Reaction Rate Law

1. Reaction Rate Law Definition: The rate, r, will always be proportional to the product of the initial concentrations of the reactants. For a general reaction … aX + bY -------> … the rate law expression is: r = k[X]m[Y]n [X] & [Y] = molar concentrations of X & Y m & n = powers to which the concentrations must be raised (can only be determined empirically) k = constant of proportionality known as the rate constant (determined empirically, only valid for a specific reaction @ specific temp.)

2. data show that kis not affected by [] changes • kdoes vary with temp. changes • values of m & n are NOT the stoichiometric numbers obtained from the balanced equation (unless it is a one-step rxn.) • m and nmay be zero, fractions or integers • sum of the exponents is called the reaction order

3. Example: H2(g) + I2(g) -----> 2 HI(g) The experimentally determined rate law eqn. is: r = k[H2][I2] This is a second order reaction b/c the sum of the exponents is 2. • the values of m & n happens to be the same as the stoichiometric numbers in the balanced eqn. • this must be a one-step rxn. (i.e. there’s only one rxn. step needed to convert reactants into products)

4. Experimental Determination of Reaction Order Example: Find the rate law and the rxn. order for the following rxn. NO + H2 -----> HNO2 (could balance eqn., but the balanced eqn. won’t necessarily give the exponential values). • before determining the value of k, we need to find the exponential value for each participant • to find the relationship of one reactant it is necessary to keep the other reactant(s) constant

5. Rates of reaction between NO and H2 at 800°C Initial Rate of Exp.  [NO] [H2 ]Rxn.Number       mol/L     mol/L      mol/Lsec1                 0.001      0.004              0.002    2                 0.002      0.004              0.008    3                 0.003      0.004              0.018    4                 0.004      0.001              0.008    5                 0.004      0.002              0.016    6                 0.004      0.003              0.024

6. From the equation we can write a partial rate law as rate = k[NO]m[H2]n Using exp. 1 & 2 …  [NO] jumps from 0.001 to 0.002mol/L (i.e. doubled)  the rate jumps from 0.002 to 0.008mol/Lsec (i.e. quadrupled)  m for the [NO] is the mathematical relationship b/t these two values i.e. 2m = 4, thus m = 2, b/c 22 = 4 Confirm this with exp. 1 & 3 The rate law expression can now be updated to:rate = k[NO]2[H2]n

7. Using exp. 4 & 5 …  [H2] doubles and the rate doubles  n for the [H2] is the mathematical relationship b/t these two values i.e. 2n = 2, thus n = 1, b/c 21 = 2 Confirm this with exp. 4 & 6 The rate law expression can be rewritten as: rate = k[NO]2[H2]1 Sum of the exponents indicates that this is a third order reaction

8. Finding k Using the rate law, fill in the values from the data table. 0.002 mol/L sec = k(0.001 mol/L)2 • (0.004 mol/L) 0.002 mol/L sec = k•(0.000001 mol2 /L2) • (0.004 mol/L) 0.002 mol/L sec = k• 0.000 000 009 mol3/L3 k =       0.002 mol/L sec .      0.000 000 004 mol3/L3 = 500,000 sec/mol2L2 or sec mol-2 L-2The rate law expression can be rewritten as: rate = 5 x 105 sec mol-2 L-2[NO mol/L]2[H2 mol/L]

9. Zeroth-Order Reaction • Graphs can be used to determine the order of reaction with respect to a particular reactant

10. First-Order Reaction

11. Second-Order Reaction