Rate Law

1. Rate Law Mathematical expression that allows calculation of reaction rate as a function of reactant concentration

2. All Chemical Reactions Show: • Reaction rate depends on (is a function of) reactant concentration • Therefore, at fixed temperature, rxn rates can be calculated for any reactant concentration

3. Example: A + B  products Product concentration ,[product], does not influence rate, but [A] and [B] do: ∴ rate ∝ [A]m[B]n Where m & n, represent the order of reaction

4. Order of Reaction • The exponent(m, n, p, …) used to describe the relationship between the initial concentration of a particular reactant and the rate of reaction • These exponents can only be solved using experimental data obtained by carrying out the reaction many times, at a fixed temperature

5. Order of Reaction Example: Example: 1) if we doubled[A], and rate also doubles, ∴ 2 = 2m and ∴ m = 1, first order reactant 2) if we doubled[A], and rate quadruples, ∴ 4= 2m and ∴ m = 2, second order reactant 3) if we doubled[A], and rate goes up 8x ∴ 8= 2m and ∴ m = 3, third order reactant 0) if we doubled[A], and rate does not change ∴ 1= 2m and ∴ m = 0, zero order reactant

6. Rate Law Equation • Since rate ∝ [A]m[B]n • ∴ rate = k[A]m[B]n • Where k = the rate constant, k must be determined experimentally and only applies to one reaction at a fixed temperature

7. Rate Law Equation (Example) • 2A + 2B + 3C  products rate = k[A]m[B]n[C]p • Given experiment data, if we find A to be first order, B to be second order & C to be zero order: rate = k[A]1[B]2[C]0 rate = k[A]1[B]2 total rate order = 1 + 2 + 0 = 3 ∴ the above reaction, is a third order reaction

8. Units for k • the rate constant k has units, but they are particular to each reaction • We use the total rate order to find these units:

10. Using Experimental Data ∴ rate = k[NO(g)]m[H2(g)]n rate = k[NO(g)]2[H2(g)]1

12. Sample Problem 1:

14. Sample Problem 2: