Reaction Rate LawDefinition: 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.)

data show that k is not affected by [] changes k does 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 n may be zero, fractions or integers sum of the exponents is called the reaction

order

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)

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

Rates of reaction between NO and H2 at 800°C

Initial Rate of

Exp. [NO] [H2 ] Rxn. Number mol/L mol/L mol/Lsec 1 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

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 0.002 mol/L (i.e. doubled) the rate jumps from 0.002 to 0.008 mol/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

Using exp. 4 & 5 … [H2] doubles and the rate doubles n for the [H2] is the mathematical relationship b/t

these two valuesi.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

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-2

The rate law expression can be rewritten as: rate = 5 x 105 sec mol-2 L-2 [NO mol/L]2[H2 mol/L]

Graphs can be used to determine the order of reaction with respect to a particular reactant

Zeroth-Order Reaction

First-Order Reaction

Second-Order Reaction