C3H8 + 5 O2  3 CO2 + 4 H2O + Energy Energy: the capacity to do work Types of Energy Kinetic Potential

6.1 Energy Energy is the capacity to do work. Potential energy is stored energy. Kinetic energy is the energy of motion. The law of conservation of energy states that the total energy in a system does not change. Energy cannot be created or destroyed.

100% PE = PEMAX 50% PE & 50% KE 100% KE = KEMAX Velocity is greatest here - Recall KE = ½ mv Height is minimum here -Recall GPE = mgh

6.1 Energy Chemical bonds store potential energy. A compound with lower potential energy is more stable than a compound with higher potential energy. Reactions that form products having lower potential energy than the reactants are favored.

6.1 Energy A. The Units of Energy A calorie (cal) is the amount of energy needed to raise the temperature of 1 g of water by 1 oC. A joule (J) is another unit of energy. 1 cal = 4.184 J Both joules and calories can be reported in the larger units kilojoules (kJ) and kilocalories (kcal). 1,000 J = 1 kJ 1,000 cal = 1 kcal 1 kcal = 4.184 kJ

6.2 Energy Changes in Reactions When molecules come together and react, bonds are broken in the reactants and new bonds are formed in the products. Bond breaking always requires an input of energy. Cl Cl Cl + Cl To cleave this bond, 58 kcal/mol must be added. Bond formation always releases energy. Cl + Cl Cl Cl To form this bond, 58 kcal/mol is released.

6.2 Energy Changes in Reactions A. Bond Dissociation Energy H is the energy absorbed or released in a reaction; it is called the heat of reaction or the enthalpy change. When energy is absorbed, the reaction is said to be endothermic and H is positive (+). When energy is released, the reaction is said to be exothermic and H is negative (−). To cleave this bond, H = +58 kcal/mol (Endothermic) Cl Cl To form this bond, H = −58 kcal/mol (Exothermic)

6.2 Energy Changes in Reactions A. Bond Dissociation Energy 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). H H H + H H = +104 kcal/mol Bond formation always has negative values, because bond formation is exothermic (H < 0). H + H H H H = −104 kcal/mol

6.2 Energy Changes in Reactions B. Calculations Involving H Values H indicates the relative strength of the bonds broken and formed in a reaction. When H is negative: More energy is released in forming bonds than is needed to break the bonds. The bonds formed in the products are stronger than the bonds broken in the reactants. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) H = −213 kcal/mol Heat is released

6.2 Energy Changes in Reactions B. Calculations Involving H Values When H is positive: More energy is needed to break bonds than is released in the formation of new bonds. The bonds broken in the reactants are stronger than the bonds formed in the products. 6 CO2(g) + 6 H2O(l) C6H12O6(aq) + 6 O2(g) ΔH = +678 kcal/mol Heat is absorbed

6.2 Energy Changes in Reactions B. Calculations Involving H Values

6.3 Energy Diagrams For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds. The orientation of the two molecules must be correct as well.

Collision Theory For a reaction to occur, there must be: Collision Orientation Energy

6.3 Energy Diagrams Ea, the energy of activation, is the difference in energy between the reactants and the transition state.

6.3 Energy Diagrams The Ea is the minimum amount of energy that the reactants must possess for a reaction to occur. Ea is called the energy barrier and the height of the barrier determines the reaction rate. When the Ea is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow. When the Ea is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

6.3 Energy Diagrams The difference in energy between the reactants and the products is the H. If H is negative, the reaction is exothermic:

6.3 Energy Diagrams If H is positive, the reaction is endothermic:

6.4 Reaction Rates A. How Concentration and Temperature Affect Reaction Rate Increasing the concentration of the reactants: Increases the number of collisions Increases the reaction rate Increasing the temperature of the reaction: Increases the average kinetic energy of the molecules Increases the reaction rate

6.4 Reaction Rates B. Catalysts A catalyst is a substance that speeds up the rate of a reaction. A catalyst is recovered unchanged in a reaction, and does not appear in the product. Catalysts accelerate a reaction by lowering Ea without affecting H.

6.4 Reaction Rates B. Catalysts The uncatalyzed reaction (higher Ea) is slower. The catalyzed reaction (lower Ea) is faster. H is the same for both reactions.

Catalysts: substances that affect the speed of a reaction without being consumed themselves Provide an alternative pathway that has a lower activation energy (Ea)

Enzymatic Hydrolysis of Sucrose

6.5 Equilibrium A reversible reaction can occur in either direction. The forward reaction proceeds to the right. CO(g) + H2O(g) CO2(g) + H2(g) The reverse reaction proceeds to the left. 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.

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

6.5 Equilibrium A. The Equilibrium Constant For the following balanced chemical equation: N2(g) + O2(g) 2 NO(g) equilibrium constant = K [NO]2 [N2] [O2] The coefficient becomes the exponent.

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

6.5 Equilibrium B. The Magnitude of the Equilibrium Constant When K is around 1 (0.01 < K < 100), [products] Both are similar in magnitude. [reactants] both reactants and products are present in similar amounts.

6.5 Equilibrium B. The Magnitude of the Equilibrium Constant For the reaction: 2 H2(g) + O2(g) 2 H2O(g) K = 2.9 x 1082 The product is favored because K > 1. The equilibrium lies to the right.

6.5 Equilibrium B. The Magnitude of the Equilibrium Constant

6.5 Equilibrium C. Calculating the Equilibrium Constant HOW TO Calculate the Equilibrium Constant for a Reaction Calculate K for the reaction between the general reactants A2 and B2. The equilibrium concentrations are as follows: Example [A2] = 0.25 M [B2] = 0.25 M [AB] = 0.50 M A2 + B2 2 AB

6.5 Equilibrium C. Calculating the Equilibrium Constant HOW TO Calculate the Equilibrium Constant for a Reaction Write the expression for the equilibrium constant from the balanced equation. Step [1] A2 + B2 2 AB [AB]2 K = [A2][B2]

6.5 Equilibrium C. Calculating the Equilibrium Constant HOW TO Calculate the Equilibrium Constant for a Reaction Substitute the given concentrations in the equilibrium expression and calculate K. Step [2] [AB]2 [0.50]2 K = = [A2][B2] [0.25][0.25] 0.25 = = 4.0 0.0625 The unit of the answer is always mol/L (or M), which is usually omitted.

Questions Write the equilibrium constant expression for the following reactions: H2 (g) + I2(g) 2 HI (g) CH3OH (g) CO(g) + 2 H2(g) N2(g) + H2(g) NH3(g)

Questions Is the following reaction reactant favored or product favored? Cd(NH3)42+(aq) Cd2+(aq) + 4 NH3(aq) K = 1.0 x 10-7 Phosgene, COCl2, is a toxic substance that is produced by the reaction of carbon monoxide and chlorine. The Kc for the reaction is 5.0. If the equilibrium concentrations for the reactio are [Cl2] = 0.25 M and [COCl2] = 0.80 M, what is the equilibrium concentration of CO(g)? CO(g) + Cl2(g) COCl2(g)

6.6 Le Châtelier’s Principle If a chemical system at equilibrium is disturbed or stressed, the system will react in a direction to minimize the disturbance or relieve the stress. Some of the possible disturbances: 1) Concentration changes 2) Temperature changes 3) Pressure changes

6.6 Le Châtelier’s Principle A. Concentration Changes 2 CO(g) + O2(g) 2 CO2(g) What happens if [CO(g)] is increased? The concentration of O2(g) will decrease. The concentration of CO2(g) will increase.

6.6 Le Châtelier’s Principle A. Concentration Changes 2 CO(g) + O2(g) 2 CO2(g) What happens if [CO2(g)] is increased? The concentration of CO(g) will increase. The concentration of O2(g) will increase.

6.6 Le Châtelier’s Principle A. Concentration Changes What happens if a product is removed? The concentration of ethanol will decrease. The concentration of the other product (C2H4) will increase.

The Effect of Pressure/Volume Changes

The Effect of Temperature Change on Equilibrium If the temp of a system at equilibrium is changed, the system will shift in a direction to counter that change A new K is established at the new temp

Catalysts Catalysts provide an alternative, more efficient mechanism Speed up rxn by lowering the Ea Result in the same ratios of products and reactants at equilibrium Do not affect the position of equilibrium K stays the same

6.6 Le Châtelier’s Principle Summary

Questions The reaction 2 SO2(g) + O2(g) Û 2 SO3(g) is exothermic. How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? Adding more O2 to the container Removing SO3 Compressing the gases Cooling the container Doubling the volume of the container Warming the mixture Adding SO3 Adding a catalyst to the mixture