Chapter 13 - Reaction Rates and Equilibrium Study Guide PDF

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A very concrete and detailed study guide of chapter 13 test that covers everything from the book and class' lecture....

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Ch. 13 Study Guide Collision Theory: assumption that chemical reactions are due to the collisions of molecules 1. The collision between the two reactant molecules must take place in the right geometric orientation. 2. The collision must occur with enough energy to break the bonds of the reactants so that new bonds can form in the products. - This minimum kinetic energy needed for the reaction is known as the activation energy (EA) Reaction Rates: measures the increase in the concentration of product or the decrease in the concentration of reactant per unit time. Factors that affect the rate of a reaction: 1. Activation Energy: Energy diagrams tell us a great deal about the amount of energy that must be overcome for a reaction to proceed from reactants to the formation of products. Be familiar with what an energy diagram looks like, what it tells you, and be able to recognize the parts of a diagram (reactants, products, activated complex, activation energy, the correct labels for the x and y axes, and whether the reaction is endothermic or exothermic). -

Activated complex: the point of maximum impact in a reaction, where the molecules are compressed together, at the high point of the activation energy. Heat of the reaction (ΔH): difference between the potential energy of the products and that of the reactants on an energy diagram (exothermic or endothermic)

2. Temperature: The higher the temperature, the higher the average kinetic energy of the molecules. The higher temperatures also cause a greater frequency of collisions between the reacting molecules. 3. Concentration of Reactants: An increase in the [reactant(s)] also increases the rate of collisions. 4. Particle Size: The more area that is exposed, the faster the reactions. Smaller particles react faster than larger particles (burning a tree vs. burning chopped wood). 5. Catalysts: a substance that provides an alternate mechanism with a lower activation energy. A catalyst increases the rate of a reaction but is not consumed in the reaction. Reactions at Equilibrium Equilibrium: a reversible process when both the forward and reverse processes proceed at the same rate so that the overall equilibrium concentrations of reactants and products remain constant. Le Chatelier’s Principle: when stress is applied to a system at equilibrium, the system reacts in such a way to counteract the stress.

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1. Increase in concentration of reactant(s) or product(s): In a system at equilibrium, the introduction of more reactant or product leads to an increase in the concentration of the species on the other side of the equation. - Increase [reactant]  leads to the system adjusting to make more product(s) - Increase [product]  leads to the system adjusting to maker more reactant(s) 2. Increase in pressure: a) decrease in volume corresponds to an increase in pressure, b) increase in the number of moles of a gaseous substance causes an increase in pressure -

Increase in pressure on a gaseous system shifts the point of equilibrium in the direction of the smaller number of moles of gas. Changes in pressure have negligible effects on liquids or solids (virtually incompressible).

3. Increase in temperature: If we add heat to the system, the reaction shifts in a direction to remove that heat. 4. A catalyst in a system that reaches a point of equilibrium increases the rate of both the forward and reverse reactions but does not change the point of equilibrium. Catalysts allow the system to reach the point of equilibrium in less time allows the reaction to be run at much lower temperatures (lower activation energy). Law of Mass Action: Equilibrium Constant (Keq): the law of mass action for a particular reaction, has a specific numerical value for a reaction at a given temperature. The reaction must contain molecules that are all in the same phase (gas or liquid). If a solid is in your equation, it is NOT included as part of the equilibrium equation. aA + bB ⇌ cC + dD Keq = [C]c [D]d [A]a [B]b

a, b, c, d = coefficients

A, B, C, D = reactants and products

A large Keq values tells us that the concentrations of products are larger than the concentrations of the reactants. A small Keq values denotes larger concentrations of the reactants than the products. Equilibria of Weak Acids and Weak Bases in Water: The ionization of weak acids produces only a small concentration of ions (usually < 5% ionization of weak acids). Therefore this is a reversible reaction where the equilibrium lies far in favor of formation of the weak acid/base. When we write the Keq equation for a reaction of a weak acid or base with water, the concentrations of all of the components vary except water, which is present in a very large excess in these reactions (water is a reactant or product and the solvent for every acid or base solution). Therefore, the concentration of H2O is essentially a constant and can be moved to the side of the equation with Keq, defining a new constant known as the acid ionization constant (Ka). Ka tells us about the strength of an acid (whereas pH tells us about the acidity of a solution).

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Example: HClO (aq) + H2O (l) ⇌ H3O+ (aq) + ClO- (aq) Keq = [H3O+] [ClO-] [HClO] [H2O] Keq [H2O] = Ka = [H3O+] [ClO-] [HClO] Similarly, in the ionization of a weak base, water is also present in excess and we define the base ionization constant (Kb). Example: NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq) Kb = [NH4+] [OH-] [NH3] The larger the Ka value, the stronger the acid. The larger the Kb value, the stronger the base. Because of the very small numbers you would be working with in dealing with the Ka and Kb values, we generally work in the log scale instead and talk about pKa and pKb values (similar to why we use the pH scale). So pKa = -log (Ka) and pKb = -log(Kb). The smaller the value of pKa, the stronger the acid. The smaller the value of pKb, the stronger the base. Henderson-Hasselbalch Equation: an equation that relates pH to pKA values for buffered solutions. pH = pKa + log ([base]/[acid]) The most effective buffers are made form solutions containing equal molar (equimolar) amounts of acid and base species. The ratio of acid to base in this situation would be 1. The log of 1 is 0, so the pH = pKa at this point. Buffers are ONLY effective if the pH of the solution is within 1 pH unit of the pKa for the weak acid.

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