Chapter 18 Reaction Rates and Equilibrium 18.1 Rates of Reaction 18.2 The Progress of Chemical Reactions 18.3 Reversible Reactions and Equilibrium 18.4 Solubility Equilibrium 18.5 Free Energy and Entropy Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How is a bicycle race like a chemical reaction? CHEMISTRY & YOU How is a bicycle race like a chemical reaction? Riders in the Tour de France bicycle race must cross steep mountains with heights of 1900 meters or more. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws Rate Laws What is the relationship between the value of the specific rate constant, k, and the speed of a chemical reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The rate of a reaction depends in part on the concentrations of the reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The rate of a reaction depends in part on the concentrations of the reactants. Suppose there were a reaction with only one reactant and one product. A  B The rate at which A forms B can be expressed as the change in A (ΔA) with time. rate = ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The rate of disappearance of A is proportional to the concentration of A. ΔA Δt [A] The proportionality can be expressed as the concentration of A, [A], multiplied by a constant, k. rate = = k × [A] ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws rate = = k × [A] ΔA Δt This equation is a rate law, an expression for the rate of a reaction in terms of the concentration of the reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws rate = = k × [A] ΔA Δt This equation is a rate law, an expression for the rate of a reaction in terms of the concentration of the reactants. The specific rate constant (k) for a reaction is a proportionality constant relating the concentrations of reactants to the rate of the reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate Laws The value of the specific rate constant, k, in a rate law is large if the products form quickly; the value is small if the products form slowly. rate = = k × [A] ΔA Δt Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

First-Order Reactions Rate Laws First-Order Reactions The order of a reaction is the power to which the concentration of a reactant must be raised to match the experimental data on concentration and rate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

First-Order Reactions Rate Laws First-Order Reactions The order of a reaction is the power to which the concentration of a reactant must be raised to match the experimental data on concentration and rate. In a first-order reaction, the rate is directly proportional to the concentration of only one reactant. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs Over time, the rate of reaction decreases because the concentration of the reactant is decreasing. The reaction A  B is a first-order reaction. If [A] is reduced by one half, the reaction rate is reduced by one half. The rate (ΔA/Δt) at any point on the graph equals the slope of the tangent to the curve at that point. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions Rate Laws Higher-Order Reactions In some reactions, two substances react to produce products. In the general equation for a double-replacement reaction below, the coefficients are represented by lowercase letters. aA + bB  cC + dD For the reaction of A with B, the rate of reaction is dependent on the concentrations of both A and B. rate = k[A]x[B]y Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions Rate Laws Higher-Order Reactions rate = k[A]x[B]y When each exponent in the rate law equals 1 (that is, x = y = 1) the reaction is said to be first order in A and first order in B. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions Rate Laws Higher-Order Reactions The overall order of a reaction is the sum of the exponents for the individual reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions Rate Laws Higher-Order Reactions The overall order of a reaction is the sum of the exponents for the individual reactants. A reaction that is first order in A and first order in B is thus second order overall. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Higher-Order Reactions Rate Laws Higher-Order Reactions The overall order of a reaction is the sum of the exponents for the individual reactants. A reaction that is first order in A and first order in B is thus second order overall. The actual order of a reaction must be determined by experiment. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Finding the Order of a Reaction from Experimental Data Sample Problem 18.1 Finding the Order of a Reaction from Experimental Data Consider the reaction aA  B. The rate law for this reaction is Rate = k[A]x. From the data in the table, find the order of the reaction with respect to A and the overall order of the reaction. Trial Initial concentration of A (mol/L) Initial rate (mol/(L·s)) 1 0.050 3.0  10–4 2 0.10 12  10–4 3 0.20 48  10–4 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Analyze List the knowns and the unknowns. Sample Problem 18.1 Analyze List the knowns and the unknowns. 1 Use the first two trials to calculate the order and the third to evaluate your answer. KNOWNS [A]1 = 0.050 mol/L [A]2 = 0.10 mol/L Rate1 = 3.0  10–4 mol/(L·s) Rate2 = 12  10–4 mol/(L·s) UNKNOWN Order of reaction with respect to A = ? Overall order of the reaction = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknowns. Sample Problem 18.1 Calculate Solve for the unknowns. 2 Start with the rate law for each initial concentration of A. The rate law of the reaction and the specific rate constant, k, is the same for any initial concentration of A. Rate1 = k [A1]x Rate2 = k [A2]x Divide the second expression by the first expression. = = ( ) Rate1 k [A1]x Rate2 k [A2]x [A2] [A1] x Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Calculate Solve for the unknowns. Sample Problem 18.1 Calculate Solve for the unknowns. 2 Substitute the known quantities into the equation. = ( ) 3.0  10–4 mol/(L·s) 12  10–4 mol/(L·s) 0.10 mol/L 0.050 mol/L x 4.0 = 2.0x Determine the value of x. x = 2 The reaction is second order in A. Since A is the only reactant, the reaction must be second order overall. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Evaluate Does this result make sense? Sample Problem 18.1 Evaluate Does this result make sense? 3 If the reaction was first order in A, doubling the concentration would double the rate. However, Rate2 is four times Rate1. So, the reaction is second order for A and second order overall because A is the only reactant. As a further test, look at what happens to the rate when the concentration doubles again from 0.10 mol/L to 0.20 mol/L. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

aA  bB + cC The following reaction is a second-order reaction. If the initial concentration of A is 2.0 mol/L and the initial rate is 9.6  10–7 mol/(L·s), what is the concentration when the rate is 1.2  10–7 mol/(L·s)? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

aA  bB + cC The following reaction is a second-order reaction. If the initial concentration of A is 2.0 mol/L and the initial rate is 9.6  10–7 mol/(L·s), what is the concentration when the rate is 1.2  10–7 mol/(L·s)? 1.2  10–7 mol/(L·s) 9.6  10–7 mol/(L·s) = ( ) [A2] 2.0 mol/L 2 [A2] = 0.71 mol/L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

How do most reactions progress from start to finish? Reaction Mechanisms Reaction Mechanisms How do most reactions progress from start to finish? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

A balanced equation does not tell you how a reaction occurred. Reaction Mechanisms A balanced equation does not tell you how a reaction occurred. Plants use photosynthesis to capture and store light energy. The process can be summarized by stating that carbon dioxide and water yield simple sugars and oxygen. However, the process of photosynthesis is not as simple as this summary implies. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

One-Step and Multistep Reactions Reaction Mechanisms One-Step and Multistep Reactions An elementary reaction is a reaction in which reactants are converted to products in a single step. This type of reaction has only one activation-energy peak and one activated complex. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

One-Step and Multistep Reactions Reaction Mechanisms One-Step and Multistep Reactions Most chemical reactions consist of two or more elementary reactions. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

One-Step and Multistep Reactions Reaction Mechanisms One-Step and Multistep Reactions Most chemical reactions consist of two or more elementary reactions. The series of elementary reactions or steps that take place during the course of a complex reaction is called a reaction mechanism. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

One-Step and Multistep Reactions Reaction Mechanisms One-Step and Multistep Reactions An intermediate is a product of one step in a reaction mechanism and a reactant in the next step. An intermediate has a more stable structure and longer lifetime than an activated complex. Intermediates do not appear in the overall chemical equation for a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Interpret Graphs The figure below shows a reaction progress curve for a complex chemical reaction. A reaction progress curve shows all the energy changes that occur as reactants are converted to products. The graph has a peak for each activated complex and a valley for each intermediate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU In the mountain stage of the Tour de France, a rider encounters a series of peaks and valleys. In terms of energy, how does the trip through the mountains compare to what happens during a multistep reaction? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

CHEMISTRY & YOU In the mountain stage of the Tour de France, a rider encounters a series of peaks and valleys. In terms of energy, how does the trip through the mountains compare to what happens during a multistep reaction? Riders need extra energy each time they must climb a peak. This extra energy compares to the activation energy needed in a chemical reaction. Each time they ride down into a valley, they are in an intermediate state. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate-Determining Steps Reaction Mechanisms Rate-Determining Steps In a multistep chemical reaction, the steps do not all progress at the same rate. One step will be slower than the others. The slowest step will determine, or limit, the rate of the overall reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate-Determining Steps Reaction Mechanisms Rate-Determining Steps Consider the reaction mechanism for the decomposition of nitrous oxide (N2O). Experiments have shown that the mechanism consists of the two steps shown below. N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate-Determining Steps Reaction Mechanisms Rate-Determining Steps N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) In the first step, nitrous oxide decomposes into nitrogen gas and oxygen atoms. The oxygen atoms are an intermediate. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Rate-Determining Steps Reaction Mechanisms Rate-Determining Steps N2O(g) N2(g) +O(g) N2O(g) + O(g) N2(g) + O2(g) 2N2O (g) 2N2(g) + O2(g) For the decomposition of nitrous oxide, the first step is the rate-determining step. To increase the rate of the overall reaction, you would need to increase the rate of the first step. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In the following reaction mechanism, which is an intermediate? A2 2A 2A + B2 2AB A2 + B2 2AB A. A2 C. A B. B2 D. AB Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

In the following reaction mechanism, which is an intermediate? A2 2A 2A + B2 2AB A2 + B2 2AB A. A2 C. A B. B2 D. AB Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Key Concepts and Key Equations The value of the specific rate constant, k, in a rate law is large if the products form quickly; the value is small if the products form slowly. Most chemical reactions consist of two or more elementary reactions. Rate = = k  [A] DA Dt Rate = k[A]x[B]y Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms rate law: an expression relating the rate of a reaction to the concentration of the reactants specific rate constant: a proportionality constant relating the concentrations of reactants to the rate of the reaction first-order reaction: a reaction in which the reaction rate is proportional to the concentration of only one reactant Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

Glossary Terms elementary reaction: a reaction in which reactants are converted to products in a single step reaction mechanism: a series of elementary reactions that take place during the course of a complex reaction intermediate: a product of one of the steps in a reaction mechanism; it becomes a reactant in the next step Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.

END OF 18.2 Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.