Reaction Rates Equilibrium 1 Reaction Rates Balanced reactions

Reaction Rates & Equilibrium 1

Reaction Rates Balanced reactions tell us how much product will be produced. Reaction rates tell us how fast the products will be produced. 2

Activation Energy All reactions need a little extra energy to get started. At higher temperatures, more reactants have the energy needed to get the reaction started. 3

Exothermic Reactions (Energy Released) 4

Endothermic Reactions (Energy Absorbed) 5

How do Reactants become Products? Reactants must hit each other hard enough to break their old bonds. (Collision Theory) They must hit each other in the right orientation to form new bonds. (Collision Theory) Any situation that makes these two things happen more often will speed up the reaction. (Frequency) Concentration Temperature Surface Area Catalysts 6

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Concentration If you increase the concentration of the reactants, there are more reactants around to run into. Thus, more reactions happen faster and the products are formed faster. If you decrease the concentration of the reactants, the reaction slows down. 8

Temperature If you increase the temperature of the reaction, the speed of the molecules increases. The reactants hit each other more often and harder making it easier to break bonds. Thus, more reactions happen and the products are formed faster. If you decrease the temperature you decrease the speed of the molecules and the reaction slows down. 9

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Surface Area The more surface area of a solid substance that is available to react, the faster the reaction rate. For example, a crushed solid would have more surface area than a single cube of the same mass, making the crushed version able to react faster. 11

Catalysts A catalyst is a substance that lowers the activation energy of a reaction but is not used up by the reaction. Enzymes are organic catalysts. Catalysts always increase the speed of the reaction. 12

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Reaction with and without a catalyst Higher activation energy Lower activation energy Reaction occurs slower Reaction occurs faster 14

Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

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As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

Depicting Equilibrium Since, in a system at equilibrium, both the forward and reverse reactions are being carried out, we write its equation with a double arrow. N 2 O 4 (g) 2 NO 2 (g)

Law of Chemical Equilibrium Consider the following general reaction: → c. C+d. D a. A+b. B ← The law of chemical equilibrium states that the molar concentrations of the products (raised to the powers c and d), divided by the molar concentrations of the reactants (raised to the powers a and b), equals a constant.

Equilibrium Constant Keq • Mathematically, we express the law of chemical equilibrium as follows: [C]c[D]d Keq = [A]a[B]b Do NOT include pure solids or liquids in equilibrium constant expressions • The constant, Keq, is the general equilibrium constant. • The value of Keq varies with temperature. So a given value of Keq is valid only for a specific temperature.

Equilibrium Constant Expression PCl 5(s) + H 2 O(g) ↔ 2 HCl(g) + POCl 3 (g) Keq = [C]c[D]d [A]a[B]b Do NOT include pure solids or liquids in equilibrium constant expressions [HCl]2[POCl 3] Keq = [H 2 O] 22

Calculating Keq PCl 5(s) + H 2 O(g) ↔ 2 HCl(g) + POCl 3 (g) [H 2 O] = 0. 050 M [HCl] = 0. 750 M [HCl]2[POCl 3] Keq = [H 2 O] Keq = 5. 6 = [POCl 3] = 0. 500 M (0. 750)2(. 500). 050 23

At equilibrium, are products OR reactants favored? PCl 5(s) + H 2 O(g) ↔ 2 HCl(g) + POCl 3 (g) Keq = 5. 6 • Keq >1: more products than reactants at equilibrium • Keq <1: more reactants than products at equilibrium 24

Solubility Product Equilibrium Insoluble salts are really very slightly soluble. If we add Ag 2 SO 4 to water, some of the slightly soluble Ag 2 SO 4 dissolves: Ag 2 SO 4(s) ↔ 2 Ag+(aq) + SO 42 -(aq) We can write the solubility product equilibrium constant, Ksp, for the reaction: Ksp = [Ag+]2 [SO 42 -] Recall, we don’t include pure solids or liquids in equilibrium constant expressions.

Experimental Determination of Ksp We can calculate the numerical value of Ksp if we know the concentrations of all the species in the reaction. Mg(OH)2(s) ↔ Mg 2+(aq) + 2 OH-(aq) If the concentrations at equilibrium are [Mg 2+] = 0. 00016 M, and [OH-] = 0. 00033 M, what is Ksp? Ksp = [Mg 2+][OH-]2 = (0. 00016)(0. 00033)2 Ksp = 1. 7 × 10 -11