barrier that must be overcome for the reaction to occur.

15.  When the E a is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow.  When the E a is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

16.  ΔH is negative,  the reaction is exothermic:

17.  If  H is positive, the reaction is endothermic

18.  Increasing the concentration of the reactants: o Increases the number of collisions o Increases the reaction rate  Increasing the temperature of the reaction: o Increases the kinetic energy of the molecules o Increases the reaction rate  A catalyst is a substance that speeds up the rate of a reaction and can be recovered unchanged. o Catalysts lower Activation Energy, E a

19.  A catalyst is a substance that speeds up the rate of reaction  A catalyst is recovered unchanged in a reaction, and does not appear in the product  A catalyst lowers Ea  With a catalyst  H is the same

20.  The uncatalyzed reaction (higher E a ) is slower  The catalyzed reaction (lower E a ) is faster

21.  Enzymes (usually protein molecules) are biological catalysts held together in a very specific three- dimensional shape  The active site binds a reactant, which then under- goes a very specific reaction with an enhanced rate.  The enzyme lactase converts the carbohydrate lactose into the two sugars glucose and galactose  People who lack adequate amounts of lactase suffer from abdominal cramping and diarrhea because they cannot digest lactose when it is ingested.

22.  A reversible reaction can occur in either direction  The system is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction  The net concentrations of reactants and products do not change at equilibrium CO(g) + H 2 O(g)CO 2 (g) + H 2 (g) The forward reaction proceeds to the right. The forward reaction proceeds to the right. The reverse reaction proceeds to the left. The reverse reaction proceeds to the left.

23.  The relationship between the concentration of the products and the concentration of the reactants is the equilibrium constant, K. a A + b Bc C + d D equilibrium constant =K = [C] c [D] d [A] a [B] b = [products] [reactants] *Brackets, [ ], are used to symbolize concentration in moles per liter (mol/L).

24. N 2 (g) + O 2 (g) 2 NO(g) equilibrium constant =K= [N 2 ] [O 2 ] [NO] 2 *The coefficient becomes the exponent.

25.  Magnitude of the equilibrium constant  When K is much greater than 1, equilibrium lies to the right and favors the products  When K is much less than 1, equilibrium lies to the left and favors the reactants [products] [reactants] The numerator is larger. [products] [reactants] The denominator is larger.

26.  When K is around 1 (0.01 < K < 100), both reactants and products are present in similar amounts [products] [reactants] Both are similar in magnitude.

27. HOW TO Calculate the Equilibrium Constant for a Reaction A 2 + B 2 2 AB Step [1] Write the expression for the equilibrium constant from the balanced equation. [AB] 2 [A 2 ][B 2 ] K = Step [2] Substitute the given concentrations in the equilibrium expression and calculate K. [AB] 2 [A 2 ][B 2 ] K == [0.50] 2 [0.25][0.25] = 0.25 0.0625 = 4.0

28.  If a chemical system at equilibrium is disturbed or stressed, the system will react in a direction that counteracts the disturbance or relieves the stress  Some of the possible disturbances: ◦ Concentration changes ◦ Temperature changes ◦ Pressure changes

29.  What happens if [CO(g)] is increased? ◦ The concentration of O 2 (g) will decrease. ◦ The concentration of CO 2 (g) will increase. 2 CO(g) + O 2 (g)2 CO 2 (g)

30.  What happens if [CO 2 (g)] is increased? ◦ The concentration of CO(g) will increase. ◦ The concentration of O 2 (g) will increase. 2 CO(g) + O 2 (g)2 CO 2 (g)

31. 31 What happens if a product is removed? The concentration of ethanol will decrease. The concentration of the other product (C 2 H 4 ) will increase.

32. 32 When the temperature is increased, the reaction that absorbs heat is favored. An endothermic reaction absorbs heat, so increasing the temperature favors the forward reaction.

33. 33 Conversely, when the temperature is decreased, the reaction that adds heat is favored. An exothermic reaction releases heat, so increasing the temperature favors the reverse reaction.

34. 34 When pressure increases, equilibrium shifts in the direction that decreases the number of moles in order to decrease pressure.

35. 35 When pressure decreases, equilibrium shifts in the direction that increases the number of moles in order to increase pressure.