Presentation Transcript

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