Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI(g)  H 2 (g) + I 2 (g) Rate = − 1212  [HI]  t =  [I 2 ]  t  [H 2 ]  t =

Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t = Sample Exercise 14.3 p. 563

Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

Sulfur Clock Demonstration Data FLASK LABELREACTION TIME (SECONDS) A B C D

Concentration and Rate If we compare Experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O(l)

Concentration and Rate Likewise, when we compare Experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O(l)

What is Rate Law? aA + bBproducts

Concentration and Rate This means theRate  [NH 4 + ] & Rate  [NO 2 − ] Rate  [NH 4 + ] [NO 2 − ] which, when written as an equation, becomes Rate Law Eqn = k [NH 4 + ] m [NO 2 − ] n This equation is called the rate law, and k is the rate constant. m and n are called reaction orders. Therefore,

What are m and n exponents? Exponents indicate how the rate is affected by each reactant concentration – termed reaction orders. Rate Law must be determined experimentally.

Overall Reaction Order Rate = k [NH 4 + ] m [NO 2 − ] n The overall reaction order can be found by adding the exponents on the reactants in the rate law … so m + n = ? = 2 This reaction is second-order overall.

Rate Constant (k) Reveals how fast or slow rxn proceeds. LARGE k (~ 10 9 ) = FAST RATE small k (< or = 10 1 ) = SLOW RATE

Sample Exercise 14.6 p.568

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