Chapter 13 Integrated Rate Laws

1. Chapter 13 Integrated Rate Laws I. Integrated Rate Laws A. Preview: So far, we have looked at rate as a function of concentration Next, we look at concentration as a function of time aA products • The Integrated First Order Rate Law 1. Integration is a calculus operation • Integration of the first order rate law: • Gives the Integrated first-order rate law: • Integration and Differentiation are connected. If we know the integrated rate law, we could differentiate it to get the rate law • If we know the initial concentration [A]0 and k, we can now calculate [A] at any time (t).

2. How is the integrated rate law of any more help? • The integrated first order rate law is in the form of a straight line • A plot of ln[A] vs t gives a straight line with slope = -k and intercept = ln[A]0 • If plot is a straight line, reaction is1st order • If not a straight line, not 1st order • Example: 2 N2O5 4 NO2 + O2 Find k

3. Half-life of a first order reaction • The integrated first order rate law can be rewritten as: • The half-life of a reaction = time it takes for [A] = [A]0/2

4. [N2O5] 0.1000.050 = 100 s [N2O5] 0.0500.025 = 100 s [N2O5] 0.0250.0125 = 100 s d) The half life does not depend on concentration for a first order reaction e) t1/2 = 100 s from the graph. Is there another way to calculate it? • The new form of the integrated rate law helps: • Example: A first order reaction has t1/2 = 20 min. k? How much time until reaction reaches 75%completion?

5. The Integrated Second Order Rate Law • For aA product that is second order in [A]: • The integrated form is: • This is also in the form of a straight line. A plot of 1/[A] vs t gives a straight line with slope = k, and intercept = 1/[A]0

6. Half-life of Second Order Reactions • Now, the half-life depends on the concentration • Every half life is twice as long as the previous one 5) Example: 2 C4H6 C8H12 Order? k? t1/2?

7. The Integrated Zero-Order Rate Law • Some reactions don’t depend on concentration at all • These are often reactions that occur on a surface • Once the surface is full, more reactant doesn’t help • Example: 2 N2O 2 N2 + O2

8. The rate law: • Integrated Zero Order Rate Law: [A] = -kt + [A]0 • A plot of [A] vs. t gives a straight line with slope = -k and intercept = [A]0 • Half-life of a zero order reaction:

9. Pseudo-First-Order Rate Laws • So far, we have only had one reactant aA products • How do we develop the rate law and the integrated rate law if we have more than one reactant? • Example: BrO3- + 5 Br- + 6 H+ 3 Br2 + 3 H2O • Rate Law: • Integration would be difficult for this complicated rate law • We do the reaction with Pseudo-First-Order Conditions: • [Br-] = very large ~ constant = [Br-]0 • [H+] = very large ~ constant = [H+]0 • [BrO3-] = very small • Rate law can be expressed more simply:

10. Rate law is now just like the first order case: • Integrated Pseudo-First-Order rate law: • This can be graphed the same way as first order law to get k’ • Then we can find k from k’ • Once we have k we can calculate anything else we need to know ln[BrO3-]