Experimental Rate Laws

Learning Objective The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. The student is able to analyze concentration vs. time data to determine the rate law for a zeroth, first, or second order reaction.

Generally, chemical reactions happen faster when the concentrations of the reactants are higher. More concentrated particles, more collisions. More collisions, more effective collisions.

Each reactant could have a different effect on the reaction rate. For some reactants, doubling the concentration could have no effect on the rate of the reaction. (Zero order) For some reactants, doubling the concentration could cause the reaction rate to double. (First order) For some reactants, doubling the concentration could cause the reaction rate to quadruple. (Second order) The only way to know is by experiment!

For any reaction, the experimental rate law takes the following form: Rate = k[X]x[Y]y[Z]z… The units for rate will be in M/s or mol/L∙s or mol∙L-1∙s-1 Depending on the reaction, other time units (min, h, day) are possible

For any reaction, the experimental rate law takes the following form: Rate = k[X]x[Y]y[Z]z… [X], [Y], and [Z] represent the molar concentration of a reactant (or a catalyst). The units for molar concentration are M or mol/L or mol∙L-1. A reaction may have only one reactant that affects its rate. It may have multiple reactants that affect the rate. Only experimentation can determine which reactants have an effect on the rate of the reaction.

For any reaction, the experimental rate law takes the following form: Rate = k[X]x[Y]y[Z]z… x, y, and z are exponents that represent the order of the rate, relative to the specific reactant concentration.

For any reaction, the experimental rate law takes the following form: Rate = k[X]x[Y]y[Z]z… k is the rate law constant. As with any proportional relationship, an experimentally determined constant is needed for the relationship to become an equality. The units of k depend on the overall order (sum of the exponents) of the rate law.

H2O2 + 3 I- + 2 H+  I3- + 2 H2O Initial [H2O2], (M) [I-], (M) [H+], (M) Initial Rate of Appearance of I3- (M/s) Exp. 1 0.010 0.00050 1.15 x 10-6 Exp. 2 0.020 2.30 x 10-6 Exp. 3 Exp. 4 0.00100

H2O2 + 3 I- + 2 H+  I3- + 2 H2O Notice: Our rate law does not include all of the reactants. The rate was not affected by [H+]. (We’ll explain why in a later lesson!) The exponents in our rate law did not correspond to the coefficients for the reactants in the balanced equation.

Learning Objective The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. The student is able to analyze concentration vs. time data to determine the rate law for a zeroth, first, or second order reaction.