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**The Progress of Chemical Reactions**

Chemistry 122

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Rate Laws One of the variables influencing the rate, or speed of a reaction, is concentration Due to effective collisions, the number of reactant particles eventually decrease and the reaction slows down When a reactant’s concentration decreases, the products concentration increases This makes the rate of a reaction always positive

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A → B When there is only one reactant (decomposition), its concentration can be written into the formula Rate = k[A] Ex. H2O2 → H2 + O2 Rate = k[A]m = k[H2O2]1 But what does it mean?

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**The significance of ‘k’**

When ‘k’ is large, the rate of reaction is fast and products form quickly Units include L/(mol∙s), L2/(mol2∙s), and s-1 k is unique for every reaction and must be supplied if we are to determine the rate of the reaction

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Reaction order Rate = k[A]; it is understood that the notation [A] means the same as [A]1 This exponent ‘1’ is referred to as its reaction order, not the coefficient It is used to define how the rate is affected by the concentration of the reactant When it is in the first order, or exponent = 1, the rate changes proportionally to the concentration

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**As the concentration changes, so does the reaction rate**

First order reactions Rate = k[A]m As the concentration changes, so does the reaction rate If the concentration decreases by ¼, the reaction rate slows down by ¼. If the concentration doubles, so does the rate of reaction Page 577

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**Second order reactions**

In double replacement reactions, there are two reactants Therefore, both concentrations have to be considered when determining the rate of reaction The rate is calculated as follows: Rate = k[A]m[B]n The rate has to be determined through experimentation since it is not always the case that the exponents in the rate law correspond to the coefficients in a chemical equation

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**Finding the order of a reaction**

2NO(g) + 2H2(g) → N2(g) + 2H2O(g) This reaction occurs in more than one step Rate = k[NO]2[H2]1 This has been determined experimentally, it is not a reflection of the balanced chemical equation The reaction is described as second order in NO, first order in H2, and third order overall because the sum of the orders for the individual reactants is = 3.

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**In the real world of chemistry**

Most reactions are more complex than one-step reactions As a result, reactions cannot be predicted solely on the basis of a balanced equation; it must be determined experimentally This is because the rate depends primarily on the slowest step in the reaction mechanism

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**Using a graph to determine the rate of reaction**

When the rate of reaction produces a curve, you must draw a tangent to the curve in order to determine k Take the slope of the line at two points along the curve and determine the rate (concentration/time)

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**Determining reaction order**

The method of initial rates determines reaction order by comparing the initial rates of reaction carried out with varying reactant concentrations Use the expression aA + bB → products as well as rate = k[A]m[B]n

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**Example Trial Initial [A] (M) Initial [B] (M) Initial Rate 1 0.100**

Compare the concentration of Trial 1 and 2 to their respective rates. Compare Trial 2 and 3 to their respective rates. What do you notice? Which reactant is first order? Which is second? What is the order of the overall reaction? Write the rate in the form of an equation. Trial Initial [A] (M) Initial [B] (M) Initial Rate 1 0.100 2.00 x 10-3 2 0.200 4.00 x 10-3 3 16.0 x 10-3 Example

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Questions 1. Write the rate law for the reaction aA →bB if the reaction is third order in A. [B] is not part of the rate law. 2. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction aA + bB → products. Hint: any number to the zero power equals one. For example, (0.22)0 = 1 and (55.6)0 = 1. Trial Initial [A] (M) Initial [B] (M) Initial Rate 1 0.100 2.00 x 10-3 2 0.200 3 4.00 x 10-3

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Questions 3. Given the following experimental data, use the method of initial rates to determine the rate law for the reaction CH3CHO(g) → CH4(g) + CO(g) Trial Initial [CH3CHO] (M) Initial Rate (mol/(L∙s) 1 2.00 x 10-3 2.70 x 10-11 2 4.00 x 10-3 10.8 x 10-11 3 8.00 x 10-3 43.2 x 10-11

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Reaction mechanisms If we were to know all the changes that occur as reactants turn into products, we would be able to graph the transformation This is known as a reaction progress curve The simplest curve is for an elementary reaction Such reactions convert reactants to products in a single step There is only one activated complex

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Reaction mechanism Most chemical reactions occur as a series of steps which produces many peaks and valleys The series of reactions that take place constitute a reaction mechanism The peaks are the individual activated complexes and the valleys are the energies of the intermediates An intermediate is a temporary product; becoming a reactant in the next step of the overall reaction Figure 18.28, p. 578

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**For the remainder of class**

Guided reading for section 18.5 Questions 36, 37, p. 577 Important points to consider: Rate in most cases is mol/L/s or mol∙s/L, Anything is square brackets [A] is typically in molarity (mol/L). Q. 38 – 42, p. 579.

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13.2-13.3 The Rate of Chemical Reactions – The Rate Law.

13.2-13.3 The Rate of Chemical Reactions – The Rate Law.

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