Chapter 13: Chemical Kinetics CHE 124: General Chemistry II Dr. Jerome Williams, Ph.D. Saint Leo University

Overview Rate Laws & Reaction Orders Method of Initial Rates

The Rate Law Rate Law of a reaction is mathematical relationship between rate of reaction and the concentrations of the reactants – and homogeneous catalysts as well Rate Laws are determined experimentally!! Rate of a reaction is directly proportional to the concentration of each reactant raised to a power for the reaction aA + bB  products the rate law would have the form given below – n and m are called the orders for each reactant – k is called the rate constant

Reaction Order the exponent on each reactant in the rate law is called the order with respect to that reactant the sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction: 2 NO(g) + O 2 (g)  2 NO 2 (g) Rate = k[NO] 2 [O 2 ] The reaction is second order with respect to [NO], first order in [O 2 ], and third order overall.

Sample Rate Laws The bottom reaction is autocatalytic because a product affects the rate. Hg 2+ is a negative catalyst; increasing its concentration slows the reaction. 5Tro: Chemistry: A Molecular Approach, 2/e

Rate = k[NO][O 3 ] Solve: Conceptual Plan: Relationships: [NO] = 1.00 x 10 −6 M, [O 3 ] = 3.00 x 10 −6 M, Rate = 6.60 x 10 −6 M/s k, M −1  s −1 Given: Find: Example: The rate equation for the reaction of NO with ozone is Rate = k[NO][O 3 ]. If the rate is 6.60 x 10 −5 M/sec when [NO] = 1.00 x 10 −6 M and [O 3 ] = 3.00 x 10 −6 M, calculate the rate constant Rate, [NO], [O 3 ]k 6Tro: Chemistry: A Molecular Approach, 2/e

Finding the Rate Law: the Initial Rate Method Rate laws are determined experimentally The rate law shows how the rate of a reaction depends on the concentration of the reactants Changing the initial concentration of a reactant will therefore affect the initial rate of the reaction 7 if for the reaction A → Products then doubling the initial concentration of A doubles the initial reaction rate then doubling the initial concentration of A does not change the initial reaction rate then doubling the initial concentration of A quadruples the initial reaction rate Tro: Chemistry: A Molecular Approach, 2/e

Rate = k[A] n If a reaction is Zero Order, the rate of the reaction is always the same – doubling [A] will have no effect on the reaction rate If a reaction is First Order, the rate is directly proportional to the reactant concentration – doubling [A] will double the rate of the reaction If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentration – doubling [A] will quadruple the rate of the reaction 8Tro: Chemistry: A Molecular Approach, 2/e

Determining the Rate Law When there Are Multiple Reactants Changing each reactant will effect the overall rate of the reaction By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant 9Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below Comparing Expt #1 and Expt #2, the [NO 2 ] changes but the [CO] does not 10Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 11Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 12Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below 13Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below n = 2, m = 0 14Tro: Chemistry: A Molecular Approach, 2/e

Example 13.2: Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below Substitute the concentration s and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) Tro: Chemistry: A Molecular Approach, 2/e

Practice – Determine the rate law and rate constant for the reaction NH NO 2 −  N H 2 O given the data below 16Tro: Chemistry: A Molecular Approach, 2/e