1. DIFFERENTIAL RATE LAW A few things first… Reactions are reversible and the reverse reaction is important. When the rate of the forward reaction equals the rate of the reverse reaction the reaction is at EQUILIBRIUM Though important, the reverse reaction is complicated…so study reaction soon after mixing to minimize reverse reaction (method of initial rates) Rate law (or rate expression) Rate law (or rate expression) – expression which shows how the rate depends on the concentration of reactant(s)  Depends only on the concentrations of the reactants (reverse reaction ignored)  Example: 2NO 2 (g)  O 2 (g) + 2NO(g) has the rate law format…  k, rate constant – proportionality constant that is temperature dependent, concentration independent & must be determined experimentally  n, reactant order – (generally) a positive integer 0, 1, 2 & must be determined experimentally. NOT balancing coefficients (more below) The rate constant, k, is temperature dependent!

2.  For additional reactants n, m & o must be determined experimentally!  A + B + C  products rate = k[A] n [B] m [C] o  Order of reaction – sum of reactant orders (n+m+o) Reactant order (n, m,…)  Zero order (n = 0)  A change in reactant concentration [A] has no effect on the reaction rate  Not very common: rate = k[A] 0 or rate = k  First order (n = 1)  A reaction rate is directly proportional to the reactant concentration [A]  Doubling [A] doubles reaction rate  Very common: rate = k[A] 1 or rate = k[A]  Nuclear decay usually fit into this category  Second order (n = 2)  A reaction rate is quadrupled when reactant concentration [A] doubles  Reaction rate increase by a factor of 9 when [A] is tripled, etc.  Common, particularly in gas-­‐phase reactions. rate = k[A] 2  Fractional orders are rare, but do exist! I’m confused…which reaction rate?  For the reaction 2NO 2 (g)  O 2 (g) + 2NO(g) you could write two reaction rates for NO 2 …these are equivalent!

3. Determining the rate law (rate equation)  Two techniques – depends on what the problem gives you!  Differential rate law  Given reactant(s) concentration(s) and initial reaction rate  Use “table logic” and method of initial rates  Integrated rate law  Given reactant concentration and time.  Find linear graph  Use AP equation sheet Differential rate law  BIG IDEA – find rate law…rate = k[A] n [B] m  Use “table logic”  Compare two experiments where one reactant concentration is constant. See how other reactant effects initial rate EXAMPLE 1 – Find rate law and rate constant, k for the following reaction. NH 4 + (aq) + NO 2 -­‐ (aq)  N 2 (g) + 2H 2 O(l) Experiment [NH 4 + ][NO 2 -­‐ ] Initial rate (M/s) 10.1000.0050 1.35  10 -­‐7 20.1000.010 2.70  10 -­‐7 30.2000.010 5.40  10 -­‐7

4. EXAMPLE 2 – Find the rate law and rate constant, k for the following reaction: BrO 3 -­‐ (aq) + 5Br -­‐ (aq) + 6H + (aq)  3Br 2 (l) + 3H 2 O(l) EXAMPLE 3 – The reaction below was studied and the following data were obtained: I -­‐ (aq) + OCl -­‐ (aq)  IO -­‐ (aq) + Cl -­‐ (aq) Determine the rate law, value for the rate constant and rate for an experiment where both [I -­‐ ] and [OCl -­‐ ] is initially 0.15 M. Experiment [BrO 3 -­‐ ] [Br -­‐ ][H+][H+]Initial rate (M/s) 10.10 8.0  10 -­‐4 20.200.10 1.6  10 -­‐3 30.20 0.10 3.2  10 -­‐3 40.10 0.20 3.2  10 -­‐3 Experiment[I -­‐ ][OCl -­‐ ]Initial rate (M/s) 10.120.18 7.91  10 -­‐2 20.0600.18 3.95  10 -­‐2 30.0300.090 9.88  10 -­‐3 40.240.090 7.91  10 -­‐2